Question

A particle is moving on the curve x^y=1. As it passes through the point (2,1/4), the xcoordinate is increasing at a rate of 2 units per second. At what rate is the y-coordinate changing at that time?

Answer 1

Answer:

y-coordinate is decreasing at the rate of $\dfrac{1}{2}$ unit/sec.

Step-by-step explanation:

Given that:

The curve of the particle $x^2y = 1$

Then:

$y = \dfrac{1}{x^2}$

Taking the differential of y with respect to t

$\dfrac{dy}{dt}= \dfrac{dx^{-2}}{dx} * \dfrac{dx}{dt}$

$= -2x^{-3} \dfrac{dx}{dt}$

At (2, 1/4)

$\dfrac{dx}{dt} = 2$

This implies that:

$\implies \dfrac{dy}{dt} = -\dfrac{2}{8}(2)$

$\dfrac{dy}{dt} = -\dfrac{1}{2} \ \ unit/sec$

Thus, y-coordinate is decreasing at the rate of $\dfrac{1}{2}$ unit/sec.