Past Exam Questions

Answer: The asymptotes are the vertical line \(x=-\lambda\) and the oblique line \(y=x-\lambda\).

For \(\lambda\gt 0\), the curve has turning points at \((0,0)\) and \((-2\lambda,-4\lambda)\).

For \(\lambda\lt 0\), the same turning points occur; the sketch is drawn with the vertical asymptote on the opposite side of the \(y\)-axis from the \(\lambda\gt 0\) case.

We first rewrite the curve by division:

\(y=\frac{x^{2}}{x+\lambda}=x-\lambda+\frac{\lambda^{2}}{x+\lambda}.\)

This shows immediately that as \(x\to\pm\infty\), the term \(\frac{\lambda^{2}}{x+\lambda}\to 0\), so the oblique asymptote is

\(y=x-\lambda.\)

The denominator is zero when \(x+\lambda=0\), so there is also a vertical asymptote

\(x=-\lambda.\)

To find the turning points, differentiate:

\(y=x-\lambda+\lambda^{2}(x+\lambda)^{-1}.\)

Hence

\(\frac{dy}{dx}=1-\frac{\lambda^{2}}{(x+\lambda)^{2}}.\)

Set \(\frac{dy}{dx}=0\):

\(1=\frac{\lambda^{2}}{(x+\lambda)^{2}}\), so \(x+\lambda=\pm\lambda\).

This gives \(x=0\) or \(x=-2\lambda\).

Now find the corresponding \(y\)-values:

• if \(x=0\), then \(y=\frac{0^{2}}{\lambda}=0\), so one turning point is \((0,0)\);
• if \(x=-2\lambda\), then \(y=\frac{(-2\lambda)^2}{-2\lambda+\lambda}=\frac{4\lambda^2}{-\lambda}=-4\lambda\), so the other turning point is \((-2\lambda,-4\lambda)\).

So the asymptotes are \(x=-\lambda\) and \(y=x-\lambda\), and the turning points are \((0,0)\) and \((-2\lambda,-4\lambda)\).

For the sketch:

• when \(\lambda\gt 0\), the vertical asymptote is to the left of the \(y\)-axis;
• when \(\lambda\lt 0\), the vertical asymptote is to the right of the \(y\)-axis.

In each case, draw the curve approaching both asymptotes and passing through the turning points \((0,0)\) and \((-2\lambda,-4\lambda)\).

Answer: (a) Since \(x=\mathrm{e}^{u}\), we have \(u=\ln x\) and therefore \(\dfrac{du}{dx}=\dfrac{1}{x}\).

So

\(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{dy}{du}\),

hence \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\).

Differentiating again with respect to \(x\),

\(\dfrac{d}{dx}\left(x\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)\).

The left-hand side is \(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}\), while the right-hand side is

\(\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).

Therefore

\(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\),

and multiplying by \(x\) gives

\(x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=\dfrac{d^2y}{du^2}\).

Using \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\), this becomes

\(x^2\dfrac{d^2y}{dx^2}=\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\).

(b) Substitute these relations into \(x^2\dfrac{d^2y}{dx^2}+5x\dfrac{dy}{dx}+3y=30x^2\):

\(\left(\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\right)+5\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).

Hence

\(\dfrac{d^2y}{du^2}+4\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).

To solve, first find the complementary function from

\(m^2+4m+3=0\), so \((m+1)(m+3)=0\) and \(m=-1,-3\).

Thus

\(y_c=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}\).

For a particular integral, try \(y_p=Ce^{2u}\). Then

\(y_p'=2Ce^{2u}\), \(y_p''=4Ce^{2u}\).

Substituting gives

\((4C+8C+3C)e^{2u}=30e^{2u}\), so \(15C=30\) and \(C=2\).

Therefore the general solution in terms of \(u\) is

\(y=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}+2\mathrm{e}^{2u}\).

Finally, since \(\mathrm{e}^{u}=x\), we have \(\mathrm{e}^{-u}=x^{-1}\), \(\mathrm{e}^{-3u}=x^{-3}\), and \(\mathrm{e}^{2u}=x^2\). So

\(y=\dfrac{A}{x}+\dfrac{B}{x^3}+2x^2\).

Let \(x=\mathrm{e}^{u}\), so \(u=\ln x\) and \(\dfrac{du}{dx}=\dfrac{1}{x}\).

Then

\(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{dy}{du}\),

so \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\).

Differentiating \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\) with respect to \(x\),

\(\dfrac{d}{dx}\left(x\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)\).

The left-hand side is \(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}\).

For the right-hand side, use the chain rule:

\(\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).

So

\(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).

Multiplying by \(x\) gives

\(x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=\dfrac{d^2y}{du^2}\).

Since \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\), we obtain

\(x^2\dfrac{d^2y}{dx^2}=\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\).

Now substitute into \(x^2\dfrac{d^2y}{dx^2}+5x\dfrac{dy}{dx}+3y=30x^2\):

\(\left(\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\right)+5\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\),

hence

\(\dfrac{d^2y}{du^2}+4\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).

To solve this differential equation, first find the complementary function from

\(m^2+4m+3=0\).

So \((m+1)(m+3)=0\), giving \(m=-1,-3\).

Therefore

\(y_c=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}\).

For a particular solution, try \(y_p=Ce^{2u}\). Then

\(y_p'=2Ce^{2u}\), \(y_p''=4Ce^{2u}\).

Substituting into the equation gives

\((4C+8C+3C)e^{2u}=30e^{2u}\), so \(15C=30\) and \(C=2\).

Thus

\(y=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}+2\mathrm{e}^{2u}\).

Finally, since \(\mathrm{e}^{u}=x\),

\(\mathrm{e}^{-u}=\dfrac{1}{x}\), \(\mathrm{e}^{-3u}=\dfrac{1}{x^3}\), and \(\mathrm{e}^{2u}=x^2\).

So the general solution in terms of \(x\) is

\(y=\dfrac{A}{x}+\dfrac{B}{x^3}+2x^2\).

Answer: (a) \(1-\operatorname{sech}^2 t=\tanh^2 t\).

(b) \(\frac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2 t\).

(c) The point is \(\left(\frac{8}{25}+\ln\frac{3}{5},\;\frac{881}{625}\right)\).

(a) Using the exponential definitions,

\(\operatorname{sech} t=\frac{2}{e^t+e^{-t}}\), \(\tanh t=\frac{e^t-e^{-t}}{e^t+e^{-t}}\).

