Sec 4 A-Math Applications of differentiation worksheet with answers

A free Sec 4 A-Math worksheet on applications of differentiation: 6 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 A-Math worksheet: Applications of differentiation

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Find the equation of the tangent to the curve

y = 2x^{2} - \dfrac{2}{x}

at the point where

x = -1

Answer[3]

(a)A point moves along the curve

y = 2x^{2} + x - 5

so that its

x

-coordinate is increasing at a rate of

0.2

units per second. Find the rate of change of its

y

-coordinate when

x = 2

Answer (a)[3]

(b)The radius of a circle is increasing at a rate of

4

cm/s. Find the rate of increase of its area when the radius is

10

cm, leaving your answer in terms of

\pi

Answer (b)[3]

(a)Find the range of values of

x

for which

y = 2x^{3} - 6x^{2} - 48x + 5

is increasing.

Answer (a)[3]

(b)Find the range of values of

x

for which

y = 2x^{3} - 18x^{2} + 30x + 3

is increasing.

Answer (b)[3]

Section B

A point

P

moves along the curve

y = \dfrac{3x + 1}{x + 1}

, for

x > -1

, in such a way that its

x

-coordinate is increasing at a constant rate of

3

units per second.

(a)Find

\dfrac{dy}{dx}

Answer (a)[2]

(b)Find the rate of change of the

y

-coordinate of

P

at the instant when

x = 0

Answer (b)units per second[2]

(c)Find the

x

-coordinate of

P

at the instant when its

y

-coordinate is increasing twice as fast as its

x

-coordinate.

Answer (c)[3]

The equation of a curve is

y = 3e^{x} - 2e^{-x}

(a)Find

\dfrac{dy}{dx}

Answer (a)[1]

(b)Explain why

y

is an increasing function of

x

for all real values of

x

[2]

(c)Find the exact

x

-coordinate of the point where the curve crosses the

x

-axis.

Answer (c)[2]

(d)Find the exact

x

-coordinates of the points on the curve at which the gradient is

7

Answer (d)[3]

An open cylindrical container (it has a base but no lid) has radius

r

cm and height

h

cm. Its total external surface area is

27\pi

cm² and its volume is

V

cm³.

(a)Show that

V = \dfrac{1}{2}\pi(27r - r^{3})

[2]

(b)Given that

r

can vary, find the value of

r

for which

V

has a stationary value.

Answer (b)cm[3]

(c)Determine whether this stationary value of

V

is a maximum or a minimum.

Answer (c)[2]

(d)Find this value of

V

, leaving your answer in terms of

\pi

Answer (d)cm³[2]

(e)Find the corresponding height of the container.

Answer (e)cm[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

y = -2x + 2

[3]

y = 2x^{2} - 2x^{-1}

\dfrac{dy}{dx} = 2(2)x - 2(-1)x^{-2}

= 4x + 2x^{-2}

\dfrac{dy}{dx} = 4(-1) + 2(-1)^{-2}

= -2

y = 2(-1)^{2} - 2(-1)^{-1}

= 4

y - y_{1} = m(x - x_{1})

y - 4 = -2(x + 1)

y = -2x + 2

Question 2

(a)Answer:

= \dfrac{9}{5}

units/s[3]

\dfrac{dy}{dx} = 2(2)x + 1

= 4x + 1

\dfrac{dy}{dx} = 4(2) + 1

= 9

\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}

\dfrac{dy}{dt} = 9 \times 0.2

= \dfrac{9}{5}

units/s

(b)Answer:

= 80\pi

cm²/s[3]

A = \pi r^{2}

\dfrac{dA}{dr} = 2\pi r

\dfrac{dA}{dr} = 2\pi (10)

= 20\pi

\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}

\dfrac{dA}{dt} = 20\pi \times 4

= 80\pi

cm²/s

Question 3

(a)Answer:

x < -2 \text{ or } x > 4

[3]

\dfrac{dy}{dx} = 2(3)x^{2} - 6(2)x - 48(1)

= 6x^{2} - 12x - 48

6x^{2} - 12x - 48 > 0

x^{2} - 2x - 8 > 0

(x + 2)(x - 4) 0

The curve is above the x-axis outside the roots.

x < -2 \text{ or } x > 4

(b)Answer:

x < 1 \text{ or } x > 5

[3]

\dfrac{dy}{dx} = 2(3)x^{2} - 18(2)x + 30(1)

= 6x^{2} - 36x + 30

6x^{2} - 36x + 30 > 0

x^{2} - 6x + 5 > 0

(x - 1)(x - 5) 0

The curve is above the x-axis outside the roots.

