Question

A point is moving along the graph of the given function such that \(d x / d t\) is 2 centimeters per second. Find \(d y / d t\) for the given values of \(x\). \(y=\tan x\) (a) \(x=-\frac{\pi}{3}\) (b) \(x=-\frac{\pi}{4}\) (c) \(x=0\)

Short answer

(a) \(\frac{dy}{dt}\) is 8 cm/s when \(x = -\frac{\pi}{3}\). (b) \(\frac{dy}{dt}\) is 4 cm/s when \(x = -\frac{\pi}{4}\). (c) \(\frac{dy}{dt}\) is 2 cm/s when \(x = 0\).

Step-by-step solution

Find the Derivative of y

The function given is \(y = \tan x\). The derivative of \(y\) with respect to \(x\) is then given by \(y' = \frac{dy}{dx} = \sec^2 x\).

Utilize the Formula for Related Rates

The formula for related rates is given by \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). We know that \(\frac{dx}{dt} = 2\) cm/s and \(\frac{dy}{dx} = \sec^2 x\), so we can substitute these into our formula to obtain \(\frac{dy}{dt} = 2\sec^2x\).

Find \(\frac{dy}{dt}\) for \(x = -\frac{\pi}{3}\)

Substitute \(x = -\frac{\pi}{3}\) into \(\frac{dy}{dt} = 2\sec^2x\). Given that \(\sec(-\frac{\pi}{3}) = -2\), we then have \(\frac{dy}{dt} = 2(-2)^2 = 8\) cm/s.

Find \(\frac{dy}{dt}\) for \(x = -\frac{\pi}{4}\)

Substitute \(x = -\frac{\pi}{4}\) into \(\frac{dy}{dt} = 2\sec^2x\). Given that \(\sec(-\frac{\pi}{4}) = -\sqrt{2}\), we then have \(\frac{dy}{dt} = 2(-\sqrt{2})^2 = 4\) cm/s.

Find \(\frac{dy}{dt}\) for \(x = 0\)

Substitute \(x = 0\) into \(\frac{dy}{dt} = 2\sec^2x\). Given that \(\sec(0) = 1\), we then have \(\frac{dy}{dt} = 2(1)^2 = 2\) cm/s.