Answer: \(\frac{\mathrm{d}S}{\mathrm{d}t}=\frac{4\pi}{3}\text{ cm}^2\text{ s}^{-1}\).

For a sphere,

\(V=\frac43\pi r^3.\)

When \(V=36\pi\),

\(\frac43\pi r^3=36\pi,\)

so \(r^3=27\), and therefore \(r=3\).

Differentiate \(V=\frac43\pi r^3\) with respect to \(r\):

\(\frac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^2.\)

Using \(\frac{\mathrm{d}V}{\mathrm{d}t}=2\pi\),

\(2\pi=4\pi r^2\frac{\mathrm{d}r}{\mathrm{d}t}.\)

At \(r=3\),

\(\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{2\pi}{36\pi}=\frac{1}{18}.\)

The surface area is \(S=4\pi r^2\), so

\(\frac{\mathrm{d}S}{\mathrm{d}r}=8\pi r.\)

At \(r=3\), \(\frac{\mathrm{d}S}{\mathrm{d}r}=24\pi\). Hence

\(\frac{\mathrm{d}S}{\mathrm{d}t}=\frac{\mathrm{d}S}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}=24\pi\cdot\frac{1}{18}=\frac{4\pi}{3}.\)

Answer: \(S=\dfrac{9}{8\pi}\).

The cylinder is open at both ends, so its curved surface area is

\(S=2\pi rh.\)

Since \(h=4r\),

\(S=2\pi r(4r)=8\pi r^2.\)

Differentiate with respect to \(t\):

\(\frac{\mathrm dS}{\mathrm dt}=16\pi r\frac{\mathrm dr}{\mathrm dt}.\)

It is given that

\(\frac{\mathrm dS}{\mathrm dt}=6\frac{\mathrm dr}{\mathrm dt}.\)

So

\(16\pi r\frac{\mathrm dr}{\mathrm dt}=6\frac{\mathrm dr}{\mathrm dt}.\)

At the instant considered, \(\frac{\mathrm dr}{\mathrm dt}\neq0\), so

\(16\pi r=6.\)

Hence

\(r=\frac{3}{8\pi}.\)

Therefore

\(S=8\pi\left(\frac{3}{8\pi}\right)^2 =\frac{9}{8\pi}.\)

This completes the solution and gives the required result.

Answer: \(\displaystyle \frac{dr}{dt}=\frac{1}{6\pi}\text{ cm s}^{-1}\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

Differentiate

\(V=\frac43\pi r^3\)

with respect to \(r\):

\(\frac{dV}{dr}=4\pi r^2.\)

At \(r=6\),

\(\frac{dV}{dr}=4\pi(6)^2=144\pi.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}.\)

Since \(\frac{dV}{dt}=24\),

\(24=144\pi\frac{dr}{dt}.\)

Hence

\(\frac{dr}{dt}=\frac{24}{144\pi}=\frac{1}{6\pi}.\)

Answer: (b) \(\displaystyle \frac{\sqrt3}{2}\) m min\(^{-1}\), or \(0.866\) m min\(^{-1}\).

Answer: (b) \(\displaystyle \frac{\sqrt3}{2}\) m min\(^{-1}\), or \(0.866\) m min\(^{-1}\).

(a) In an equilateral triangle, the height is \(\frac{\sqrt3}{2}\) times the side length. For the water cross-section,

\(h=\frac{\sqrt3}{2}x\).

Hence

\(x=\frac{2h}{\sqrt3}\).

The area of the triangular cross-section of water is

\(\frac12 xh\).

Substitute \(x=\frac{2h}{\sqrt3}\):

The container has length 5 m, so the volume is

(b) Differentiate \(V=\frac{5\sqrt3}{3}h^2\) with respect to \(h\):

\(\displaystyle \frac{dV}{dh}=\frac{10\sqrt3}{3}h\).

Using the chain rule,

When \(h=0.1\),

Given \(\frac{dV}{dt}=0.5\),

\(\displaystyle 0.5=\frac{\sqrt3}{3}\frac{dh}{dt}\).

