IntegrationP1^
6
EXAMPLE 6.2 A curve is such that
d
dy
x
x
x=+ (^382). Given that the point (4, 20) lies on the curve,
find the equation of the curve.
SOLUTION
Rewrite the gradient function as d
d
y
x
=+ 38 xx
(^12) – 2
.
By integration, yx=× + x c(^3) − +
2
3
8
1
32 –^1yx=− (^2) x^8 +c
(^32)
Since the curve passes through the point (4, 20),
20 24
(^328)
=−() 4 +c
⇒ 20 = 16 − 2 + c
⇒ c = 6
So the equation of the curve is yx
x
=− 2 8 + 6
(^32)
.
EXAMPLE 6.3 The gradient function of a curve is ddyx^ = 4 x − 12.
(i) The minimum y value is 16. By considering the gradient function, find the
corresponding x value.
(ii) Use the gradient function and your answer from part (i) to find the equation of
the curve.
SOLUTION
(i) At the minimum, the gradient of the curve must be zero,
4 x − 12 = 0 ⇒ x = 3.
(ii) d
d
y
x
(^) = 4 x − 12
⇒ y = 2 x 2 − 12 x + c.
At the minimum point, x = 3 and y = 16
⇒ 16 = 2 × 32 − 12 × 3 + c
⇒ c = 34
So the equation of the curve is y = 2 x^2 − 12 x + 34.
Dividing by^32 is the same
as multiplying by^23.