Edexcel International A Level Maths · P4 · Pure Mathematics 4
Implicit differentiation and rates of change
28 free, high-quality past-paper-style questions on Implicit differentiation and rates of change for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 249 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-4-B-01P4 · A2Easy6 marks
The curve C has equation 3x2 − 2xy + y2 = 11. The point P(1, 4) lies on C.
(a)
Find dy/dx in terms of x and y.
(3)
(b)
Find the gradient of C at P.
(1)
(c)
Find an equation of the tangent to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(2)
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Answer
Mark
Notes
Part (a)
6x and 2ydy/dx from y2 (chain rule seen)
M1
−2xy differentiated as −2y − 2xdy/dx (product rule)
M1
dy/dx = (y − 3x)/(y − x) or equivalent
A1
Part (b)
(4 − 3)/(4 − 1) = 1/3
B1
Part (c)
y − 4 = ⅓(x − 1), using their gradient
M1
x − 3y + 11 = 0 (or any integer multiple)
A1
Model sketch
Figure: C: 3x² − 2xy + y² = 11 and the tangent at P
Common error: Treating y as a constant in −2xy. It is a product of two functions of x, so it gives −2y − 2x dy/dx.
Ink spilled on a sheet of fabric forms a circular stain. The area of the stain increases at a constant rate of 6 cm2 s−1. The stain is modelled as a circle of radius r cm.
(a)
Find the rate at which the radius is increasing at the instant when r = 4.
(3)
(b)
Find the rate at which the circumference of the stain is increasing at this instant.
(2)
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Answer
Mark
Notes
Part (a)
dA/dr = 2πr
B1
dr/dt = (dA/dt) ÷ (dA/dr) = 6/(8π)
M1
= 3/(4π) ≈ 0.239 cm s−1
A1
Part (b)
dC/dt = 2π × dr/dt, using C = 2πr and their dr/dt
M1
2π × 3/(4π) = 1.5 cm s−1
A1
Exam tip: Keep dr/dt as an exact fraction such as 3/(4π) when you use it again: the π cancels in the second part and gives an exact answer.
A pond has a surface area of 100 m2. Water lilies cover A m2 of the surface t days after they were planted. The rate at which the covered area increases is proportional to the product of the area covered and the area not yet covered.
(a)
Write down a differential equation connecting A and t, using k for the constant of proportionality.
(2)
(b)
When 20 m2 of the pond is covered, the covered area is increasing at 4 m2 per day. Find the value of k.
(2)
(c)
Find the rate at which the covered area is increasing when half of the pond is covered.
(1)
(d)
Describe what the model predicts about the rate of increase as A approaches 100.
(1)
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Answer
Mark
Notes
Part (a)
dA/dt on the left with a proportionality constant
B1
dA/dt = kA(100 − A), k a positive constant
B1
Part (b)
4 = k × 20 × 80
M1
k = 1/400 (= 0.0025)
A1
Part (c)
(1/400) × 50 × 50 = 6.25 m2 per day
B1
Part (d)
100 − A → 0, so the rate of increase tends to zero: the spread slows down as the pond becomes fully covered
B1
Exam tip: “The area not yet covered” is 100 − A, not 100. Translate each phrase of the description into a factor before you write the equation.
Figure 1 shows the curve C with equation x2 − xy + y2 = 9. The point P(3, 3) lies on C, and C crosses the positive x-axis at the point A.
(a)
Find dy/dx in terms of x and y.
(3)
(b)
Find the gradient of C at P, and show that the normal to C at P passes through O.
(2)
(c)
Write down the coordinates of A. Find an equation of the normal to C at A, and the coordinates of the point where this normal meets the y-axis.
(3)
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Answer
Mark
Notes
Part (a)
2x from x2 and 2y dy/dx from y2 (chain rule)
M1
−xy differentiated as −y − x dy/dx (product rule)
M1
dy/dx = (2x − y)/(x − 2y) or equivalent
A1
Part (b)
gradient at P = (6 − 3)/(3 − 6) = −1
M1
normal gradient 1: y − 3 = 1(x − 3), i.e. y = x, which passes through O(0, 0)
A1
Part (c)
A(3, 0)
B1
gradient at A = 6/3 = 2, so normal gradient −½ and y = −½(x − 3)
M1
meets the y-axis at (0, 3/2)
A1
Common error: Using the tangent gradient in the equation of the normal. The normal is perpendicular to the tangent, so its gradient is −1/m: at A the tangent gradient 2 gives a normal gradient of −½.
