AQA A-Level Maths: Calculus Applications

A food company makes a closed cylindrical tin (a circular base, a circular top and a curved side) to hold a fixed volume of $250\pi\ \text{cm}^3$250π cm3 of soup. The tin has base radius $r\ \text{cm}$r cm and height $h\ \text{cm}$h cm. To keep material costs down, the company wants to minimise the total external surface area $S\ \text{cm}^2$S cm2 of the tin.

(a) Using the fixed-volume condition to eliminate $h$h, show that $S = 2\pi r^2 + \frac{500\pi}{r}.$S=2πr2+r500π​. (4 marks)

(b) Find $\dfrac{dS}{dr}$drdS​ and hence find the value of $r$r for which $S$S is stationary. (5 marks)

(c) Justify that this value of $r$r gives a minimum surface area, find the corresponding height $h$h, and state the minimum surface area as an exact multiple of $\pi$π. (3 marks)

The curve $C$C has equation $y = x^2$y=x2 and the line $l$l has equation $y = x + 2$y=x+2.

(a) Find the coordinates of the two points where $l$l meets $C$C. (3 marks)

(b) The line $l$l and the curve $C$C enclose a finite region $R$R. Use integration to find the exact area of $R$R. (4 marks)

(c) A second region $T$T is bounded by the curve $C$C, the $x$x-axis and the line $x = 2$x=2. The region $T$T is rotated through $2\pi$2π radians about the $x$x-axis. Find the exact volume of the solid generated. (3 marks)

(a) Using integration by parts with $u = \ln x$u=lnx, find $\displaystyle\int \ln x\,dx$∫lnxdx. (4 marks)

(b) Hence show that $\displaystyle\int_1^e \ln x\,dx = 1$∫1e​lnxdx=1. (4 marks)

Water is poured into an inverted right circular cone (vertex pointing downwards) at a constant rate of $8\ \text{cm}^3\,\text{s}^{-1}$8 cm3s−1. The cone is shaped so that, for the water surface, the radius is always half the depth; that is, when the water has depth $h\ \text{cm}$h cm its surface radius is $r = \tfrac{1}{2}h\ \text{cm}$r=21​h cm.

The volume of a cone of base radius $r$r and height $h$h is $V = \dfrac{1}{3}\pi r^2 h$V=31​πr2h.

Find the rate at which the depth of the water is increasing at the instant when $h = 4\ \text{cm}$h=4 cm, giving your answer as an exact value in $\text{cm}\,\text{s}^{-1}$cms−1.

(6 marks)

A curve passes through the point $(0,\,3)$(0,3) and, at every point $(x,\,y)$(x,y) on the curve with $y > 0$y>0, its gradient satisfies the differential equation $\frac{dy}{dx} = \frac{x}{y}.$dxdy​=yx​.

By separating the variables, find $y$y as a function of $x$x, giving your answer in the form $y = f(x)$y=f(x).

(5 marks)

The table below gives values of $y = \sqrt{x}$y=x​, correct to four decimal places, for five equally spaced values of $x$x.

$x$x	$0$0	$1$1	$2$2	$3$3	$4$4
$y$y	$0$0	$1.0000$1.0000	$1.4142$1.4142	$1.7321$1.7321	$2.0000$2.0000

(a) Use the trapezium rule with all five values to find an estimate for $\displaystyle\int_0^4 \sqrt{x}\,dx$∫04​x​dx, giving your answer to three decimal places. (2 marks)

(b) State, with a reason, whether the trapezium rule gives an over-estimate or an under-estimate of the true value of this integral. (2 marks)