What is 6.2: Gradient of exponential functions?

Differentiating gives d/dxe^kx=ke^kx, so the gradient of an exponential is a constant multiple of the function itself.

How many marks are 6.2 questions worth?

6.2 practice drills on this page are worth 1–7 marks each; marks in a real exam vary by question.

Which past exam papers has 6.2 appeared in?

We have not yet indexed a verified past-paper appearance for 6.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.2

Know that the gradient of e^(kx) is equal to k·e^(kx) and hence understand why the exponential model is suitable in many applications.

Draft — not yet indexed

Gradient of exponential functions

Worked answers and methods for 6.2 on Edexcel A-level Maths 9MA0.

Explanation

Differentiating gives $\dfrac{d}{dx}e^{kx}=ke^{kx}$ , so the gradient of an exponential is a constant multiple of the function itself.
For $Q=Q_0e^{kt}$ , the rate satisfies $dQ/dt=kQ$ : positive $k$ models growth and negative $k$ models decay.
A tangent at $x=x_0$ uses the point $(x_0,e^{kx_0})$ and gradient $ke^{kx_0}$ in $y-y_0=m(x-x_0)$ .
This proportional-rate property is the mathematical reason for choosing an exponential model.
Changing conditions can make its constant proportionality assumption unsuitable, so compare the model's rate with observed changes before accepting it as a valid model.

Worked example

Find the equation of the tangent to $y=e^{2x}$ at $x=1/2$ .

1.At $x=1/2$ , the point is $(1/2,e)$ .
2.Since $dy/dx=2e^{2x}$ , the gradient there is $2e$ .
3.Thus $y-e=2e(x-1/2)$ , which simplifies to $y=2ex$ .

Answer: $y=2ex$

Common mistakes

Don't differentiate $e^{kx}$ as $e^{kx}$ and omit the chain-rule factor $k$ .
Don't use an exponential model when the rate of change is not proportional to the current amount.

Exam tip

Show both $\mathrm dy/\mathrm dx=ky$ and the constant of proportionality when justifying an exponential model.

Worked practice

Tier 1 · Easy

Differentiate

y=e^{4x}

with respect to

x

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=4e^{4x}$	1
Notes
For $e^{kx}$ , differentiation multiplies the function by $k$ . Here $k=4$ , so $dy/dx=4e^{4x}$ .

(1 mark)

Tier 2 · Standard

For the curve

y=e^{2x}

, find the exact coordinates of the point where the gradient is

10

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\left(\dfrac12\ln5,5\right)$	3
Notes
Differentiate to obtain $dy/dx=2e^{2x}$ . Setting the gradient equal to $10$ gives $2e^{2x}=10$ , so $e^{2x}=5$ and $x=\tfrac12\ln5$ . At this value, $y=e^{2x}=5$ , giving the point $\left(\tfrac12\ln5,5\right)$ .

(3 marks)

Tier 3 · Hard

A culture is modelled by

P=320e^{0.18t}

, where

t

is in hours. Find its instantaneous growth rate when

P=500

, and explain the feature of the model that makes this calculation direct.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$90$ units per hour The growth rate is proportional to the current population.	4
Notes
Differentiate: $dP/dt=0.18(320e^{0.18t})=0.18P$ . When $P=500$ , $dP/dt=0.18(500)=90$ units per hour. The exponential model makes the derivative a constant multiple of the current value.

(4 marks)

Tier 1 · Easy

Given that

y=7e^{-0.4x}

, show that

\dfrac{\mathrm dy}{\mathrm dx}=-0.4y

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{\mathrm dy}{\mathrm dx}=-2.8e^{-0.4x}=-0.4y$	2
Notes
Differentiating gives $\dfrac{\mathrm dy}{\mathrm dx}=7(-0.4)e^{-0.4x}=-2.8e^{-0.4x}$ . Since $y=7e^{-0.4x}$ , this is $-0.4y$ .

(2 marks)

Tier 2 · Standard

The curve

y=Ae^{3x}

passes through

(\ln2,40)

. Determine, in exact form, its tangent at this point.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$y-40=120(x-\ln2)$ (or $y=120x+40-120\ln2$ )	4
Notes
Using the point, $40=Ae^{3\ln2}=8A$ , so $A=5$ . Also $\dfrac{\mathrm dy}{\mathrm dx}=3Ae^{3x}=3y$ , so the gradient at the point is $3(40)=120$ . The tangent is therefore $y-40=120(x-\ln2)$ .

