What is 7.5: Implicit and parametric differentiation?

For an implicit relation, differentiate every term with respect to x and attach a factor dy/ dx whenever a differentiated term contains y.

How many marks are 7.5 questions worth?

7.5 practice drills on this page are worth 3–7 marks each; marks in a real exam vary by question.

Which past exam papers has 7.5 appeared in?

We have not yet indexed a verified past-paper appearance for 7.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.5

Differentiate simple functions and relations defined implicitly or parametrically, for first derivative only.

Draft — not yet indexed

Implicit and parametric differentiation

Worked answers and methods for 7.5 on Edexcel A-level Maths 9MA0.

Explanation

For an implicit relation, differentiate every term with respect to $x$ and attach a factor $\frac{\mathrm dy}{\mathrm dx}$ whenever a differentiated term contains $y$ .
For parametric equations $x=x(t)$ and $y=y(t)$ , use $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$ where $\mathrm dx/\mathrm dt\ne0$ .
After finding the gradient, use the parameter or relation to obtain the actual point before writing a tangent or normal equation.
A common error in implicit differentiation is to write $\frac{\mathrm d}{\mathrm dx}(xy)=y+x$ instead of $y+x\frac{\mathrm dy}{\mathrm dx}$ .

Worked example

A curve has parametric equations $x=t^2+1$ and $y=t^3-3t$ . Find the equation of its tangent when $t=2$ .

1. $\frac{\mathrm dx}{\mathrm dt}=2t$ and $\frac{\mathrm dy}{\mathrm dt}=3t^2-3$ , so $\frac{\mathrm dy}{\mathrm dx}=\frac{3t^2-3}{2t}$ .
2.At $t=2$ , the gradient is $\frac94$ .
3.The point is $(5,2)$ , so the tangent is $y-2=\frac94(x-5)$ .

Answer: $y-2=\frac94(x-5)$

Common mistakes

Don't substitute the parameter before forming $\mathrm dy/\mathrm dx$ , concealing a zero value of $\mathrm dx/\mathrm dt$ .
Don't calculate the two parametric derivatives but use their product instead of dy/dx equals (dy/dt)/(dx/dt).

Exam tip

For a parametric tangent, evaluate both parameter derivatives at the stated parameter before forming the gradient and line equation.

Worked practice

Tier 1 · Easy

The curve obeys

x^2+y^2=25

. By implicit differentiation, obtain its gradient at a general point

(x,y)

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=-\frac{x}{y}$ for $y\ne0$	3
Notes
Differentiate both sides with respect to $x$ : $2x+2y\frac{\mathrm dy}{\mathrm dx}=0$ . For $y\ne0$ , rearranging gives $\frac{\mathrm dy}{\mathrm dx}=-\frac{x}{y}$ . At $y=0$ the curve instead has a vertical tangent.

(3 marks)

Tier 2 · Standard

The curve

x^2+xy+y^2=7

passes through

(1,2)

. Obtain a formula for its gradient, then determine the tangent at

(1,2)

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2x+y}{x+2y}$ Tangent $y-2=-\dfrac45(x-1)$ , or any algebraically equivalent equation such as $4x+5y-14=0$	5
Notes
Differentiate implicitly: $2x+(x\,dy/dx+y)+2y\,dy/dx=0$ . Collecting derivative terms gives $(x+2y)dy/dx=-(2x+y)$ , so $dy/dx=-(2x+y)/(x+2y)$ . At $(1,2)$ the gradient is $-(2+2)/(1+4)=-4/5$ , hence the tangent is $y-2=-\tfrac45(x-1)$ .

(5 marks)

Tier 3 · Hard

The curve has parametric equations

x=t+\frac1t

and

y=t-\frac1t

, where

t>0

. Find the exact equations of the tangent and normal at

t=2

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Tangent $y-\frac32=\frac53\left(x-\frac52\right)$ Normal $y-\frac32=-\frac35\left(x-\frac52\right)$	7
Notes
$\frac{\mathrm dx}{\mathrm dt}=1-\frac1{t^2}$ and $\frac{\mathrm dy}{\mathrm dt}=1+\frac1{t^2}$ . At $t=2$ , these are $\frac34$ and $\frac54$ , so the tangent gradient is $\frac53$ and the normal gradient is $-\frac35$ . The point is $(\frac52,\frac32)$ , giving the two stated equations.

(7 marks)

Tier 1 · Easy

A curve has parametric equations

x=t^2+1

and

y=5t-2

. Find

\dfrac{\mathrm dy}{\mathrm dx}

in terms of

t

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{5}{2t}$ for $t\neq0$	3
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=2t$ and $\dfrac{\mathrm dy}{\mathrm dt}=5$ . Therefore $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}=5/(2t)$ , provided $t\neq0$ .