x < 1 \text{ or } x > 5

Question 4

(a)Answer:

\dfrac{dy}{dx} = \dfrac{2}{(x + 1)^{2}}

[2]

\dfrac{dy}{dx} = \dfrac{(x + 1)(3) - (3x + 1)(1)}{(x + 1)^{2}}

= \dfrac{2}{(x + 1)^{2}}

(b)Answer:

6

units per second[2]

\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}

= \dfrac{2}{(0 + 1)^{2}} \times 3

= 6

units per second

(c)Answer:

x = 0

[3]

\dfrac{dy}{dt} = 2\dfrac{dx}{dt}

, so

\dfrac{dy}{dx} = 2

\dfrac{2}{(x + 1)^{2}} = 2

(x + 1)^{2} = 1

x + 1 = \pm1

x = 0

x = -2

(rejected, as

x > -1

)

Question 5

(a)Answer:

\dfrac{dy}{dx} = 3e^{x} + 2e^{-x}

[1]

\dfrac{dy}{dx} = 3e^{x} - 2(-e^{-x})

= 3e^{x} + 2e^{-x}

(b)[2]

e^{x} > 0

and

e^{-x} > 0

for every real

x

.So

\dfrac{dy}{dx} = 3e^{x} + 2e^{-x} > 0

for all

x

, and

y

is an increasing function.

(c)Answer:

x = \dfrac{1}{2}\ln \dfrac{2}{3}

[2]

3e^{x} - 2e^{-x} = 0

3e^{2x} = 2

e^{2x} = \dfrac{2}{3}

2x = \ln \dfrac{2}{3}

x = \dfrac{1}{2}\ln \dfrac{2}{3}

(As

y

is increasing, this is the only point where the curve meets the

x

-axis.)

(d)Answer:

x = -\ln 3

x = \ln 2

[3]

3e^{x} + 2e^{-x} = 7

Multiply by

e^{x}

3e^{2x} - 7e^{x} + 2 = 0

(3e^{x} - 1)(e^{x} - 2) = 0

e^{x} = \dfrac{1}{3}

e^{x} = 2

x = -\ln 3

x = \ln 2

Question 6

(a)[2]

Base and curved surface:

\pi r^{2} + 2\pi rh = 27\pi

h = \dfrac{27 - r^{2}}{2r}

V = \pi r^{2}h

= \pi r^{2}\left(\dfrac{27 - r^{2}}{2r}\right)

= \dfrac{1}{2}\pi(27r - r^{3})

(shown)

(b)Answer:

r = 3

cm[3]

\dfrac{dV}{dr} = \dfrac{1}{2}\pi(27 - 3r^{2})

27 - 3r^{2} = 0

r^{2} = 9

r = 3

, since

r > 0

(c)Answer: Maximum[2]

\dfrac{d^{2}V}{dr^{2}} = -3\pi r

When

r = 3

\dfrac{d^{2}V}{dr^{2}} = -9\pi < 0

, so

V

is a maximum.

(d)Answer:

27\pi

cm³[2]

V = \dfrac{1}{2}\pi(27 \times 3 - 3^{3})

= \dfrac{1}{2}\pi(81 - 27)

= 27\pi

(e)Answer:

h = 3

cm[1]

h = \dfrac{27 - 3^{2}}{2(3)}

= 3

Practice questions by skill

Sec 4 A-Math Applications of differentiation worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Applications of differentiation notes. Or practise it online with step-by-step help when you get one wrong.

Section B

A point

P

moves along the curve

y = \dfrac{3x + 1}{x + 1}

, for

x > -1

, in such a way that its

x

-coordinate is increasing at a constant rate of

3

units per second.

(a)Find

\dfrac{dy}{dx}

Answer (a)[2]

(b)Find the rate of change of the

y

-coordinate of

P

at the instant when

x = 0

Answer (b)units per second[2]

(c)Find the

x

-coordinate of

P

at the instant when its

y

-coordinate is increasing twice as fast as its

x

-coordinate.