Then

\(1-\operatorname{sech}^2 t=1-\left(\frac{2}{e^t+e^{-t}}\right)^2\)

\(=\frac{(e^t+e^{-t})^2-4}{(e^t+e^{-t})^2}\)

\(=\frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2}\)

\(=\frac{e^{2t}-2+e^{-2t}}{(e^t+e^{-t})^2}\)

\(=\frac{(e^t-e^{-t})^2}{(e^t+e^{-t})^2}\)

\(=\left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2=\tanh^2 t\).

So \(1-\operatorname{sech}^2 t=\tanh^2 t\).

(b) Given

\(x=\frac12\tanh^2 t+\ln(\operatorname{sech} t), \qquad y=1+\tanh^4 t\).

Differentiate with respect to \(t\):

\(\frac{\mathrm{d}y}{\mathrm{d}t}=4\tanh^3 t\,\operatorname{sech}^2 t\).

Also,

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac12\cdot 2\tanh t\,\operatorname{sech}^2 t+\frac{1}{\operatorname{sech} t}\cdot \frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech} t)\).

Since \(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech} t)=-\operatorname{sech} t\tanh t\), this gives

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\tanh t\,\operatorname{sech}^2 t-\tanh t\)

\(=\tanh t(\operatorname{sech}^2 t-1)\).

Using part (a), \(\operatorname{sech}^2 t-1=-\tanh^2 t\), so

\(\frac{\mathrm{d}x}{\mathrm{d}t}=-\tanh^3 t\).

Hence

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{4\tanh^3 t\,\operatorname{sech}^2 t}{-\tanh^3 t}=-4\operatorname{sech}^2 t\).

(c) From part (b),

\(\frac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2 t\).

Differentiate with respect to \(t\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-4\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^2 t)=8\operatorname{sech}^2 t\tanh t\).

Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\mathrm{d}y}{\mathrm{d}x})}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{8\operatorname{sech}^2 t\tanh t}{-\tanh^3 t}=-8\frac{\operatorname{sech}^2 t}{\tanh^2 t}\).

We are told that

\(-8\frac{\operatorname{sech}^2 t}{\tanh^2 t}=-\frac92\).

So

\(8\operatorname{sech}^2 t=\frac92\tanh^2 t\).

Using \(\operatorname{sech}^2 t=1-\tanh^2 t\),

\(8(1-\tanh^2 t)=\frac92\tanh^2 t\).

Multiply by 2:

\(16-16\tanh^2 t=9\tanh^2 t\), so \(16=25\tanh^2 t\).

Hence \(\tanh^2 t=\frac{16}{25}\). Since \(t\gt 0\), \(\tanh t=\frac45\).

Then

\(\operatorname{sech}^2 t=1-\frac{16}{25}=\frac{9}{25}\), so \(\operatorname{sech} t=\frac35\).

Now

\(x=\frac12\tanh^2 t+\ln(\operatorname{sech} t)=\frac12\cdot\frac{16}{25}+\ln\frac35=\frac{8}{25}+\ln\frac35\),

\(y=1+\tanh^4 t=1+\left(\frac{16}{25}\right)^2=1+\frac{256}{625}=\frac{881}{625}\).

So the required point is \(\left(\frac{8}{25}+\ln\frac35,\;\frac{881}{625}\right)\).

Answer: The exact length of the curve is \(\frac{1}{2}e^2-\frac{1}{6}\).

For a curve with parametric equations \(x(t)\) and \(y(t)\), the arc length from \(t=0\) to \(t=1\) is

\(\displaystyle s=\int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here,

\(x=\frac{1}{2}e^{2t}-\frac{1}{3}t^3-\frac{1}{2}, \qquad y=2e^t(t-1).\)

Differentiating,

\(\displaystyle \frac{dx}{dt}=e^{2t}-t^2\)

and

\(\displaystyle \frac{dy}{dt}=2e^t(t-1)+2e^t=2te^t.\)

Then

\(\begin{aligned} \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 &=(e^{2t}-t^2)^2+4t^2e^{2t}\\ &=e^{4t}-2t^2e^{2t}+t^4+4t^2e^{2t}\\ &=e^{4t}+2t^2e^{2t}+t^4\\ &=(e^{2t}+t^2)^2. \end{aligned}\)

So, for \(0\le t\le 1\),

\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=e^{2t}+t^2.\)

Hence

\(\begin{aligned} s&=\int_0^1 (e^{2t}+t^2)\,dt\\ &=\left[\frac{1}{2}e^{2t}+\frac{1}{3}t^3\right]_0^1\\ &=\left(\frac{1}{2}e^2+\frac{1}{3}\right)-\frac{1}{2}\\ &=\frac{1}{2}e^2-\frac{1}{6}. \end{aligned}\)

Therefore, the exact length is \(\boxed{\frac{1}{2}e^2-\frac{1}{6}}\).

Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t^2}{\sqrt{1-t^2}}\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-t^3(1-t^2)^{-\frac32}(2-t^2)\).

Hence \(a=3\) and \(b=-\frac32\).

(a) Since \(x\) and \(y\) are given in terms of \(t\), use parametric differentiation:

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}.\)

Now

\(\displaystyle y=\cos^{-1}t \implies \frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{\sqrt{1-t^2}}.\)

Also

\(\displaystyle x=1+\frac1t \implies \frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{1}{t^2},\)

so

\(\displaystyle \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2.\)

Therefore

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-t^2}}\times(-t^2)=\frac{t^2}{\sqrt{1-t^2}}.\)

(b) Differentiate \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t^2}{\sqrt{1-t^2}}\) with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{t^2}{\sqrt{1-t^2}}\right).\)

This can be written as

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{t^2}{\sqrt{1-t^2}}\right)=\frac{2t\sqrt{1-t^2}+t^3(1-t^2)^{-1/2}}{1-t^2}.\)

Now use

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot\frac{\mathrm{d}t}{\mathrm{d}x}.\)

Since \(\displaystyle \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2t\sqrt{1-t^2}+t^3(1-t^2)^{-1/2}}{1-t^2}\,(-t^2).\)

So

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-2t^3(1-t^2)^{-1/2}-t^5(1-t^2)^{-3/2}.\)

Factorising gives

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-t^3(1-t^2)^{-3/2}\bigl(2(1-t^2)+t^2\bigr).\)

Hence

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-t^3(1-t^2)^{-3/2}(2-t^2).\)

Comparing with \(\,-t^a(1-t^2)^b(2-t^2)\), we get

\(\displaystyle a=3,\quad b=-\frac32.\)

Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\mathrm{e}^t(t^3+t^2)\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=t^3\mathrm{e}^t(t^2+4t+2)\).