The curve shown in Figure 1 has equation y = kax, where k and a are positive constants. The curve passes through the points (0, 4) and (2, 1), which are marked on the figure.
(a)
Find the value of k and the value of a.
(2)
(b)
By taking natural logarithms, find dy/dx in terms of x.
(3)
(c)
Find an equation of the normal to the curve at the point (0, 4), and show that this normal crosses the x-axis where x = −16 ln 2.
(2)
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Answer
Mark
Notes
Part (a)
k = 4, from the point (0, 4)
B1
4a2 = 1 with a > 0, so a = ½
B1
Part (b)
ln y = ln 4 + x ln ½
M1
(1/y) dy/dx = ln ½
M1
dy/dx = 4(½)x ln ½ = −4 ln 2 × 2−x or equivalent
A1
Part (c)
gradient at x = 0 is −4 ln 2, so normal gradient 1/(4 ln 2) and y − 4 = x/(4 ln 2)
M1
y = 0 gives x = −16 ln 2, as required
A1
Exam tip: A gradient of the form ax ln a is negative when 0 < a < 1, because ln a < 0. A falling curve must give a negative gradient, which is a quick check on your sign.
A bowl of soup is left to cool in a room where the temperature is a constant 20 °C. The temperature of the soup is θ °C at time t minutes after it is served. The rate at which the temperature of the soup decreases is proportional to the difference between the temperature of the soup and the temperature of the room. Figure 1 shows the graph of θ against t, together with the tangent to the graph at t = 0, which passes through the marked points (0, 80) and (20, 20).
(a)
Write down a differential equation connecting θ and t, using k for a positive constant.
(2)
(b)
Use the tangent shown in Figure 1 to find the value of dθ/dt when t = 0, and hence find the value of k.
(3)
(c)
Find the temperature of the soup at the instant when it is cooling at a rate of 0.9 °C per minute.
(2)
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Answer
Mark
Notes
Part (a)
dθ/dt with a factor (θ − 20)
B1
dθ/dt = −k(θ − 20), k > 0
B1
Part (b)
gradient of tangent = (20 − 80)/(20 − 0)
M1
dθ/dt = −3 (°C per minute) at t = 0
A1
−3 = −k(80 − 20), so k = 0.05
A1
Part (c)
0.9 = 0.05(θ − 20)
M1
θ = 38 °C
A1
Common error: Using the gradient of a chord of the curve instead of the tangent. The rate at an instant is the gradient of the tangent at that instant, so read two well-separated points on the tangent line.
A tank is in the shape of an inverted square-based pyramid with its top horizontal. The top is a square of side 6 m and the vertex is 4 m vertically below the centre of the top, as shown in Figure (a). Figure (b) shows the vertical cross-section through the vertex, parallel to one side of the top. Water is poured into the empty tank at a constant rate of 0.6 m3 per minute. When the depth of water above the vertex is h m, the volume of water is V m3.
(a)
Show that V = ¾h3.
(2)
(b)
Find the rate at which the depth of water is increasing when h = 2.
(3)
(c)
Find the rate at which the area of the water surface is increasing at this instant.
(2)
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Answer
Mark
Notes
Part (a)
by similar triangles in Figure (b), the water surface is a square of side 6h/4 = 1.5h
M1
V = ⅓ × (1.5h)2 × h = ¾h3, as required
A1
Part (b)
dV/dh = (9/4)h2
B1
dh/dt = 0.6 ÷ ((9/4) × 4) = 0.6 ÷ 9
M1
= 1/15 ≈ 0.0667 m per minute
A1
Part (c)
area S = 2.25h2, so dS/dt = 4.5h × dh/dt
M1
= 9 × 1/15 = 0.6 m2 per minute
A1
Exam tip: In a pyramid or cone, write every length of the water's shape in terms of the depth before you differentiate. Then V is a function of h alone and dV/dh is a single derivative.
The circle C has equation x2 + y2 + 4x − 10y + 4 = 0. The point A(1, 9) lies on C.
(a)
Use implicit differentiation to find dy/dx in terms of x and y.
(3)
(b)
Find an equation of the tangent to C at A, in the form ax + by + c = 0, where a, b and c are integers.