Answer (c)[3]

Answers and worked solutions

Question 1

Answer:

y = -2x + 2

[3]

y = 2x^{2} - 2x^{-1}

\dfrac{dy}{dx} = 2(2)x - 2(-1)x^{-2}

= 4x + 2x^{-2}

\dfrac{dy}{dx} = 4(-1) + 2(-1)^{-2}

= -2

y = 2(-1)^{2} - 2(-1)^{-1}

= 4

y - y_{1} = m(x - x_{1})

y - 4 = -2(x + 1)

y = -2x + 2

Question 2

(a)Answer:

= \dfrac{9}{5}

units/s[3]

\dfrac{dy}{dx} = 2(2)x + 1

= 4x + 1

\dfrac{dy}{dx} = 4(2) + 1

= 9

\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}

\dfrac{dy}{dt} = 9 \times 0.2

= \dfrac{9}{5}

units/s

(b)Answer:

= 80\pi

cm²/s[3]

A = \pi r^{2}

\dfrac{dA}{dr} = 2\pi r

\dfrac{dA}{dr} = 2\pi (10)

= 20\pi

\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}

\dfrac{dA}{dt} = 20\pi \times 4

= 80\pi

cm²/s

Question 3

(a)Answer:

x < -2 \text{ or } x > 4

[3]

\dfrac{dy}{dx} = 2(3)x^{2} - 6(2)x - 48(1)

= 6x^{2} - 12x - 48

6x^{2} - 12x - 48 > 0

x^{2} - 2x - 8 > 0

(x + 2)(x - 4) 0

The curve is above the x-axis outside the roots.

x < -2 \text{ or } x > 4

(b)Answer:

x < 1 \text{ or } x > 5

[3]

\dfrac{dy}{dx} = 2(3)x^{2} - 18(2)x + 30(1)

= 6x^{2} - 36x + 30

6x^{2} - 36x + 30 > 0

x^{2} - 6x + 5 > 0

(x - 1)(x - 5) 0

The curve is above the x-axis outside the roots.

x < 1 \text{ or } x > 5

Question 4

(a)Answer:

\dfrac{dy}{dx} = \dfrac{2}{(x + 1)^{2}}

[2]

\dfrac{dy}{dx} = \dfrac{(x + 1)(3) - (3x + 1)(1)}{(x + 1)^{2}}

= \dfrac{2}{(x + 1)^{2}}

(b)Answer:

6

units per second[2]

\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}

= \dfrac{2}{(0 + 1)^{2}} \times 3

= 6

units per second

(c)Answer:

x = 0

[3]

\dfrac{dy}{dt} = 2\dfrac{dx}{dt}

, so

\dfrac{dy}{dx} = 2

\dfrac{2}{(x + 1)^{2}} = 2

(x + 1)^{2} = 1

x + 1 = \pm1

x = 0

x = -2

(rejected, as

x > -1

)

Question 5

(a)Answer:

\dfrac{dy}{dx} = 3e^{x} + 2e^{-x}

[1]

\dfrac{dy}{dx} = 3e^{x} - 2(-e^{-x})

= 3e^{x} + 2e^{-x}

(b)[2]

e^{x} > 0

and

e^{-x} > 0

for every real

x

.So

\dfrac{dy}{dx} = 3e^{x} + 2e^{-x} > 0

for all

x

, and

y

is an increasing function.

(c)Answer:

x = \dfrac{1}{2}\ln \dfrac{2}{3}

[2]

3e^{x} - 2e^{-x} = 0

3e^{2x} = 2

e^{2x} = \dfrac{2}{3}

2x = \ln \dfrac{2}{3}

x = \dfrac{1}{2}\ln \dfrac{2}{3}

(As

y

is increasing, this is the only point where the curve meets the

x

-axis.)

(d)Answer:

x = -\ln 3

x = \ln 2

[3]

3e^{x} + 2e^{-x} = 7

Multiply by

e^{x}

3e^{2x} - 7e^{x} + 2 = 0

(3e^{x} - 1)(e^{x} - 2) = 0

e^{x} = \dfrac{1}{3}

e^{x} = 2

x = -\ln 3

x = \ln 2

Question 6

(a)[2]

Base and curved surface:

\pi r^{2} + 2\pi rh = 27\pi

h = \dfrac{27 - r^{2}}{2r}

V = \pi r^{2}h

= \pi r^{2}\left(\dfrac{27 - r^{2}}{2r}\right)

= \dfrac{1}{2}\pi(27r - r^{3})

(shown)

(b)Answer:

r = 3

cm[3]

\dfrac{dV}{dr} = \dfrac{1}{2}\pi(27 - 3r^{2})

27 - 3r^{2} = 0

r^{2} = 9

r = 3

, since

r > 0

(c)Answer: Maximum[2]

\dfrac{d^{2}V}{dr^{2}} = -3\pi r

When

r = 3

\dfrac{d^{2}V}{dr^{2}} = -9\pi < 0

, so

V

is a maximum.

(d)Answer:

27\pi

cm³[2]

V = \dfrac{1}{2}\pi(27 \times 3 - 3^{3})

= \dfrac{1}{2}\pi(81 - 27)

= 27\pi

(e)Answer:

h = 3

cm[1]

h = \dfrac{27 - 3^{2}}{2(3)}

= 3