Further Applications of Differentiation (Edexcel International A Level (IAL) Maths: Pure 4): Exam Questions

2 hours19 questions
1a
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3 marks

In a computer animation, the side length, s mm, of a square is increasing at a constant rate of 2 millimetres per second.

(i) Write down the value of fraction numerator d s over denominator d t end fraction, where t is time and measured in seconds.

(ii) Write down a formula for the area, A mm2, of the square and hence findfraction numerator d A over denominator d s end fraction.

1b
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3 marks

Use the chain rule to find an expression for fraction numerator d A over denominator d t end fraction in terms of s and hence find the rate at which the area is increasing when s = 10.

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2a
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3 marks

The side length, x cm, of a cube increases at a constant rate of 0.1 cm s-1.

(i) Write down the value of fraction numerator d x over denominator d t end fraction, where t is time and measured in seconds.

(ii) Write down a formula for the volume, V cm3, of the cube and hence findfraction numerator d V over denominator d x end fraction.

2b
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3 marks

Use the chain rule to find an expression for fraction numerator d V over denominator d t end fraction in terms of x and hence find the rate at which the volume is increasing when the side length of the cube is 4 cm.

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3a
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2 marks

The rate at which the radius, r cm, of a sphere increases over time (t seconds) is directly proportional to the temperature (To C) of its immediate surroundings.

Write down an equation linkingfraction numerator d r over denominator d t end fraction, T and the constant of proportionality, k.

3b
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3 marks

When the surrounding temperature is 20 oC, the radius of the sphere is increasing at a rate of 0.4 cm s-1.

Find the value of k.

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4a
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5 marks

(i) Write down a formula for the volume of a cube, V cm3, and the surface area, S cm2, of a cube, in terms of the side length of a cube, x cm.

(ii) Show that fraction numerator d x over denominator d V end fraction equals fraction numerator 1 over denominator 3 x squared end fraction and find an expression for fraction numerator d S over denominator d x end fraction.

4b
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5 marks

The volume of a cube is decreasing at a constant rate of 0.6 cm3 s-1.

(i) Explain why fraction numerator d V over denominator d t end fractionwhere t is time in seconds, has the value of -0.6.

(ii) Use the chain rule to find an expression for fraction numerator d S over denominator d t end fraction in terms of fraction numerator d S over denominator d x end fractionfraction numerator d x over denominator d V end fraction and fraction numerator d V over denominator d t end fraction

(iii) Hence write fraction numerator d S over denominator d t end fraction in terms of x and find the rate at which the area of the cube is decreasing at the instant when its side length is 5 cm.

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1
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4 marks

In a computer animation, the radius of a circle increases at a constant rate of 1 millimetre per second. Find the rate, per second, at which the area of the circle is increasing at the time when the radius is 8 millimetres. Give your answer as a multiple of straight pi.

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2
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5 marks

The side length of a cube increases at a rate of 0.1 cm s-1.
Find the rate of change of the volume of the cube at the instant the side length is 5 cm.
You may assume that the cube remains cubical at all times.

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3a
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5 marks

In the production process of a glass sphere, hot glass is blown such that the radius, r cm, increases over time (t seconds) in direct proportion to the temperature (T oC) of the glass.

Find an expression, in terms of r and T, for the rate of change of the volume (V cm3) of a glass sphere.

3b
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3 marks

When the temperature of the glass is 1200 oC, a glass sphere has a radius of 2 cm and its volume is increasing at a rate of 5 cm3s-1.

Find the rate of increase of the radius at this time.

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4
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6 marks

An ice cube, of side length x cm, is melting at a constant rate of 0.8 cm3 s-1.
Assuming that the ice cube remains in the shape of a cube whilst it melts, find the rate at which its surface area is melting at the point when its side length is 2 cm2.

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5
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5 marks

A bowl is in the shape of a hemisphere of radius 8 cm.

The volume of liquid in the bowl is given by the formula

V space equals space 8 pi h squared space minus space 1 third pi h cubed

where h cm is the depth of the liquid (ie the height between the bottom of the bowl and the level of the liquid).