(2)
(c)
The tangent meets the x-axis at B and the y-axis at D. Find the exact area of triangle OBD, where O is the origin.
(2)
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Answer
Mark
Notes
Part (a)
2x + 2ydy/dx + 4 − 10dy/dx = 0
M1
collects: (2y − 10)dy/dx = −2x − 4
M1
dy/dx = −(x + 2)/(y − 5)
A1
Part (b)
gradient −3/4 and y − 9 = −¾(x − 1)
M1
3x + 4y − 39 = 0
A1
Part (c)
B(13, 0) and D(0, 39/4) from their tangent
M1
area = ½ × 13 × 39/4 = 507/8
A1
Model sketch
Figure: the circle, the tangent at A and triangle OBD
Exam tip: Check your implicit derivative with geometry: the centre is (−2, 5), so the radius to A has gradient 4/3, and a tangent must have gradient −3/4.
Explain why y ≠ 0 at every point of C, and hence show that at any stationary point of C, y = −2x.
(2)
(c)
Find the coordinates of the stationary point of C.
(2)
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Answer
Mark
Notes
Part (a)
x2y → 2xy + x2dy/dx
M1
xy2 → y2 + 2xydy/dx
M1
(x2 + 2xy)dy/dx = −(2xy + y2)
M1
factorises both sides to reach the given result with no errors
A1
Part (b)
if y = 0 the left-hand side is 0 ≠ 16, so y ≠ 0
B1
dy/dx = 0 requires y(2x + y) = 0, so 2x + y = 0, i.e. y = −2x
B1
Part (c)
substitutes y = −2x: −2x3 + 4x3 = 16 ⇒ x3 = 8
M1
(2, −4)
A1
Exam tip: A fraction is zero when its numerator is zero and its denominator is not. Rule out each factor of the numerator that cannot happen (here y = 0), and check the denominator at your answer: x(x + 2y) = −12 ≠ 0.
A cube of ice of edge x cm melts so that it remains a cube. Its volume V cm3 decreases at a rate proportional to its surface area, so that dV/dt = −k × (surface area), where t is the time in minutes and k is a positive constant.
(a)
Show that dx/dt = −2k.
(4)
(b)
The edge of the cube decreases from 3 cm to 2 cm in 20 minutes. Find the value of k.
(2)
(c)
Find the rate at which the volume is decreasing when the edge is 2.5 cm.
(2)
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Answer
Mark
Notes
Part (a)
V = x3 so dV/dx = 3x2, and surface area = 6x2
B1
dV/dt = 3x2 dx/dt (chain rule)
M1
3x2 dx/dt = −6kx2
M1
dx/dt = −2k, as required
A1
Part (b)
the edge decreases at a constant rate: dx/dt = −1/20
M1
k = 1/40 (= 0.025)
A1
Part (c)
dV/dt = −(1/40) × 6 × 2.52, or 3 × 2.52 × (−1/20)
M1
decreasing at 0.938 cm3 per minute (accept 0.9375)
A1
Exam tip: A “show that” with a surprisingly simple answer, like a constant rate, is a strong hint about the next part: here a constant dx/dt means you can find it from two edge lengths and the time between them.
A water tank is a vertical cylinder with internal radius 0.5 m. Water drains from a tap at the base. When the depth of water is h m, the volume of water is V m3 and the water drains so that dV/dt = −c√h, where t is the time in minutes and c is a positive constant.
(a)
Show that dh/dt = −k√h, where k = 4c/π.
(3)
(b)
When the depth is 1.44 m, it is falling at 0.03 m per minute. Find the value of k.
(2)
(c)
Find the rate at which the volume of water is decreasing when the depth is 0.81 m.
(2)
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Answer
Mark
Notes
Part (a)
V = π(0.5)2h = πh/4, so dV/dh = π/4
B1
dh/dt = (dV/dt) ÷ (dV/dh) = −c√h ÷ (π/4)
M1
= −(4c/π)√h, so k = 4c/π
A1
Part (b)
−0.03 = −k√1.44 = −1.2k
M1
k = 0.025
A1
Part (c)
dV/dt = (π/4) × (−0.025 × 0.9), or c = 0.025π/4 used with √0.81 = 0.9
M1
decreasing at 0.0177 m3 per minute
A1
Common error: Using the depth rate as if it were the volume rate. In the last part convert back with dV/dt = (π/4) dh/dt; the two rates differ by the cross-sectional area.