Therefore

So the depth is increasing at \(\frac{\sqrt3}{2}\) m min\(^{-1}\), or approximately \(0.866\) m min\(^{-1}\).

Answer: (a) \(2.50\text{ cm s}^{-1}\); (b) \(240\text{ cm}^2\text{ s}^{-1}\).

For a cube of side \(x\),

\(V=x^3.\)

When \(V=512\),

\(x^3=512,\)

so

\(x=8.\)

(a) Differentiate \(V=x^3\) with respect to \(x\):

\(\frac{dV}{dx}=3x^2.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}.\)

So at \(x=8\),

\(480=3(8^2)\frac{dx}{dt}.\)

Thus

\(\frac{dx}{dt}=\frac{480}{192}=2.5.\)

The rate of increase of \(x\) is

\(2.50\text{ cm s}^{-1}.\)

(b) The surface area is

\(S=6x^2.\)

Hence

\(\frac{dS}{dx}=12x.\)

Using the chain rule,

\(\frac{dS}{dt}=\frac{dS}{dx}\frac{dx}{dt}.\)

At \(x=8\),

\(\frac{dS}{dt}=12(8)(2.5)=240.\)

So the rate of increase of \(S\) is

\(240\text{ cm}^2\text{ s}^{-1}.\)

Answer: \(30\pi\text{ cm}^2\text{ s}^{-1}\).

The area of the circle is

\(A=\pi r^2.\)

Differentiate with respect to \(r\):

\(\frac{dA}{dr}=2\pi r.\)

The radius is increasing at

\(\frac{dr}{dt}=5.\)

Using the chain rule,

\(\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}.\)

So

\(\frac{dA}{dt}=2\pi r\cdot5=10\pi r.\)

When \(r=3\),

\(\frac{dA}{dt}=10\pi(3)=30\pi.\)

The area is increasing at

\(30\pi\text{ cm}^2\text{ s}^{-1}.\)

Answer: \(\dfrac{\pi}{8}\text{ cm}^3\text{ s}^{-1}\).

The volume is

\(V=\frac43\pi r^3.\)

Differentiate with respect to \(r\):

\(\frac{dV}{dr}=4\pi r^2.\)

Also

\(\frac{dr}{dt}=0.5.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}.\)

So

\(\frac{dV}{dt}=4\pi r^2(0.5)=2\pi r^2.\)

When \(r=0.25=\frac14\),

\(\frac{dV}{dt}=2\pi\left(\frac14\right)^2 =2\pi\cdot\frac1{16} =\frac{\pi}{8}.\)

Therefore the volume is increasing at

\(\frac{\pi}{8}\text{ cm}^3\text{ s}^{-1}.\)

Answer: \(\frac{dy}{du}=\operatorname{sec}^2u\), and \(\frac{dy}{dx}=\dfrac{\operatorname{sec}^2\left(\sqrt[3]{x-1}\right)}{3\sqrt[3]{(x-1)^2}}\).

Since \(y=\tan u\),

\(\frac{dy}{du}=\boxed{\operatorname{sec}^2u}.\)

Also,

\(x=u^3+1.\)

Differentiate with respect to \(u\):

\(\frac{dx}{du}=3u^2.\)

Using the chain rule,

\(\frac{dy}{dx}=\frac{\frac{dy}{du}}{\frac{dx}{du}} =\frac{\operatorname{sec}^2u}{3u^2}.\)

From \(x=u^3+1\),

\(u=\sqrt[3]{x-1}.\)

Therefore

\(\frac{dy}{dx} =\boxed{\frac{\operatorname{sec}^2\left(\sqrt[3]{x-1}\right)}{3\sqrt[3]{(x-1)^2}}}.\)

Answer: \(\dfrac{3\pi}{100}\text{ cm}^2\text{ s}^{-1}\).