Since \(x\) and \(y\) are both given in terms of \(t\), use parametric differentiation:

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\).

(a) Differentiate with respect to \(t\):

From \(y=t\mathrm{e}^t\),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{e}^t+t\mathrm{e}^t=\mathrm{e}^t(t+1)\).

From \(x=1+\frac{1}{t}=1+t^{-1}\),

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2}\).

Hence

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{e}^t(t+1)}{-t^{-2}}=-t^2\mathrm{e}^t(t+1)=-\mathrm{e}^t(t^3+t^2)\).

(b) For the second derivative, use

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}\).

From part (a),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\mathrm{e}^t(t^3+t^2)\).

Differentiate with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\mathrm{e}^t(3t^2+2t)-\mathrm{e}^t(t^3+t^2)=-\mathrm{e}^t(t^3+4t^2+2t)\).

Also, since \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2}\), we have

\(\displaystyle \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2\).

Therefore

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\mathrm{e}^t(t^3+4t^2+2t)(-t^2)=t^2\mathrm{e}^t(t^3+4t^2+2t)\).

Factorising,

\(\displaystyle t^2\mathrm{e}^t(t^3+4t^2+2t)=t^3\mathrm{e}^t(t^2+4t+2)\).

So \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=t^3\mathrm{e}^t(t^2+4t+2)\).

Answer: (a) The exact length of the curve is \(4\sqrt{3}\).

(b) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) for \(0\lt t\lt 1\).

(a) For a parametric curve, its length is

\(\displaystyle \int_0^3 \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t.\)

Differentiate:

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{3}t^{3/2}-2t^{1/2}\right)=t^{1/2}-t^{-1/2},\qquad \frac{\mathrm{d}y}{\mathrm{d}t}=2.\)

So

\(\displaystyle \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2=(t^{1/2}-t^{-1/2})^2+4=t+2+t^{-1}=(t^{1/2}+t^{-1/2})^2.\)

Since \(t\gt 0\),

\(\displaystyle \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}=t^{1/2}+t^{-1/2}.\)

Hence the length is

\(\displaystyle \int_0^3 (t^{1/2}+t^{-1/2})\,\mathrm{d}t=\left[\frac{2}{3}t^{3/2}+2t^{1/2}\right]_0^3=4\sqrt{3}.\)

Therefore the exact length is \(\boxed{4\sqrt{3}}\).

(b) First,

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{2}{t^{1/2}-t^{-1/2}}.\)

Differentiate with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{t^{1/2}-t^{-1/2}}\right)=\frac{-2\left(\frac12 t^{-1/2}+\frac12 t^{-3/2}\right)}{(t^{1/2}-t^{-1/2})^2}.\)

Now use

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot \frac{\mathrm{d}t}{\mathrm{d}x},\)

so

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{-2\left(\frac12 t^{-1/2}+\frac12 t^{-3/2}\right)}{(t^{1/2}-t^{-1/2})^3}.\)

Simplifying,

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{t+1}{(t-1)^3}.\)

We want

\(\displaystyle -\frac{t+1}{(t-1)^3}\gt 0.\)

For \(0\lt t\le 3\), we have \(t+1\gt 0\), so this is positive exactly when \((t-1)^3\lt 0\), i.e. when \(t\lt 1\).

Given the domain, the required set is \(\boxed{0\lt t\lt 1}\).

Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{t^2+1+\sqrt{t^2+1}}{t^2}\).

(b) When \(t=\frac34\), \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{80}{3}\).

(a) Differentiate both equations with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-1+\frac{1}{1+t^2}, \qquad \frac{\mathrm{d}y}{\mathrm{d}t}=1+\frac{1}{\sqrt{1+t^2}}.\)

Now

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}.\)

Simplify each derivative:

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-1+\frac{1}{1+t^2}=-\frac{t^2}{1+t^2},\)

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=1+\frac{1}{\sqrt{1+t^2}}=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}}.\)

Hence

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}}\times\frac{1+t^2}{-t^2}.\)

Since \(1+t^2=(\sqrt{1+t^2})^2\),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{(\sqrt{1+t^2}+1)\sqrt{1+t^2}}{t^2}=-\frac{1+t^2+\sqrt{1+t^2}}{t^2}.\)

(b) Let

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1+t^2+\sqrt{1+t^2}}{t^2}.\)

Write \(N=1+t^2+\sqrt{1+t^2}\), so that \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{N}{t^2}\).

Then

\(\displaystyle N'=2t+\frac{t}{\sqrt{1+t^2}}.\)

Using the quotient rule,

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\frac{t^2N'-2tN}{t^4}.\)

So

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\frac{t^2\left(2t+\frac{t}{\sqrt{1+t^2}}\right)-2t\left(1+t^2+\sqrt{1+t^2}\right)}{t^4}.\)

This simplifies to

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\frac{t^2(1+t^2)^{-1/2}-2(1+\sqrt{1+t^2})}{t^3}.\)

For parametric equations,

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}{\frac{\mathrm{d}x}{\mathrm{d}t}}.\)

Using \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{t^2}{1+t^2}\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{t^2(1+t^2)^{-1/2}-2(1+\sqrt{1+t^2})}{t^5}(1+t^2).\)

Now set \(t=\frac34\). Then

\(\displaystyle 1+t^2=\frac{25}{16}, \qquad \sqrt{1+t^2}=\frac54, \qquad t^5=\left(\frac34\right)^5=\frac{243}{1024}.\)

Also,

\(\displaystyle t^2(1+t^2)^{-1/2}=\frac{9}{16}\cdot\frac45=\frac{9}{20},\)

\(\displaystyle 2(1+\sqrt{1+t^2})=2\left(1+\frac54\right)=\frac92.\)

So

\(\displaystyle \frac{t^2(1+t^2)^{-1/2}-2(1+\sqrt{1+t^2})}{t^5}(1+t^2)=\left(\frac{9}{20}-\frac92\right)\cdot\frac{25}{16}\cdot\frac{1024}{243}.\)

Since \(\displaystyle \frac{9}{20}-\frac92=-\frac{81}{20}\), this gives

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{81}{20}\cdot\frac{25}{16}\cdot\frac{1024}{243}=-\frac{80}{3}.\)

Answer: The length of the curve is \(\mathrm{e}^2+\frac{5}{3}\).

