Past Exam Questions

Variables \(s\) and \(t\) are such that \(s=4t+3e^{-t}\).

(i) Find the value of \(s\) when \(t=0\).

(ii) Find the exact value of \(t\) when \(\dfrac{ds}{dt}=2\).

(iii) Find the approximate increase in \(s\) when \(t\) increases from \(\ln5\) to \(\ln5+h\), where \(h\) is small.

A curve has equation

\(y=6x-x\sqrt{x}.\)

(i) Find the coordinates of the stationary point of the curve.

(ii) Determine the nature of this stationary point.

(iii) Find the approximate change in \(y\) when \(x\) increases from \(4\) to \(4+h\), where \(h\) is small.

The variables \(x\) and \(y\) are such that \(y=\ln(x^2+1)\).

(i) Find an expression for \(\dfrac{dy}{dx}\).

(ii) Hence, find the approximate change in \(y\) when \(x\) increases from \(3\) to \(3+h\), where \(h\) is small.

The volume, \(V\), of a sphere is increasing at the constant rate of \(2 \pi \mathrm{~cm}^{3} \mathrm{~s}^{-1}\). Find the rate of change of the surface area, \(S\), of this sphere when the volume of the sphere is \(36 \pi \mathrm{~cm}^{3}\).

A cylinder, open at both ends, has base radius \(r \mathrm{~cm}\) and height \(4 r \mathrm{~cm}\). Its curved surface area is \(S \mathrm{~cm}^{2}\). Given that \(r\) varies with time \(t\), find \(S\) at the instant when \(\frac{\mathrm{d} S}{\mathrm{~d} t}=6 \frac{\mathrm{~d} r}{\mathrm{~d} t}\).

The volume \(V\text{ cm}^3\) of a sphere of radius \(r\text{ cm}\) is given by

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

At the instant when \(r=6\), the volume is increasing at a rate of \(24\text{ cm}^3\text{s}^{-1}\). Find the rate at which the radius is increasing at this instant, giving your answer in terms of \(\pi\).

The diagram shows a container in the shape of a triangular prism. The container has length 5 m and the cross-section is an equilateral triangle. The container is being filled with water. At the instant when the depth of water is \(h\) m, the surface of the water has width \(x\) m.

(a) Show that the volume of water in the container is given by \(\displaystyle V=\frac{5\sqrt3}{3}h^2\).

(b) Water is pumped into the container at a rate of \(0.5\) m\(^3\) min\(^{-1}\). Find the rate at which the depth of the water is increasing when \(h=0.1\).

A cube of side \(x\text{ cm}\) has surface area \(S\text{ cm}^2\). The volume, \(V\text{ cm}^3\), of the cube is increasing at a rate of \(480\text{ cm}^3\text{ s}^{-1}\). Find, at the instant when \(V=512\),

(a) the rate of increase of \(x\),

(b) the rate of increase of \(S\).

The radius, \(r\) cm, of a circle is increasing at the rate of \(5\text{ cm s}^{-1}\). Find, in terms of \(\pi\), the rate at which the area of the circle is increasing when \(r=3\).

A sphere has volume \(V\text{ cm}^3\), where

\(V=\frac{4}{3}\pi r^3,\)

and \(r\) cm is the radius of the sphere.

The radius is increasing at a rate of \(0.5\text{ cm s}^{-1}\). Find, in terms of \(\pi\), the rate of change of the volume when \(r=0.25\).

The variables \(x\), \(y\) and \(u\) are such that \(y=\tan u\) and \(x=u^3+1\).

(i) State the rate of change of \(y\) with respect to \(u\).

(ii) Hence find the rate of change of \(y\) with respect to \(x\), giving your answer in terms of \(x\).

A circle has diameter \(x\) cm. The diameter is increasing at a constant rate of \(0.01\text{ cm s}^{-1}\). Find the exact rate of change of the area of the circle when \(x=6\).

The volume, \(V\), and surface area, \(S\), of a sphere of radius \(r\) are given by \(V=\dfrac43\pi r^3\) and \(S=4\pi r^2\), respectively.

The volume of a sphere increases at a rate of \(200\text{ cm}^3\) per second. At the instant when the radius of the sphere is \(10\) cm, find

(i) the rate of increase of the radius of the sphere,

(ii) the rate of increase of the surface area of the sphere.

In this question all lengths are in metres.

A water container is in the shape of a triangular prism. The cross-section of the water in the container is an isosceles triangle \(ABC\), with \(\angle ABC=\angle BAC=30^\circ\). The length of \(AB\) is \(x\) and the depth of water is \(h\). The length of the container is \(5\).

(i) Show that \(x=2\sqrt3h\) and hence find the volume of water in the container in terms of \(h\).

(ii) The container is filled at a rate of \(0.5\text{ m}^3\) per minute. At the instant when \(h=0.25\text{ m}\), find

(a) the rate at which \(h\) is increasing,

(b) the rate at which \(x\) is increasing.

A water cup is in the shape of a cone with its axis vertical. The vertical angle of the cone is \(\frac{\pi}{6}\) radians. The depth of water in the cup is \(h\), and the surface of the water is a circle of radius \(r\).

The volume of a cone of height \(h\) and base radius \(r\) is \(V=\frac13\pi r^2h\).

It is known that

\(\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}, \qquad \cos\frac{\pi}{12}=\frac{\sqrt6+\sqrt2}{4}, \qquad \tan\frac{\pi}{12}=2-\sqrt3.\)

(i) Find an expression for \(r\) in terms of \(h\) and show that the volume of water in the cup is

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

(ii) Water is poured into the cup at a rate of \(30\text{ cm}^3\text{ s}^{-1}\). Find, correct to 2 decimal places, the rate at which the depth of water is increasing when \(h=5\).

In this question all lengths are in metres.

A water container is in the shape of a triangular prism. The cross-section of the water in the container is an isosceles triangle \(ABC\), with \(\angle ABC=\angle BAC=30^\circ\). The length of \(AB\) is \(x\) and the depth of water is \(h\). The length of the container is \(5\).

(i) Show that \(x=2\sqrt3h\) and hence find the volume of water in the container in terms of \(h\).

(ii) The container is filled at a rate of \(0.5\text{ m}^3\) per minute. At the instant when \(h=0.25\text{ m}\), find

(a) the rate at which \(h\) is increasing,

(b) the rate at which \(x\) is increasing.

It is given that \(y=m x^{2}+\frac{x}{2}+n\), where \(m\) and \(n\) are non-zero constants. It is also given that \(3\left(\frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}\right)=\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^{2}-y\) for all values of \(x\). Find the values of \(m\) and \(n\).

A curve \(y=f(x)\) is such that \(\displaystyle \frac{d^2y}{dx^2}=(2-3x)^{-1/3}\). The curve passes through the point \((-2,10.2)\). The gradient of the tangent to the curve at \((-2,10.2)\) is \(-6\). Find \(f(x)\).

A curve has equation \(y=(3x-5)^3-2x\).

(i) Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\).

(ii) Find the exact value of the \(x\)-coordinate of each of the stationary points of the curve.

(iii) Use the second derivative test to determine the nature of each of the stationary points.

A curve is such that its gradient at the point \((x,y)\) is given by \((5x-2)^{\frac13}\). The curve passes through the point \(\left(2,\frac{32}{5}\right)\).

Find the coordinates of the stationary point on the curve.