If the diameter is \(x\), then the radius is \(\frac{x}{2}\). Hence the area is

\(A=\pi\left(\frac{x}{2}\right)^2=\frac{\pi x^2}{4}.\)

Differentiate with respect to \(x\):

\(\frac{dA}{dx}=\frac{\pi x}{2}.\)

Using the chain rule,

\(\frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt}.\)

Given \(\frac{dx}{dt}=0.01=\frac{1}{100}\), when \(x=6\),

\(\frac{dA}{dt}=\frac{\pi(6)}{2}\cdot\frac{1}{100} =\frac{3\pi}{100}.\)

So the exact rate of change of area is

\(\boxed{\frac{3\pi}{100}\text{ cm}^2\text{ s}^{-1}}.\)

Answer: (i) \(\dfrac1{2\pi}\text{ cm s}^{-1}\approx0.159\text{ cm s}^{-1}\). (ii) \(40\text{ cm}^2\text{ s}^{-1}\).

We are given

\(\dfrac{dV}{dt}=200.\)

Since

\(V=\dfrac43\pi r^3,\)

differentiate with respect to \(r\):

\(\dfrac{dV}{dr}=4\pi r^2.\)

By the chain rule,

\(\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}.\)

At \(r=10\),

\(200=4\pi(10)^2\dfrac{dr}{dt}=400\pi\dfrac{dr}{dt}.\)

So

\(\dfrac{dr}{dt}=\dfrac{200}{400\pi}=\dfrac1{2\pi}.\)

Therefore the radius is increasing at

\(\boxed{\dfrac1{2\pi}\text{ cm s}^{-1}}\), approximately \(0.159\text{ cm s}^{-1}\).

Now

\(S=4\pi r^2,\)

so

\(\dfrac{dS}{dr}=8\pi r.\)

Again using the chain rule,

\(\dfrac{dS}{dt}=\dfrac{dS}{dr}\cdot\dfrac{dr}{dt}.\)

At \(r=10\),

\(\dfrac{dS}{dt}=8\pi(10)\cdot\dfrac1{2\pi}=40.\)

Thus the surface area is increasing at

\(\boxed{40\text{ cm}^2\text{ s}^{-1}}.\)

Answer: (i) \(V=5\sqrt3h^2\). (ii)(a) \(\dfrac{dh}{dt}=\dfrac1{5\sqrt3}=0.115\text{ m min}^{-1}\). (ii)(b) \(\dfrac{dx}{dt}=\dfrac25\text{ m min}^{-1}\).

(i) The perpendicular depth \(h\) bisects the base \(AB\), so each half of the base has length \(\frac{x}{2}\).

In one of the right-angled triangles,

\(\tan30^\circ=\frac{h}{x/2}.\)

Since \(\tan30^\circ=\frac1{\sqrt3}\),

\(\frac1{\sqrt3}=\frac{2h}{x}.\)

Therefore

\(x=2\sqrt3h.\)

The cross-sectional area of water is

\(\frac12xh.\)

The length of the container is \(5\), so

\(V=5\left(\frac12xh\right).\)

Substitute \(x=2\sqrt3h\):

\(V=5\left(\frac12(2\sqrt3h)h\right) =5\sqrt3h^2.\)

(ii)(a) Differentiate \(V=5\sqrt3h^2\) with respect to \(h\):

\(\frac{dV}{dh}=10\sqrt3h.\)

At \(h=0.25\),

\(\frac{dV}{dh}=10\sqrt3(0.25)=2.5\sqrt3.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}.\)

The filling rate is \(\frac{dV}{dt}=0.5\), so

\(0.5=2.5\sqrt3\frac{dh}{dt}.\)

Hence

\(\frac{dh}{dt} =\frac{0.5}{2.5\sqrt3} =\frac1{5\sqrt3} =0.115\ldots.\)

So \(h\) is increasing at \(0.115\text{ m min}^{-1}\).