Figure 1 shows the curve C with equation y2 = x(x − 3)2. C is symmetrical about the x-axis and has no points with x < 0. Starting from O, C forms a closed loop for 0 ≤ x ≤ 3; the loop closes at the point D(3, 0), where C crosses itself. For x > 3, C continues as two separate arms, one above and one below the x-axis, leaving D in different directions. At the points A and B on the loop the tangent is parallel to the x-axis. The point P(4, 2) lies on the upper arm of C.
(a)
Show that, for y ≠ 0, dy/dx = 3(x − 1)(x − 3)/(2y).
(3)
(b)
Find the coordinates of A and of B.
(3)
(c)
Find an equation of the tangent to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(3)
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Answer
Mark
Notes
Part (a)
2y dy/dx on the left
M1
product rule on the right: (x − 3)2 + 2x(x − 3)
M1
= (x − 3)(3x − 3) = 3(x − 1)(x − 3), giving the result
A1
Part (b)
dy/dx = 0 needs x = 1, since x = 3 gives y = 0, which is the point D
M1
y2 = 1 × (1 − 3)2 = 4
M1
A(1, 2) and B(1, −2)
A1
Part (c)
gradient at P = 3 × 3 × 1/(2 × 2) = 9/4
B1
y − 2 = (9/4)(x − 4)
M1
9x − 4y − 28 = 0 (or any integer multiple)
A1
Exam tip: Where y = 0 the formula for dy/dx has a zero denominator and cannot be used directly. Check such points separately: here D is where two branches of C cross, not a stationary point.
Figure 1 shows the closed curve C with equation sin x + sin y = 1, for 0 ≤ x ≤ π and 0 ≤ y ≤ π. The points P(π/6, π/6) and Q(π/6, 5π/6) lie on C.
(a)
Find dy/dx in terms of x and y.
(2)
(b)
Find the coordinates of the points on C at which the tangent is parallel to the x-axis, and of the points at which the tangent is parallel to the y-axis.
(4)
(c)
The tangents to C at P and at Q meet at the point T. Find the exact coordinates of T.
(3)
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Answer
Mark
Notes
Part (a)
cos x + cos y dy/dx = 0
M1
dy/dx = −cos x/cos y
A1
Part (b)
parallel to the x-axis: cos x = 0, so x = π/2 and then sin y = 0
M1
(π/2, 0) and (π/2, π)
A1
parallel to the y-axis: cos y = 0, so y = π/2 and then sin x = 0
M1
(0, π/2) and (π, π/2)
A1
Part (c)
gradient −1 at P and +1 at Q
B1
tangents y = −x + π/3 and y = x + 2π/3
M1
T(−π/6, π/2)
A1
Common error: Writing the derivative of sin y as cos y. Every term in y picks up a factor dy/dx by the chain rule: the derivative of sin y with respect to x is cos y dy/dx.
An observer stands at the point O on level ground. A drone takes off from the point F on the ground, 40 m from O, and rises vertically at a constant speed of 3 m s−1, as shown in Figure 1. When the drone D is at height h m, the distance OD is s m and the angle of elevation of D from O is θ radians.
(a)
Find the rate at which s is increasing when h = 30.
(3)
(b)
Find the rate at which θ is increasing when h = 30.
(3)
(c)
Find the exact height of the drone at the instant when s is increasing at 2.4 m s−1.
(2)
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Answer
Mark
Notes
Part (a)
s2 = 402 + h2, so 2s ds/dt = 2h dh/dt
M1
h = 30 gives s = 50
B1
ds/dt = 30 × 3/50 = 1.8 m s−1
A1
Part (b)
tan θ = h/40, so sec2θ dθ/dt = (1/40) dh/dt
M1
cos θ = 40/50 = 0.8 at this instant
B1
dθ/dt = 0.64 × 3/40 = 0.048 rad s−1
A1
Part (c)
3h/√(1600 + h2) = 2.4
M1
9h2 = 5.76(1600 + h2) gives h = 160/3 (≈ 53.3)
A1
Exam tip: When two lengths are tied together by Pythagoras, differentiate the whole equation with respect to t. You avoid a square root and get the connection between the rates in one line.
Figure 1 shows the cross-section of a trough. The trough is a prism of length 2 m whose cross-section is an isosceles trapezium with a horizontal base of width 0.4 m, a horizontal top of width 1.0 m and a depth of 0.5 m. When the depth of water in the trough is h m, the volume of water is V m3.