For a parametric curve, the length from \(t=0\) to \(t=2\) is

\(\displaystyle s=\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here

\(x=\mathrm{e}^t-\frac13 t^3\), so

\(\displaystyle \frac{dx}{dt}=\mathrm{e}^t-t^2.\)

Also

\(y=4\mathrm{e}^{t/2}(t-2)\), so using the product rule,

\(\displaystyle \frac{dy}{dt}=4\left(\frac12\mathrm{e}^{t/2}(t-2)+\mathrm{e}^{t/2}\right)=2t\mathrm{e}^{t/2}.\)

Hence

\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\sqrt{\left(\mathrm{e}^t-t^2\right)^2+\left(2t\mathrm{e}^{t/2}\right)^2}.\)

Simplifying,

\(\displaystyle \left(\mathrm{e}^t-t^2\right)^2+4t^2\mathrm{e}^t=\mathrm{e}^{2t}-2t^2\mathrm{e}^t+t^4+4t^2\mathrm{e}^t=\mathrm{e}^{2t}+2t^2\mathrm{e}^t+t^4=(\mathrm{e}^t+t^2)^2.\)

So

\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\mathrm{e}^t+t^2.\)

Therefore

\(\displaystyle s=\int_0^2 (\mathrm{e}^t+t^2)\,dt=\left[\mathrm{e}^t+\frac13 t^3\right]_0^2=\left(\mathrm{e}^2+\frac83\right)-1=\mathrm{e}^2+\frac53.\)

So the length of the curve is \(\boxed{\mathrm{e}^2+\frac53}\).

the length is \(\dfrac{15}{8}+2\ln2\), and \(\dfrac{d^2y}{dx^2}=\dfrac{-2(1+t^{-2})}{(t-t^{-1})^3}\).

Differentiate the parametric equations: \(\dot{x}=t-t^{-1}\) and \(\dot{y}=2\).

Then \(\dot{x}^2+\dot{y}^2=(t-t^{-1})^2+4=t^2+2+t^{-2}=(t+t^{-1})^2\).

Since \(t>0\), the speed is \(t+t^{-1}\). Hence the length is \(\displaystyle \int_{1/2}^{2}(t+t^{-1})\,dt=\left[\dfrac12t^2+\ln t\right]_{1/2}^{2}=\dfrac{15}{8}+2\ln2\).

Also \(\dfrac{dy}{dx}=\dfrac{\dot{y}}{\dot{x}}=\dfrac{2}{t-t^{-1}}\).

Differentiating with respect to \(t\), \(\dfrac{d}{dt}\left(\dfrac{2}{t-t^{-1}}\right)=\dfrac{-2(1+t^{-2})}{(t-t^{-1})^2}\).

Therefore \(\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\dfrac{dt}{dx}=\dfrac{-2(1+t^{-2})}{(t-t^{-1})^3}\).

the length is \(\dfrac{85}{12}\), and in part (b) \(\dfrac{d^2y}{dx^2}=-25\) when \(t=1\).

For \(a=\dfrac23\), \(b=-\dfrac12\), \(\dot{x}=3-2t^{-2}+2t^2\) and \(\dot{y}=4+\dfrac32t^{-2}-\dfrac32t^2\).

Expanding and simplifying, \(\dot{x}^2+\dot{y}^2=\dfrac{25}{4}(t^2+t^{-2})^2\).

Since \(1\le t\le2\), the speed is \(\dfrac52(t^2+t^{-2})\). Hence the exact length is \(\dfrac52\int_1^2(t^2+t^{-2})\,dt=\dfrac52\left[\dfrac{t^3}{3}-t^{-1}\right]_1^2=\dfrac{85}{12}\).

For part (b), put \(a=b=0\). Then \(\dot{x}=3-2t^{-2}\) and \(\dot{y}=4+\dfrac32t^{-2}\).

So \(\dfrac{dy}{dx}=\dfrac{4+\frac32t^{-2}}{3-2t^{-2}}\).

Differentiate with respect to \(t\) and multiply by \(dt/dx\): \(\dfrac{d^2y}{dx^2}=\dfrac{-25t^{-3}}{(3-2t^{-2})^3}\).

At \(t=1\), this equals \(-25\).

Answer:

We have \(x=\sin t\) and \(y=\sin 2t\), for \(0\leqslant t\leqslant \pi\).

First differentiate with respect to \(t\): \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=\cos t\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=2\cos 2t\).

So

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\dfrac{2\cos 2t}{\cos t}.\)

Now differentiate again with respect to \(t\):

\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{2\cos 2t}{\cos t}\right).\)

Using the quotient rule,

\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{2\cos 2t}{\cos t}\right)=\dfrac{(-4\sin 2t)\cos t-(2\cos 2t)(-\sin t)}{\cos^2 t} = \dfrac{-4\sin 2t\cos t+2\cos 2t\sin t}{\cos^2 t}.\)

Then

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\dfrac{\mathrm{d}}{\mathrm{d}t}(\mathrm{d}y/\mathrm{d}x)}{\mathrm{d}x/\mathrm{d}t}=\dfrac{-4\sin 2t\cos t+2\cos 2t\sin t}{\cos^3 t}.\)

This can be simplified using \(\sin 2t=2\sin t\cos t\):

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{-8\sin t\cos^2 t+2\cos 2t\sin t}{\cos^3 t}.\)

Since \(\cos 2t=\cos^2 t-\sin^2 t\), this becomes

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{-8\sin t\cos^2 t+2\sin t(\cos^2 t-\sin^2 t)}{\cos^3 t}=\dfrac{4\sin^3 t-6\sin t}{\cos^3 t}.\)

Now for stationary points, set \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\). Since \(\cos t\neq 0\) at a stationary point of the curve, we need \(2\cos 2t=0\), so \(\cos 2t=0\).

Hence \(2t=\dfrac{\pi}{2}\) or \(\dfrac{3\pi}{2}\), giving \(t=\dfrac{\pi}{4}\) or \(\dfrac{3\pi}{4}\).

Find the coordinates:

• When \(t=\dfrac{\pi}{4}\), \(x=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\) and \(y=\sin\dfrac{\pi}{2}=1\).
• When \(t=\dfrac{3\pi}{4}\), \(x=\sin\dfrac{3\pi}{4}=\dfrac{1}{\sqrt2}\) and \(y=\sin\dfrac{3\pi}{2}=-1\).

So the stationary points are \(\left(\dfrac{1}{\sqrt2},1\right)\) and \(\left(\dfrac{1}{\sqrt2},-1\right)\).