(4 marks)

Tier 3 · Hard

A quantity is modelled by

Q=Ce^{-kt}

, where

C>0

and

k>0

. When

Q=180

, it is decreasing at

27

units per hour. Given also that

Q=600

when

t=0

, find the time when

Q=100

, giving your answer to

3

significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=0.15$ $t=11.9$ hours	6
Notes
Differentiating gives $\dfrac{\mathrm dQ}{\mathrm dt}=-kQ$ . Hence $-27=-180k$ , so $k=0.15$ . Since $Q(0)=C=600$ , setting $Q=100$ gives $100=600e^{-0.15t}$ . Thus $e^{-0.15t}=1/6$ and $t=\ln6/0.15=11.945\ldots$ , so $t=11.9$ hours to $3$ significant figures.

(6 marks)

Tier 2 · Standard

The point

P

on the curve

y=5e^{2x}

has

x

-coordinate

p

. Show that the tangent at

P

meets the

x

-axis at a point whose

x

-coordinate is

p-\dfrac12

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	The tangent meets the $x$ -axis at $\left(p-\dfrac12,0\right)$	4
Notes
At $P$ , $y=5e^{2p}$ and $\dfrac{\mathrm dy}{\mathrm dx}=10e^{2x}$ , so the tangent gradient is $10e^{2p}$ . Its equation is $y-5e^{2p}=10e^{2p}(x-p)$ . At the $x$ -axis, $y=0$ , so $-5e^{2p}=10e^{2p}(x-p)$ . Dividing by the non-zero factor $5e^{2p}$ gives $-1=2(x-p)$ , hence $x=p-1/2$ .

(4 marks)

Tier 3 · Hard

The curve

y=Ae^{kx}

passes through

(2,12)

and has gradient

9

there. Find

A

and

k

in exact form. The tangent at this point meets the coordinate axes at

R

and

S

. Find the exact area of triangle

ORS

, where

O

is the origin.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$k=\dfrac34$ and $A=12e^{-3/2}$ Area of triangle $ORS=2$ square units	6
Notes
Since $\dfrac{\mathrm dy}{\mathrm dx}=kAe^{kx}=ky$ , the data at $(2,12)$ give $9=12k$ , so $k=3/4$ . Then $12=Ae^{(3/4)(2)}$ , giving $A=12e^{-3/2}$ . The tangent is $y-12=9(x-2)$ , or $y=9x-6$ . Its intercepts are $R=(2/3,0)$ and $S=(0,-6)$ , so the area of triangle $ORS$ is $\tfrac12(2/3)(6)=2$ square units.

(6 marks)

Tier 3 · Hard

The curve

y=Ae^{kx}

, where

A>0

and

k>0

, passes through

P(1,8)

. The normal to the curve at

P

passes through

Q(5,6)

. Find

A

and

k

in exact form. Hence find the exact coordinates of the point on the curve where the gradient is

5

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$k=\dfrac14$ and $A=8e^{-1/4}$ $\left(1+4\ln\!\left(\dfrac52\right),20\right)$	6
Notes
The normal gradient is $(6-8)/(5-1)=-1/2$ , so the tangent gradient at $P$ is $2$ . Since $\mathrm dy/\mathrm dx=ky$ , $2=8k$ and $k=1/4$ . Using $P$ gives $8=Ae^{1/4}$ , so $A=8e^{-1/4}$ . Where the gradient is $5$ , $ky=5$ , hence $y=20$ . Writing the curve as $y=8e^{(x-1)/4}$ gives $20=8e^{(x-1)/4}$ , so $x=1+4\ln(5/2)$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A curve has equation

y=Ae^{kx}+c

, where

A>0

and

k>0

. At

x=0

its value is

7

and its gradient is

6

. At

x=\ln3

its value is

19

and its gradient is

18

. Determine

A

k

and

c

. Hence find the exact value of

x

for which

y=37

, and find the tangent at

x=0

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$A=6$ , $k=1$ and $c=1$ $x=\ln6$ Tangent $y=6x+7$	7
Notes
For $y=Ae^{kx}+c$ , $\mathrm dy/\mathrm dx=kAe^{kx}=k(y-c)$ . The two point-gradient pairs give $6=k(7-c)$ and $18=k(19-c)$ . Subtracting gives $12=12k$ , so $k=1$ ; then $c=1$ . At $x=0$ , $7=A+1$ , hence $A=6$ . Thus $y=6e^x+1$ . Setting $y=37$ gives $e^x=6$ and $x=\ln6$ . The tangent uses point $(0,7)$ and gradient $6$ , so it is $y=6x+7$ .

(7 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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