(a)
Show that V = 0.8h + 1.2h2.
(3)
(b)
Water flows into the trough at a constant rate of 0.05 m3 per minute. Find the rate at which the depth of water is increasing when h = 0.25.
(3)
(c)
Find the depth of water at the instant when the depth is increasing at 0.03125 m per minute.
(2)
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Answer
Mark
Notes
Part (a)
the width of the water surface is 0.4 + 1.2h, since each sloping side moves out 0.3 m over the 0.5 m depth
M1
area of the wet cross-section = ½h(0.4 + 0.4 + 1.2h) = 0.4h + 0.6h2
Common error: Taking the width of the water surface as the full top width 1.0 m at every depth. The width grows with the depth: it is 0.4 m at the bottom and reaches 1.0 m only when the trough is full.
Figure 1 shows the curve with equation y = 2x. The tangent to the curve at the point P, where x = p, passes through the origin O.
(a)
By taking natural logarithms, show that dy/dx = 2x ln 2.
(2)
(b)
Show that p = 1/ln 2, and hence show that the y-coordinate of P is e.
(4)
(c)
The normal to the curve at P meets the x-axis at the point Q. Find the exact x-coordinate of Q.
(3)
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Answer
Mark
Notes
Part (a)
ln y = x ln 2, so (1/y) dy/dx = ln 2
M1
dy/dx = y ln 2 = 2x ln 2, as required
A1
Part (b)
tangent at P: y − 2p = 2p ln 2 (x − p)
M1
through O: −2p = −p × 2p ln 2
M1
2p ≠ 0, so p ln 2 = 1 and p = 1/ln 2
A1
y = 21/ln 2 = e(ln 2)/(ln 2) = e, as required
A1
Part (c)
gradient of the tangent at P is e ln 2, so the normal gradient is −1/(e ln 2)
M1
0 − e = −(x − 1/ln 2)/(e ln 2)
M1
x = 1/ln 2 + e2 ln 2
A1
Exam tip: The tangent at x = p has gradient 2p ln 2 for any p; only the condition that it passes through O fixes p. Write the general tangent first, then substitute the point it must pass through.
Water flows into a storage tank at a constant rate of 30 m3 per hour. At the same time, water leaks out of the tank at a rate proportional to the volume V m3 of water in the tank. The tank is empty at time t = 0, where t is measured in hours. Figure 1 shows the graph of V against t; the graph approaches the dashed line V = 600.
(a)
Write down a differential equation connecting V and t, using k for the positive constant of proportionality.
(2)
(b)
Explain, with reference to Figure 1, why k = 0.05.
(2)
(c)
The tank is a vertical cylinder of radius 5 m. Find the exact rate at which the depth of water is increasing when the depth is 2 m, and give its value to 3 significant figures.
(3)
(d)
Find the volume of water in the tank at the instant when the depth is increasing at 0.2 m per hour.
(2)
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Answer
Mark
Notes
Part (a)
rate in 30 and rate out kV
B1
dV/dt = 30 − kV
B1
Part (b)
the graph levels off as V approaches 600, so dV/dt = 0 when V = 600
M1
30 − 600k = 0 gives k = 30/600 = 0.05
A1
Part (c)
V = 25πh, so dV/dh = 25π, and V = 50π when h = 2
B1
dh/dt = (30 − 0.05 × 50π) ÷ 25π
M1
= (12 − π)/(10π) ≈ 0.282 m per hour
A1
Part (d)
30 − 0.05V = 25π × 0.2 = 5π
M1
V = 600 − 100π ≈ 286 m3
A1
Exam tip: A level line that the graph approaches marks the volume at which the rates in and out balance, so dV/dt = 0 there. Reading that limit from the graph is the quickest way to the constant.
y2 = −1 has no real solutions, so there are no such points
A1
Model sketch
Figure: C with the tangent at A and the two horizontal tangents
Exam tip: Parallel to the x-axis: numerator of dy/dx = 0. Parallel to the y-axis: denominator = 0. In both cases the resulting line must be solved together with the curve; “no real solutions” is a complete answer when you say why.
A water trough is a prism of length 3 m. Its cross-section is an isosceles triangle with a horizontal top edge of width 0.8 m and its vertex 0.6 m vertically below the top edge. The trough stands on its vertex edge with its top horizontal. When the depth of water is h m, the volume of water is V m3.