To determine their nature, substitute into \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\):

At \(t=\dfrac{\pi}{4}\), \(\sin t=\dfrac{1}{\sqrt2}\) and \(\cos t=\dfrac{1}{\sqrt2}\), so

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{4\left(\frac{1}{2\sqrt2}\right)-6\left(\frac{1}{\sqrt2}\right)}{\left(\frac{1}{\sqrt2}\right)^3}=-8,\)

which means a maximum.

At \(t=\dfrac{3\pi}{4}\), \(\sin t=\dfrac{1}{\sqrt2}\) and \(\cos t=-\dfrac{1}{\sqrt2}\), so

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=+8,\)

which means a minimum.

Answer:

r For the first part, write r \(I_n=\int_0^{\frac{\pi}{2}}\cos^{n-1}x\,\cos x\,dx\). Integrating by parts with \(u=\cos^{n-1}x\) and \(dv=\cos x\,dx\), we get \(du=-(n-1)\cos^{n-2}x\sin x\,dx\) and \(v=\sin x\).

So \(I_n=\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\). The bracketed term is zero at both limits, hence \(I_n=(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x(1-\cos^2x)\,dx\).

This gives \(I_n=(n-1)\left(I_{n-2}-I_n\right)\), so \(nI_n=(n-1)I_{n-2}\). Therefore \(I_n=\frac{n-1}{n}I_{n-2}\) for \(n\ge 2\).

For the curve, the mean value of \(y\) over \(0\le x\le a\) is \(m=\frac{1}{a}\int_0^a y\,dx\).

Using the parametric equations \(x=a\sin^3t\), \(y=a\cos^3t\) with \(0\le t\le \frac{\pi}{2}\), we have \(\frac{dx}{dt}=3a\sin^2t\cos t\). Hence \(m=\frac{1}{a}\int_0^{\frac{\pi}{2}} a\cos^3t\,(3a\sin^2t\cos t)\,dt=3a\int_0^{\frac{\pi}{2}}\cos^4t\sin^2t\,dt\).

Now \(\sin^2t=1-\cos^2t\), so \(m=3a\int_0^{\frac{\pi}{2}}\left(\cos^4t-\cos^6t\right)dt\), as required.

To find the exact value, use the reduction formula.

Let \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\). Then \(I_0=\frac{\pi}{2}\), and \(I_2=\frac{1}{2}I_0=\frac{\pi}{4}\), \(I_4=\frac{3}{4}I_2=\frac{3\pi}{16}\), \(I_6=\frac{5}{6}I_4=\frac{5\pi}{32}\).

Therefore \(m=3a(I_4-I_6)=3a\left(\frac{3\pi}{16}-\frac{5\pi}{32}\right)=3a\cdot\frac{\pi}{32}=\frac{3\pi a}{32}\).

For \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\), write \(\cos^n x=\cos^{n-1}x\cdot \cos x\) and integrate by parts with \(u=\cos^{n-1}x\), \(dv=\cos x\,dx\).

Then \(du=-(n-1)\cos^{n-2}x\sin x\,dx\) and \(v=\sin x\), so

\(I_n=\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\).

The boundary term is zero because \(\sin 0=0\) and \(\cos\left(\frac{\pi}{2}\right)=0\). Hence

\(I_n=(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\).

Using \(\sin^2x=1-\cos^2x\),

\(I_n=(n-1)\int_0^{\frac{\pi}{2}}\left(\cos^{n-2}x-\cos^n x\right)dx=(n-1)(I_{n-2}-I_n)\).

So \(nI_n=(n-1)I_{n-2}\), and therefore

\(I_n=\frac{n-1}{n}I_{n-2}\), for \(n\ge 2\).

For the curve, the mean value of \(y\) over \(0\le x\le a\) is

\(m=\frac{1}{a}\int_0^a y\,dx\).

Now \(x=a\sin^3t\) and \(y=a\cos^3t\), with \(0\le t\le \frac{\pi}{2}\). Also

\(\frac{dx}{dt}=3a\sin^2t\cos t\).

So

\(m=\frac{1}{a}\int_0^{\frac{\pi}{2}} a\cos^3t\cdot 3a\sin^2t\cos t\,dt=3a\int_0^{\frac{\pi}{2}}\cos^4t\sin^2t\,dt\).

Since \(\sin^2t=1-\cos^2t\), this becomes

\(m=3a\int_0^{\frac{\pi}{2}}\left(\cos^4t-\cos^6t\right)dt\).

To evaluate it exactly, use the reduction formula for \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\).

First, \(I_0=\frac{\pi}{2}\). Then

\(I_2=\frac{1}{2}I_0=\frac{\pi}{4}\),

\(I_4=\frac{3}{4}I_2=\frac{3\pi}{16}\),

\(I_6=\frac{5}{6}I_4=\frac{5\pi}{32}\).

Hence

\(m=3a(I_4-I_6)=3a\left(\frac{3\pi}{16}-\frac{5\pi}{32}\right)=3a\cdot\frac{\pi}{32}=\frac{3\pi a}{32}\).

Answer: 9P₁: \(6x+y+4z=13\)

The acute angle between \(\Pi_1\) and \(\Pi_2\) is \(83.3^\circ\) (approximately).

A vector equation of the line of intersection is \(\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}2\\1\\0\end{array}\right)+t\left(\begin{array}{c}-1\\-6\\3\end{array}\right)\), where \(t\\in\mathbb{R}\).

The plane \(\Pi_1\) is given by \(\mathbf r=(2,-3,1)+\lambda(1,-2,-1)+\mu(1,2,-2)\).

So two direction vectors in the plane are \(\mathbf a=(1,-2,-1)\) and \(\mathbf b=(1,2,-2)\). A normal vector is their cross product:

\(\mathbf n_1=\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\\\1&-2&-1\\\\1&2&-2\end{vmatrix}=(6,1,4)\).

Since \(\mathbf n_1\) is normal to \(\Pi_1\), the cartesian equation has the form \(6x+y+4z=c\). Using the point \((2,-3,1)\) on the plane:

\(6(2)+(-3)+4(1)=12-3+4=13\).

Hence \(\Pi_1: 6x+y+4z=13\).

For the acute angle between the planes, we use the angle between their normal vectors. A normal to \(\Pi_2: 3x-2y-3z=4\) is \(\mathbf n_2=(3,-2,-3)\).

So

\(\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}=\frac{|6\cdot3+1\cdot(-2)+4\cdot(-3)|}{\sqrt{6^2+1^2+4^2}\,\sqrt{3^2+(-2)^2+(-3)^2}}=\frac{4}{\sqrt{53}\sqrt{22}}\).