(3 marks)

Tier 2 · Standard

A curve has parametric equations

x=t^3+1

and

y=t^2-4t

. Find the value of

t

at which the curve has a horizontal tangent, and find the equations of the tangent and normal there.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$t=2$ Tangent $y=-4$ Normal $x=9$	5
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=3t^2$ and $\dfrac{\mathrm dy}{\mathrm dt}=2t-4$ . A horizontal tangent requires $2t-4=0$ , so $t=2$ ; here $\mathrm dx/\mathrm dt=12\neq0$ . The point is $(2^3+1,2^2-4(2))=(9,-4)$ . Hence the horizontal tangent is $y=-4$ and the normal is the vertical line $x=9$ .

(5 marks)

Tier 3 · Hard

The curve

x^2+xy+2y^2=8

passes through

P(2,1)

. The normal at

P

meets the coordinate axes at

A

and

B

. Find the exact area of triangle

OAB

, where

O

is the origin.

(7)

(Total for Question 6 is 7 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Area of triangle $OAB=\dfrac{49}{60}$ square units	7
Notes
Implicit differentiation gives $2x+y+(x+4y)\dfrac{\mathrm dy}{\mathrm dx}=0$ . At $(2,1)$ , the tangent gradient is $-5/6$ , so the normal gradient is $6/5$ . Its equation is $y-1=\frac65(x-2)$ . Setting $y=0$ gives the $x$ -intercept $A=(7/6,0)$ ; setting $x=0$ gives $B=(0,-7/5)$ . Therefore the area is $\frac12\left\|\frac76\cdot-\frac75\right\|=49/60$ square units.

(7 marks)

Tier 2 · Standard

A curve has parametric equations

x=2\cos t

and

y=3\sin t

. Find the exact equation of the tangent when

t=\dfrac\pi4

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$y-\dfrac{3\sqrt2}{2}=-\dfrac32(x-\sqrt2)$	4
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=-2\sin t$ and $\dfrac{\mathrm dy}{\mathrm dt}=3\cos t$ , so $\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3\cos t}{2\sin t}$ . At $t=\pi/4$ , the gradient is $-3/2$ and the point is $(\sqrt2,3\sqrt2/2)$ . Hence the tangent is $y-3\sqrt2/2=-\tfrac32(x-\sqrt2)$ .

(4 marks)

Tier 3 · Hard

A curve has parametric equations

x=t^2+t

and

y=t^3

. Find all values of

t

for which the tangent has gradient

1

, and find the equation of each tangent.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$t=1$ , with tangent $y=x-1$ $t=-\dfrac13$ , with tangent $y=x+\dfrac5{27}$	6
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=2t+1$ and $\dfrac{\mathrm dy}{\mathrm dt}=3t^2$ , so $\dfrac{\mathrm dy}{\mathrm dx}=3t^2/(2t+1)$ where $t\neq-1/2$ . Setting this equal to $1$ gives $3t^2-2t-1=0=(3t+1)(t-1)$ , so $t=1$ or $t=-1/3$ . At $t=1$ the point is $(2,1)$ , giving $y=x-1$ . At $t=-1/3$ the point is $(-2/9,-1/27)$ , giving $y+1/27=x+2/9$ , or $y=x+5/27$ .

(6 marks)

Tier 3 · Hard

A curve has parametric equations

x=t^2-1

and

y=t^3-t

. The curve passes through the origin for two distinct values of

t

. Find these values and the equation of the tangent corresponding to each value. Show that the two tangents are perpendicular.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$t=-1$ gives tangent $y=-x$ $t=1$ gives tangent $y=x$ The tangents are perpendicular	6
Notes
At the origin, $t^2-1=0$ , so $t=\pm1$ ; both also give $t^3-t=0$ . Now $\mathrm dx/\mathrm dt=2t$ and $\mathrm dy/\mathrm dt=3t^2-1$ , so $\mathrm dy/\mathrm dx=(3t^2-1)/(2t)$ . The gradients are $-1$ at $t=-1$ and $1$ at $t=1$ . Since both lines pass through the origin, their equations are $y=-x$ and $y=x$ ; their gradient product is $-1$ , so they are perpendicular.

(6 marks)

Q10

Tier 3 · Hard

10.

The curve

x^2+2xy+3y^2=12

has both horizontal and vertical tangents. By implicit differentiation, find the exact coordinates of every point with a horizontal tangent and every point with a vertical tangent.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Horizontal tangents at $(-\sqrt6,\sqrt6)$ and $(\sqrt6,-\sqrt6)$ Vertical tangents at $(-3\sqrt2,\sqrt2)$ and $(3\sqrt2,-\sqrt2)$	7
Notes
Implicit differentiation gives $2x+2y+(2x+6y)\,\mathrm dy/\mathrm dx=0$ , so $\mathrm dy/\mathrm dx=-(x+y)/(x+3y)$ . Horizontal tangents require $x+y=0$ with $x+3y\neq0$ . Substituting $x=-y$ into the curve gives $2y^2=12$ , producing the two stated points. Vertical tangents require $x+3y=0$ with $x+y\neq0$ . Substituting $x=-3y$ gives $6y^2=12$ , producing the other two points.

(7 marks)

Verified exam appearances

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