Figure: cross-section of the trough with water of depth h m; not to scale
(a)
Show that V = 2h2.
(2)
(b)
Water is pumped into the trough at a constant rate of 0.012 m3 s−1. Find the rate at which the depth is increasing when h = 0.3.
(3)
(c)
The pump is switched off and the trough is drained through a valve. Water now leaves at a rate proportional to the depth of water, with constant of proportionality k. Show that the depth decreases at a constant rate.
(3)
(d)
Draining the trough from a depth of 0.5 m takes 25 minutes. Find the value of k, with t measured in seconds.
(2)
(e)
Find the rate at which the volume of water is decreasing when the depth is 0.25 m.
(2)
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Answer
Mark
Notes
Part (a)
by similar triangles the water surface has width 0.8h/0.6 = 4h/3
M1
V = 3 × ½ × (4h/3) × h = 2h2, as required
A1
Part (b)
dV/dh = 4h
B1
dh/dt = 0.012 ÷ (4 × 0.3)
M1
= 0.01 m s−1
A1
Part (c)
dV/dt = −kh, k a positive constant
B1
4h dh/dt = −kh, using the chain rule with dV/dh = 4h
M1
dh/dt = −k/4, a constant, so the depth decreases at a constant rate
A1
Part (d)
dh/dt = −0.5/1500 = −1/3000 m s−1
M1
k = 4/3000 = 1/750 (≈ 0.00133)
A1
Part (e)
kh = 0.25/750, or 4 × 0.25 × 1/3000
M1
= 1/3000 ≈ 3.33 × 10−4 m3 s−1
A1
Exam tip: Always bring a rate of volume back to a rate of depth through dV/dh. Here dV/dh = 4h cancels the h in the leak rate, which is why the depth falls steadily.
By taking natural logarithms, show that dy/dx = xx(1 + ln x).
(4)
(b)
Find the exact coordinates of the stationary point of C.
(3)
(c)
The curve with equation y = 2x also passes through the point (2, 4). Show that at (2, 4) the gradient of C exceeds the gradient of y = 2x by exactly 4.
(3)
(d)
The tangent to C at the point where x = e meets the x-axis at B. Find the exact x-coordinate of B.
(2)
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Answer
Mark
Notes
Part (a)
ln y = x ln x
M1
(1/y)dy/dx on the left
M1
product rule on the right: ln x + x × 1/x = 1 + ln x
M1
dy/dx = y(1 + ln x) = xx(1 + ln x), as required
A1
Part (b)
xx > 0, so 1 + ln x = 0
M1
x = e−1
A1
y = (e−1)e−1 = e−1/e
A1
Part (c)
gradient of C at x = 2: 4(1 + ln 2) = 4 + 4 ln 2
M1
gradient of y = 2x: 22 ln 2 = 4 ln 2
B1
difference = 4 + 4 ln 2 − 4 ln 2 = 4, as required
A1
Part (d)
gradient ee(1 + 1) = 2ee and −ee = 2ee(x − e)
M1
x = e − ½
A1
Model sketch
Figure: C: y = xˣ with its stationary point, and y = 2ˣ through (2, 4)
Exam tip: When the variable is in both the base and the power, neither the power rule nor the rule for ax applies. Take logarithms first, then differentiate implicitly.
Figure 1 shows part of the curve C with equation x2 − 4xy + y2 + 12 = 0. C consists of two separate branches that do not meet: one lies in the first quadrant and the other is its reflection in O, lying in the third quadrant. The dashed line with equation x = 4 meets the first-quadrant branch at the points A and P, where P is above A.
(a)
Show that dy/dx = (x − 2y)/(2x − y).
(4)
(b)
Find the coordinates of the points on C at which the tangent is parallel to the x-axis.
(4)
(c)
Find the coordinates of A and of P.
(2)
(d)
The tangent to C at A and the tangent to C at P meet at the point T. Find the area of triangle ATP.
A is a point found in (b), so the tangent at A is y = 2
B1
gradient at P = (4 − 28)/(8 − 14) = 4
M1
tangent at P: y − 14 = 4(x − 4), which meets y = 2 at T(1, 2)
M1
area = ½ × 12 × 3 = 18
A1
Exam tip: Look for links between the parts. Here A turns out to be one of the points from (b), so its tangent is horizontal and needs no further working.