Therefore \(\theta\approx 83.3^\circ\).

To find the line of intersection, solve the two plane equations simultaneously:

\(6x+y+4z=13\)

\(3x-2y-3z=4\).

Eliminating \(x\) gives \(y+2z=1\), and eliminating \(y\) gives \(3x+z=6\).

Let \(z=3t\). Then \(y=1-2z=1-6t\) and \(3x+3t=6\), so \(x=2-t\).

Hence a vector equation of the line is

\(\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}2\\1\\0\end{array}\right)+t\left(\begin{array}{c}-1\\-6\\3\end{array}\right),\quad t\in\mathbb R.\)

Answer: (a) \(\tanh^2 t+\operatorname{sech}^2 t=1\).

(b) The length of \(C\) is \(1\).

(a) From the exponential definitions,

\(\tanh t=\dfrac{e^t-e^{-t}}{e^t+e^{-t}}, \qquad \operatorname{sech} t=\dfrac{2}{e^t+e^{-t}}.\)

Hence

\(\tanh^2 t+\operatorname{sech}^2 t=\left(\dfrac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2+\dfrac{4}{(e^t+e^{-t})^2}\)

\(=\dfrac{(e^t-e^{-t})^2+4}{(e^t+e^{-t})^2}\)

\(=\dfrac{e^{2t}-2+e^{-2t}+4}{(e^t+e^{-t})^2}\)

\(=\dfrac{e^{2t}+2+e^{-2t}}{(e^t+e^{-t})^2}.\)

But

\((e^t+e^{-t})^2=e^{2t}+2+e^{-2t},\)

so the fraction is equal to \(1\). Therefore

\(\tanh^2 t+\operatorname{sech}^2 t=1.\)

(b) For the parametric curve,

\(x=\ln(\cosh t), \qquad y=\tan^{-1}(\sinh t), \qquad 0\le t\le 1.\)

Differentiate:

\(\dfrac{dx}{dt}=\dfrac{1}{\cosh t}\cdot \sinh t=\tanh t.\)

Also

\(\dfrac{dy}{dt}=\dfrac{1}{1+\sinh^2 t}\cdot \cosh t=\dfrac{\cosh t}{1+\sinh^2 t}.\)

Using \(1+\sinh^2 t=\cosh^2 t\),

\(\dfrac{dy}{dt}=\dfrac{\cosh t}{\cosh^2 t}=\dfrac{1}{\cosh t}=\operatorname{sech} t.\)

The arc length is

\(s=\int_0^1 \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\)

\(=\int_0^1 \sqrt{\tanh^2 t+\operatorname{sech}^2 t}\,dt.\)

From part (a), \(\tanh^2 t+\operatorname{sech}^2 t=1\), so

\(s=\int_0^1 1\,dt=1.\)

Hence the length of \(C\) is \(1\).

Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac12\) at \(\left(4,\frac13\right)\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{29}{4}\) at \(\left(4,\frac13\right)\).

The curve is

\(9y^2-3\sinh^{-1}(xy)=1-3\ln 3\).

Differentiating implicitly with respect to \(x\),

\(18yy'-3\dfrac{xy'+y}{\sqrt{1+x^2y^2}}=0\).

(a) At \(\left(4,\frac13\right)\),

\(18\left(\frac13\right)y'-3\dfrac{4y'+\frac13}{\sqrt{1+16\left(\frac13\right)^2}}=0\)

\(6y'-3\dfrac{4y'+\frac13}{5/3}=0\)

\(6y'-\frac95\left(4y'+\frac13\right)=0\).

So

\(30y'-36y'-3=0\)

\(-6y'-3=0\)

and hence

\(y'=-\frac12\).

(b) Differentiate

\(18yy'-3\dfrac{xy'+y}{\sqrt{1+x^2y^2}}=0\)

again.

For the first term,

\(\dfrac{\mathrm{d}}{\mathrm{d}x}(18yy')=18yy''+18(y')^2\).

For the second term, using the product rule on \((xy'+y)(1+x^2y^2)^{-1/2}\),

\(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{xy'+y}{\sqrt{1+x^2y^2}}\right)=\frac{xy''+2y'}{\sqrt{1+x^2y^2}}-\frac{(xy'+y)^2(xy)}{(1+x^2y^2)^{3/2}}.\)

So

\(18yy''+18(y')^2-3\left(\dfrac{xy''+2y'}{\sqrt{1+x^2y^2}}-\dfrac{(xy'+y)^2(xy)}{(1+x^2y^2)^{3/2}}\right)=0\).

Now substitute \(x=4\), \(y=\frac13\), \(y'=-\frac12\):

\(1+x^2y^2=\frac{25}{9},\quad \sqrt{1+x^2y^2}=\frac53,\quad xy'+y=4\left(-\frac12\right)+\frac13=-\frac53,\quad xy=\frac43.\)

This gives

\(18\left(\frac13\right)y''+18\left(\frac14\right)-3\left(\frac{4y''-1}{5/3}-\frac{\left(\frac{25}{9}\right)\left(\frac43\right)}{(25/9)(5/3)}\right)=0\).

Equivalently,

\(6y''+\frac92-3\left(\frac35(4y''-1)-\frac45\right)=0\).

Simplifying,

\(6y''+\frac92-\frac{36}{5}y''+\frac{21}{5}=0\)

\(-\frac65y''+\frac{87}{10}=0\).

Hence

\(y''=\frac{29}{4}\).

Therefore, at \(\left(4,\frac13\right)\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{29}{4}\).

Answer: (a) \(1-\tanh^2 u=\operatorname{sech}^2 u\).

(b) \(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1} t)=-\frac{1}{t\sqrt{1-t^2}}\).

(c) \(\frac{\mathrm{d}y}{\mathrm{d}x}=-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1} t\).

(d) \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=t\sqrt{1-t^2}-\frac{(1-t^2)^{3/2}}{t}-2t(1-t^2)\operatorname{sech}^{-1} t\).

(a) From the definitions,

\(\tanh u=\frac{\mathrm{e}^u-\mathrm{e}^{-u}}{\mathrm{e}^u+\mathrm{e}^{-u}}\), \(\operatorname{sech}u=\frac{2}{\mathrm{e}^u+\mathrm{e}^{-u}}\).