(ii)(b) Since

\(x=2\sqrt3h,\)

differentiate with respect to \(t\):

\(\frac{dx}{dt}=2\sqrt3\frac{dh}{dt}.\)

Substitute \(\frac{dh}{dt}=\frac1{5\sqrt3}\):

\(\frac{dx}{dt} =2\sqrt3\left(\frac1{5\sqrt3}\right) =\frac25.\)

Answer: (i) \(r=h(2-\sqrt3)\) and \(V=\dfrac{\pi(7-4\sqrt3)h^3}{3}\). (ii) \(\dfrac{dh}{dt}=5.32\text{ cm s}^{-1}\).

(i) In the cross-section, the half-angle at the bottom of the cone is \(\frac{\pi}{12}\). Therefore

\(\tan\frac{\pi}{12}=\frac{r}{h}.\)

Using \(\tan\frac{\pi}{12}=2-\sqrt3\),

\(\frac{r}{h}=2-\sqrt3.\)

So

\(r=h(2-\sqrt3).\)

Now use \(V=\frac13\pi r^2h\):

\(V=\frac13\pi\left(h(2-\sqrt3)\right)^2h.\)

Therefore

\(V=\frac13\pi h^3(2-\sqrt3)^2.\)

Expand the square:

\((2-\sqrt3)^2=4-4\sqrt3+3=7-4\sqrt3.\)

Hence

\(V=\frac{\pi(7-4\sqrt3)h^3}{3}.\)

(ii) Differentiate \(V\) with respect to \(h\):

\(\frac{dV}{dh}=\pi(7-4\sqrt3)h^2.\)

Use the chain rule:

\(\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}.\)

Since \(\frac{dV}{dt}=30\),

\(30=\pi(7-4\sqrt3)h^2\frac{dh}{dt}.\)

When \(h=5\),

\(30=25\pi(7-4\sqrt3)\frac{dh}{dt}.\)

Thus

\(\frac{dh}{dt} = \frac{30}{25\pi(7-4\sqrt3)} =5.32018\ldots.\)

So the depth is increasing at

\(5.32\text{ cm s}^{-1}.\)

Answer: (i) \(V=5\sqrt3h^2\). (ii)(a) \(\dfrac{dh}{dt}=\dfrac1{5\sqrt3}=0.115\text{ m min}^{-1}\). (ii)(b) \(\dfrac{dx}{dt}=\dfrac25\text{ m min}^{-1}\).

(i) The perpendicular depth \(h\) bisects the base \(AB\), so each half of the base has length \(\frac{x}{2}\).

In one of the right-angled triangles,

\(\tan30^\circ=\frac{h}{x/2}.\)

Since \(\tan30^\circ=\frac1{\sqrt3}\),

\(\frac1{\sqrt3}=\frac{2h}{x}.\)

Therefore

\(x=2\sqrt3h.\)

The cross-sectional area of water is

\(\frac12xh.\)

The length of the container is \(5\), so

\(V=5\left(\frac12xh\right).\)

Substitute \(x=2\sqrt3h\):

\(V=5\left(\frac12(2\sqrt3h)h\right) =5\sqrt3h^2.\)

(ii)(a) Differentiate \(V=5\sqrt3h^2\) with respect to \(h\):

\(\frac{dV}{dh}=10\sqrt3h.\)

At \(h=0.25\),

\(\frac{dV}{dh}=10\sqrt3(0.25)=2.5\sqrt3.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}.\)

The filling rate is \(\frac{dV}{dt}=0.5\), so

\(0.5=2.5\sqrt3\frac{dh}{dt}.\)

Hence

\(\frac{dh}{dt} =\frac{0.5}{2.5\sqrt3} =\frac1{5\sqrt3} =0.115\ldots.\)

So \(h\) is increasing at \(0.115\text{ m min}^{-1}\).

(ii)(b) Since

\(x=2\sqrt3h,\)

differentiate with respect to \(t\):

\(\frac{dx}{dt}=2\sqrt3\frac{dh}{dt}.\)

Substitute \(\frac{dh}{dt}=\frac1{5\sqrt3}\):

\(\frac{dx}{dt} =2\sqrt3\left(\frac1{5\sqrt3}\right) =\frac25.\)