A particle is moving along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate

Solution:

Derivatives are used to find the rate of changes of a quantity with respect to the other quantity. By using the application of derivatives we can find the approximate change in one quantity with respect to the

change in the other quantity.

The equation of the curve is

6y = x³ + 2

Differentiating with respect to time, we have

⇒ 6 dy/dt = 3x² dx/dt-----(1)

⇒ 2 dy/dt = x² dx/dt

According to the question,

(dy/dt = 8 dx/dt)-----(2)

Hence,

From equation 1 and 2, we get

⇒ 2 (8 dx/dt) = x² dx/dt

⇒ 16 dx/dt = x² dx/dt

⇒ x² - 16 dx/dt = 0

⇒ x² = 16

⇒ x = ± 4

When, x = 4

Then,

y = (4³ + 2)/6

= 66/6

= 11

When x = - 4

Then,

y = ((- 4³) + 2)/6

= 62/6

= 31/3

Thus, the points on the curve are (4, 11) and (- 4, - 31/3)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 11