WMA11 Jan 2021

• WMA11/01 Edexcel IAL P1 January 2021 IAL Q1 (Differentiation & Straight Line Graph: Equations of Normals)

A curve has equation​​ 

y=2x3-5x232x+7         x>0 

• Find, in simplest form,​​ dydx​​ 

(3)

The point P lies on the curve and has x coordinate​​ 12.​​ 

• Find an equation of the normal to the curve at P, writing your answer in the form ax + by + c = 0, where a, b and c are integers to be found.​​ 

(5)

SOLUTION

a-​​ 

y=2x3-5x2-32x-1+7

dydx=6x2-10x+32x-2

dydx=6x2-10x+32x2  

b-​​ 

Let us first find the y-cordinate of point P, where the x-coordinate is given as ½.

For that, we will substitute the value of x in the equation of curve as point P lies on the curve.​​ 

y=2x3-5x2-32x-1+7

y=2123-5122-3212+7

y= 28-54-3+7

y=-44-3+7 

y= -1-3+7

y=3

So, the coordinates of point P are (1/2, 3).​​ 

Now, let us find the value of gradient of tangent to the curve at point P.​​ 

mT=dydx=6x2-10x+32x2

mT=6122-1012+32122 

mT=32-5+312

mT=32-5+6

mT=52

The gradient of normal can be found with the help of a fact that normal and tangent are perpendicular to each other.​​ 

mT×mN=-1

52×mN=-1

mN= -25

Now, that we have the point​​ P12, 3​​ and slope of normal​​ mN=-25, let us use the point-slope formula to find the equation of a line.​​ 

y-y1=mNx-x1

y-3=-25 x-12

5y-15= -2 x-12

5y-15= -2x+1

2x+5y-16=0