Then

\(1-\tanh^2u=1-\left(\frac{\mathrm{e}^u-\mathrm{e}^{-u}}{\mathrm{e}^u+\mathrm{e}^{-u}}\right)^2\)

\(=\frac{(\mathrm{e}^u+\mathrm{e}^{-u})^2-(\mathrm{e}^u-\mathrm{e}^{-u})^2}{(\mathrm{e}^u+\mathrm{e}^{-u})^2}\)

\(=\frac{4}{(\mathrm{e}^u+\mathrm{e}^{-u})^2}\)

\(=\left(\frac{2}{\mathrm{e}^u+\mathrm{e}^{-u}}\right)^2=\operatorname{sech}^2u\).

(b) Let \(u=\operatorname{sech}^{-1}t\). Then \(\operatorname{sech}u=t\).

Differentiating implicitly with respect to \(t\),

\(-\tanh u\,\operatorname{sech}u\,\frac{\mathrm{d}u}{\mathrm{d}t}=1\).

So

\(\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{1}{\tanh u\operatorname{sech}u}\).

Using part (a),

\(\tanh^2u=1-\operatorname{sech}^2u=1-t^2\).

Since \(0<t<1\), we take \(\tanh u=\sqrt{1-t^2}\), and \(\operatorname{sech}u=t\). Hence

\(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1}t)=\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{1}{t\sqrt{1-t^2}}\).

(c) Given \(x=\tanh^{-1}t\) and \(y=t\operatorname{sech}^{-1}t\),

\(\frac{\mathrm{d}y}{\mathrm{d}t}=\operatorname{sech}^{-1}t+t\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1}t)=\operatorname{sech}^{-1}t-\frac{1}{\sqrt{1-t^2}}\).

Also

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{1-t^2}\).

Therefore

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}=\left(\operatorname{sech}^{-1}t-\frac{1}{\sqrt{1-t^2}}\right)(1-t^2)\)

\(=-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1}t\).

(d) Differentiate the result from part (c) with respect to \(t\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1}t\right)\)

\(=\frac{t}{\sqrt{1-t^2}}-\frac{1-t^2}{t\sqrt{1-t^2}}-2t\operatorname{sech}^{-1}t\).

Now

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot\frac{\mathrm{d}t}{\mathrm{d}x}\), and since \(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{1-t^2}\), we have \(\frac{\mathrm{d}t}{\mathrm{d}x}=1-t^2\).

So

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=(1-t^2)\left(\frac{t}{\sqrt{1-t^2}}-\frac{1-t^2}{t\sqrt{1-t^2}}-2t\operatorname{sech}^{-1}t\right)\)

\(=t\sqrt{1-t^2}-\frac{(1-t^2)^{3/2}}{t}-2t(1-t^2)\operatorname{sech}^{-1}t\).

Answer: (a) \(s=\int_{0}^{\frac{3}{5}} \sqrt{\cosh 2t}\,dt\).

(b) \(\sqrt{\cosh 2t}=1+t^2+\cdots\), so

\(s\approx \int_0^{3/5}(1+t^2)\,dt=\left[t+\frac13 t^3\right]_0^{3/5}=\frac{84}{125}=0.672\).

(a) For a curve given parametrically by \(x=x(t)\), \(y=y(t)\), the arc length from \(t=a\) to \(t=b\) is

\(s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here

\(x=\cosh t,\qquad y=\sinh t,\)

so

\(\frac{dx}{dt}=\sinh t,\qquad \frac{dy}{dt}=\cosh t.\)

Hence

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=\sinh^2 t+\cosh^2 t.\)

Using \(\cosh 2t=\cosh^2 t+\sinh^2 t\), this becomes

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=\cosh 2t.\)

Therefore

\(s=\int_0^{3/5}\sqrt{\cosh 2t}\,dt.\)

(b) Let \(f(t)=\sqrt{\cosh 2t}=(\cosh 2t)^{1/2}.\)

We need the Maclaurin series up to and including the term in \(t^2\).

First,

\(f(0)=\sqrt{\cosh 0}=1.\)

Differentiate:

\(f'(t)=\frac12(\cosh 2t)^{-1/2}(2\sinh 2t)=\sinh 2t\,(\cosh 2t)^{-1/2}.\)

So

\(f'(0)=0.\)

Differentiate again:

\(f''(t)=2\sqrt{\cosh 2t}-\sinh^2 2t\,(\cosh 2t)^{-3/2}.\)

Hence

\(f''(0)=2.\)

Therefore the Maclaurin series is

\(f(t)=f(0)+f'(0)t+\frac{f''(0)}{2}t^2+\cdots=1+t^2+\cdots\)

So

\(\sqrt{\cosh 2t}\approx 1+t^2.\)

Thus

\(s\approx \int_0^{3/5}(1+t^2)\,dt=\left[t+\frac13 t^3\right]_0^{3/5}.\)

This gives

\(s\approx \frac35+\frac13\left(\frac35\right)^3=\frac35+\frac{9}{125}=\frac{84}{125}=0.672.\)

Answer: (a) \((\cosh x+\sinh x)^{\frac{1}{2}}=\mathrm{e}^{x/2}\).

(b) The solution satisfying the given conditions is

\(y=\mathrm{e}^{-x/2}\left(-\frac{1}{3}\cos\left(\frac{\sqrt{11}}{2}x\right)+\frac{1}{\sqrt{11}}\sin\left(\frac{\sqrt{11}}{2}x\right)\right)+\frac{4}{3}\mathrm{e}^{x/2}.\)

(a) Using \(\cosh x=\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}\) and \(\sinh x=\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}\),

\(\cosh x+\sinh x=\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}+\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}=\mathrm{e}^x.\)

Hence

\((\cosh x+\sinh x)^{1/2}=(\mathrm{e}^x)^{1/2}=\mathrm{e}^{x/2}.\)

(b) From part (a), the equation becomes

\(y''+y'+3y=5\mathrm{e}^{x/2}.\)

For the complementary function, solve

\(m^2+m+3=0.\)

This gives

\(m=\frac{-1\pm i\sqrt{11}}{2}.\)

So

\(y_c=\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{11}}{2}x\right)+B\sin\left(\frac{\sqrt{11}}{2}x\right)\right).\)

For a particular integral, try

\(y_p=k\mathrm{e}^{x/2}.\)

Then

\(y_p'=\frac12k\mathrm{e}^{x/2},\qquad y_p''=\frac14k\mathrm{e}^{x/2}.\)

Substitute into the differential equation:

\(\frac14k\mathrm{e}^{x/2}+\frac12k\mathrm{e}^{x/2}+3k\mathrm{e}^{x/2}=5\mathrm{e}^{x/2}.\)

So

\(\left(\frac14+\frac12+3\right)k=5 \Rightarrow \frac{15}{4}k=5,\)

hence

\(k=\frac43.\)

Therefore

\(y=\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{11}}{2}x\right)+B\sin\left(\frac{\sqrt{11}}{2}x\right)\right)+\frac43\mathrm{e}^{x/2}.\)

Use the conditions.