A bowl is a hemisphere of internal radius 10 cm, with its rim horizontal. Figure 1 shows a vertical cross-section through the centre O of the rim. When the depth of water in the bowl is h cm, the surface of the water is a circle of radius r cm and the volume of water is V cm3, where V = ⅓πh2(30 − h).
(a)
Use Figure 1 to show that r2 = 20h − h2.
(2)
(b)
Show that dV/dh = πr2.
(2)
(c)
Water is poured into the bowl at a constant rate of 5 cm3 s−1. Find the rate at which the depth of water is increasing when h = 4.
(2)
(d)
Find the rate at which the area of the water surface is increasing at this instant.
(2)
(e)
When h = 4 the inflow is stopped. Water then leaks out of the bowl at a rate of kh cm3 s−1, where k is a positive constant. Immediately after the inflow stops, the depth is falling at 0.01 cm s−1. Find the value of k, and the rate at which the depth is falling when h = 2.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
the line from O to the edge of the water surface is a radius of length 10, and O is (10 − h) cm above the surface
M1
r2 + (10 − h)2 = 100, so r2 = 20h − h2, as required
Exam tip: For any container, dV/dh equals the area of the water surface. Here that is πr2, which is a quick check on your derivative of the volume formula.
Figure 1 shows the curve C with equation 2x + 2y = 5. The curve meets the x-axis at A and the y-axis at B. The dashed lines have equations y = log2 5 and x = log2 5.
(a)
Show that dy/dx = −2x − y.
(3)
(b)
Find equations of the tangents to C at A and at B, and show that these tangents meet at a point T on the line y = x.
(4)
(c)
Find the exact coordinates of the point M on C at which the gradient is −1.
(3)
(d)
Explain why C does not meet the line y = log2 5.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the derivative of 2x is 2x ln 2
B1
the derivative of 2y is 2y ln 2 dy/dx (chain rule)
so 2y = 5 − 2x < 5, giving y < log2 5 at every point of C
B1
Common error: Differentiating 2y as 2y ln 2 and leaving out the factor dy/dx. A term in y differentiated with respect to x always needs the chain rule.
A tank is a cylinder of radius 1 m and length 4 m, lying with its axis horizontal. Figure 1 shows a vertical cross-section of the tank, with centre O. The surface of the water meets this cross-section in the chord AB, and angle AOB = θ radians is the angle at O facing the water, as shown. Figure 1 shows the water at a general level; it is not drawn at any particular value of θ used below, so do not measure angles or lengths from it. The depth of water is h m and the volume of water is V m3. Water is pumped into the tank at a constant rate of 0.1 m3 per minute.
(a)
Show that V = 2(θ − sin θ).
(2)
(b)
Find the rate at which θ is increasing at the instant when θ = 2π/3.
(3)
(c)
Show that h = 1 − cos(θ/2).
(2)
(d)
Find the exact rate at which the depth of water is increasing at the instant when θ = 2π/3.
(3)
(e)
Find the rate at which the width AB of the water surface is increasing at this instant.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
area of the water's cross-section = sector − triangle = ½θ − ½ sin θ, with radius 1
M1
V = 4 × ½(θ − sin θ) = 2(θ − sin θ), as required
A1
Part (b)
dV/dθ = 2(1 − cos θ)
B1
= 2(1 + ½) = 3 when θ = 2π/3, and dθ/dt = 0.1 ÷ 3
M1
= 1/30 ≈ 0.0333 rad per minute
A1
Part (c)
the perpendicular from O to AB bisects angle AOB, so O is cos(θ/2) above AB
M1
the lowest point of the tank is 1 below O, so h = 1 − cos(θ/2), as required
A1
Part (d)
dh/dθ = ½ sin(θ/2)
B1
dh/dt = ½ sin(π/3) × 1/30
M1
= √3/120 ≈ 0.0144 m per minute
A1
Part (e)
AB = 2 sin(θ/2), so d(AB)/dt = cos(θ/2) dθ/dt
M1
= ½ × 1/30 = 1/60 ≈ 0.0167 m per minute
A1
Exam tip: When a shape is described by an angle, link every quantity (volume, depth, width) to θ first. Each rate then comes from one chain rule: d(quantity)/dt = d(quantity)/dθ × dθ/dt.
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