Short Questions (Question 17 – 19)


Question 17:

In the diagram above, the straight line PR is normal to the curve   y = x 2 2 + 1 at Q. Find the value of k.

Solution:
y = x 2 2 + 1 d y d x = x At point  Q ,   x -coordinate = 2 , Gradient of the curve,  d y d x = 2 Hence, gradient of normal to the curve,  P R = 1 2 3 0 2 k = 1 2 6 = 2 + k k = 8



Question 18:
The normal to the curve y = x2 + 3x at the point P is parallel to the straight line 
y = x + 12. Find the equation of the normal to the curve at the point P.

Solution:
Given normal to the curve at point P is parallel to the straight line y = –x + 12.
Hence, gradient of normal to the curve = –1.
As a result, gradient of tangent to the curve = 1

y = x2 + 3x
d y d x = 2x + 3
2x + 3 = 1
2x = –2
x = –1
y = (–1)2+ 3(–1)
y = –2
Point P = (–1, –2).

Equation of the normal to the curve at point P is,
y – (–2) = –1 (x – (–1))
y + 2 = – x – 1
y = – x– 3




Question 19:
Given that   y = 3 4 x 2 , find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:
y = 3 4 x 2 d y d x = ( 2 ) 3 4 x = 3 2 x δ y = 47.7 48 = 0.3 Approximate change in  x  to  y δ x δ y d x d y δ x = d x d y × δ y δ x = 2 3 x × ( 0.3 ) δ x = 2 3 ( 8 ) × ( 0.3 ) y = 48 3 4 x 2 = 48 x 2 = 64 x = 8 δ x = 0.025