When \(x=0\), \(y=1\):

\(1=A+\frac43,\)

so

\(A=-\frac13.\)

Differentiate:

\(y'=\mathrm{e}^{-x/2}\left(-\frac{\sqrt{11}}{2}A\sin\left(\frac{\sqrt{11}}{2}x\right)+\frac{\sqrt{11}}{2}B\cos\left(\frac{\sqrt{11}}{2}x\right)\right)-\frac12\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{11}}{2}x\right)+B\sin\left(\frac{\sqrt{11}}{2}x\right)\right)+\frac23\mathrm{e}^{x/2}.\)

When \(x=0\), \(y'=\frac43\):

\(\frac43=\frac{\sqrt{11}}{2}B-\frac12A+\frac23.\)

Substitute \(A=-\frac13\):

\(\frac43=\frac{\sqrt{11}}{2}B+\frac16+\frac23=\frac{\sqrt{11}}{2}B+\frac56.\)

So

\(\frac12=\frac{\sqrt{11}}{2}B,\)

giving

\(B=\frac{1}{\sqrt{11}}.\)

Hence the required solution is

\(y=\mathrm{e}^{-x/2}\left(-\frac{1}{3}\cos\left(\frac{\sqrt{11}}{2}x\right)+\frac{1}{\sqrt{11}}\sin\left(\frac{\sqrt{11}}{2}x\right)\right)+\frac{4}{3}\mathrm{e}^{x/2}.\)

Answer: (a) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}-2y=4\mathrm{e}^{-x}\).

(b) \(\displaystyle u=\cosh^{-1}\!\left(\frac{28}{9}\mathrm{e}^x+\frac{5}{9}\mathrm{e}^{-2x}-2\mathrm{e}^{-x}\right)\).

(a)

Since \(y=\cosh u\),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\sinh u\,\frac{\mathrm{d}u}{\mathrm{d}x}.\)

Also, because \(u\gt 0\), we have \(\sinh u\gt 0\), so

\(\displaystyle \sqrt{\cosh^2u-1}=\sqrt{\sinh^2u}=\sinh u.\)

Differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x}=\sinh u\,\frac{\mathrm{d}u}{\mathrm{d}x}\):

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\cosh u\left(\frac{\mathrm{d}u}{\mathrm{d}x}\right)^2+\sinh u\,\frac{\mathrm{d}^2u}{\mathrm{d}x^2}.\)

Hence

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}-2y\)

\(\displaystyle =\cosh u\left(\frac{\mathrm{d}u}{\mathrm{d}x}\right)^2+\sinh u\,\frac{\mathrm{d}^2u}{\mathrm{d}x^2}+\sinh u\,\frac{\mathrm{d}u}{\mathrm{d}x}-2\cosh u\)

\(\displaystyle =\sqrt{\cosh^2u-1}\left(\frac{\mathrm{d}^2u}{\mathrm{d}x^2}+\frac{\mathrm{d}u}{\mathrm{d}x}\right)+\cosh u\left(\frac{\mathrm{d}u}{\mathrm{d}x}\right)^2-2\cosh u.\)

This is the given left-hand side, so

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}-2y=4\mathrm{e}^{-x}.\)

(b)

Solve

\(\displaystyle y''+y'-2y=4\mathrm{e}^{-x}.\)

The auxiliary equation is

\(\displaystyle m^2+m-2=0,\)

so

\(\displaystyle (m-1)(m+2)=0\)

and therefore \(m=1,-2\).

Hence the complementary function is

\(\displaystyle y_c=A\mathrm{e}^x+B\mathrm{e}^{-2x}.\)

For a particular integral, try \(\displaystyle y_p=k\mathrm{e}^{-x}\).

Then

\(\displaystyle y_p'=-k\mathrm{e}^{-x},\qquad y_p''=k\mathrm{e}^{-x}.\)

Substituting gives

\(\displaystyle k\mathrm{e}^{-x}-k\mathrm{e}^{-x}-2k\mathrm{e}^{-x}=4\mathrm{e}^{-x},\)

so \(\displaystyle -2k=4\), hence \(\displaystyle k=-2\).

Therefore

\(\displaystyle y=A\mathrm{e}^x+B\mathrm{e}^{-2x}-2\mathrm{e}^{-x}.\)

Since \(y=\cosh u\),

\(\displaystyle \cosh u=A\mathrm{e}^x+B\mathrm{e}^{-2x}-2\mathrm{e}^{-x}.\)

Also

\(\displaystyle y'=A\mathrm{e}^x-2B\mathrm{e}^{-2x}+2\mathrm{e}^{-x},\)

and from \(y=\cosh u\),

\(\displaystyle y'=\sinh u\,\frac{\mathrm{d}u}{\mathrm{d}x}.\)

When \(x=0\), \(u=\ln 3\), so

\(\displaystyle y(0)=\cosh(\ln 3)=\frac{3+\frac13}{2}=\frac53.\)

Thus

\(\displaystyle A+B-2=\frac53.\)

Also, when \(x=0\), \(\frac{\mathrm{d}u}{\mathrm{d}x}=3\), and

\(\displaystyle \sinh(\ln 3)=\frac{3-\frac13}{2}=\frac43,\)

so

\(\displaystyle y'(0)=\sinh(\ln 3)\cdot 3=\frac43\cdot 3=4.\)

Hence

\(\displaystyle A-2B+2=4.\)

So the simultaneous equations are

\(\displaystyle A+B=\frac{11}{3},\qquad A-2B=2.\)

Solving,

\(\displaystyle B=\frac59,\qquad A=\frac{28}{9}.\)

Therefore

\(\displaystyle \cosh u=\frac{28}{9}\mathrm{e}^x+\frac{5}{9}\mathrm{e}^{-2x}-2\mathrm{e}^{-x}.\)

Since \(u\gt 0\),

\(\displaystyle u=\cosh^{-1}\!\left(\frac{28}{9}\mathrm{e}^x+\frac{5}{9}\mathrm{e}^{-2x}-2\mathrm{e}^{-x}\right).\)