Liquid is leaking through a small hole in the bottom of the bowl at a constant rate of 5 cm3s-1. Find the rate of change of the depth of liquid in the bowl at the instant the height of liquid is 3 cm.

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1
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4 marks

A spherical air bubble's surface area is increasing at a constant rate of 4pi cm2s-1.

Find an expression for the rate at which the radius is increasing per second.
(The surface area of a sphere is 4pir2.)

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2
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6 marks

A cone, stood on its vertex, of radius 3 cm and height 9 cm, is being filled with sand at a constant rate of 0.2 cm3s-1.

Find the rate of change of the depth of sand in the cone at the instant the radius of the sand is 1.2 cm.

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3
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6 marks

The material used to make a spherical balloon is designed so that it can be inflated at a maximum rate of 16 cm3s-1 without bursting.

Given that the radius of the balloon is determined by the function r(t) = t over pi plus 1 half, t ≥ 0

show that the maximum time the balloon can be inflated for, without bursting, is 3 over 2 pi seconds.

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4
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6 marks

An ice lolly which is in the shape of a cylinder of radius r cm and length 8r cm is melting at a constant rate of 0.4 cm3s-1.

Assuming that the ice lolly remains in the shape of a cylinder (mathematically similar to the original cylinder) whilst it melts, find the rate at which its surface area is melting at the point when it's radius is 0.3 cm.

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5
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6 marks

The volume of liquid in a hemispherical bowl is given by the formula

v equals 1 third pi h squared left parenthesis 3 R minus h right parenthesis

where R is the radius of the bowl and h is the depth of liquid

(ie the height between the bottom of the bowl and the level of the liquid).

In a particular case, a bowl is leaking liquid through a small hole in the bottom at a rate directly proportional to the depth of liquid.

Show that the depth of liquid in the bowl is decreasing by

fraction numerator k over denominator pi space open parentheses 2 R minus h close parentheses end fraction

where k is a constant.

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1
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6 marks

A plant pot in the shape of square-based pyramid (stood on its vertex) is being filled with soil at a rate of 72 cm3s-1.

The plant pot has a height of 1 m and a base length of 40 cm.

Find the rate at which the depth of soil is increasing at the moment when the depth is 60 cm.

(The volume of a pyramid is a third of the area of the base times the height.)

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2
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6 marks

An expanding spherical air bubble has radius, r cm, at a time, t seconds, determined by the function r(t) = 0.3+ 0.1t2.

The bubble will burst if the rate of expansion of its volume exceeds 4t cm3 s-1.

Find, to one decimal place, the length of time the bubble expands for.

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3
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7 marks

A small conical pot, stood on its base, is being filled with salt via a small hole at its vertex. The cone has a height of 6 cm and a radius of 2 cm.

Salt is being poured into the pot at a constant rate of 0.3 cm3s-1.

Find, to three significant figures, the rate of change in depth of the salt at the instant when the pot is half full by volume.

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4a
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5 marks

A large block of ice used by sculptors is in the shape of a cuboid with dimensions x m by 2x m by 5x m. The block melts uniformly with its surface area decreasing at a constant rate of k m2 s-1. You may assume that as the block melts, the shape remains mathematically similar to the original cuboid.

Show that the rate of melting, by volume, is given by

fraction numerator 15 k x over denominator 34 end fractionm3 s-1.

4b
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3 marks

In the case when k = 0.2, the block of ice remains solid enough to be sculpted whilst the rate of melting, by volume, is less than 0.05 m3 s-1.
Find the value of x for the largest block of ice that can be used for ice sculpting under such conditions, giving your answer as a fraction in its lowest terms.

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5
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7 marks

The volume of liquid in a hemispherical bowl is given by the formula

V space equals space 1 third pi h squared open parentheses 3 R minus h close parentheses

where R is the radius of the bowl and h is the depth of liquid.
(ie the height between the bottom of the bowl and the level of the liquid).

In a particular case, liquid is leaking through a small hole in the bottom of a bowl at a rate directly proportional to the depth of liquid.

When the bowl is full, the rate of volume loss is equal topi.

Show that the rate of change of the depth of the liquid is inversely proportional to R(h-2R)

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