Answer to Question #267679 in Calculus for ray

Question

Find the rate at which the reciprocal of a number changes as the number increases.

Answer 1

Let the number be x and it's reciprocal be y.

So y = $\frac{1}{x}$

Differentiating with respect to time, t

$\frac{dy}{dt} = \frac{d}{dt}(\frac{1}{x})$

$=>\frac{dy}{dt} = \frac{d}{dx}(\frac{1}{x}).\frac{dx}{dt}$

$=>\frac{dy}{dt} = - \frac{1}{x²}\frac{dx}{dt}$

As the number increases, $\frac{dx}{dt} > 0$

Therefore $\frac{dy}{dt}$ < 0

So the reciprocal of the number decreases at the rate of the reciprocal of the square of the number multiplied by the rate at which the number increases.