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Chapter 16: Problem 3

Find the rate of change of \(y\) at the specified point: (a) \(y=\ln \left(3 t^{2}+5\right), t=1\) (b) \(y=\sin \left(t^{2}\right), t=2\), (c) \(y=\cos \left(t^{3}+1\right), t=1\) (d) \(y=\left(t^{3}-1\right)^{2 / 3}, t=2\) (e) \(y=4 \mathrm{e}^{\cos t}, t=\frac{\pi}{2}\)

Short Answer

Expert verified
Question: Find the rate of change of each function at the given time, t. (a) \(y=\ln \left(3 t^{2}+5\right), t=1\) (b) \(y=\sin \left(t^{2}\right), t=2\) (c) \(y=\cos \left(t^{3}+1\right), t=1\) (d) \(y=\left(t^{3}-1\right)^{2 / 3}, t=2\) (e) \(y=4 \mathrm{e}^{\cos t}, t=\frac{\pi}{2}\) Answer: (a) The rate of change at \(t = 1\) is \(\frac{3}{4}\). (b) The rate of change at \(t = 2\) is \(4\cos(4)\). (c) The rate of change at \(t = 1\) is \(-3\sin(2)\). (d) The rate of change at \(t = 2\) is \(\frac{8}{\sqrt[3]{7}}\). (e) The rate of change at \(t = \frac{\pi}{2}\) is \(-4\).

Step by step solution

01

Find the derivative of y with respect to t

Apply the chain rule: \(\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} u} * \frac{\mathrm{d} u}{\mathrm{d} t}\), where \(u = 3t^2 + 5\). We have \(\frac{\mathrm{d} y}{\mathrm{d} u} = \frac{1}{u}\) and \(\frac{\mathrm{d} u}{\mathrm{d} t} = 6t\) Multiply the two derivative to get the derivative of \(y\) with respect to \(t\): \(\frac{dy}{dt} = \frac{1}{u} * 6t = \frac{6t}{3t^2 + 5}\)
02

Evaluate the derivative at the given point

At \(t = 1\), substitute into the derivative expression: \(\frac{dy}{dt} = \frac{6(1)}{3(1)^2 + 5} = \frac{6}{8} = \frac{3}{4}\). The rate of change at \(t = 1\) is \(\frac{3}{4}\). (b) \(y=\sin \left(t^{2}\right), t=2\)
03

Find the derivative of y with respect to t

Applying the chain rule again, with \(u = t^2\). We have \(\frac{\mathrm{d} y}{\mathrm{d} u} = \cos(u)\) and \(\frac{\mathrm{d} u}{\mathrm{d} t} = 2t\). Multiply the two derivatives: \(\frac{dy}{dt} = \cos(u) * 2t = 2t\cos(t^2)\)
04

Evaluate the derivative at the given point

At \(t = 2\), substitute into the derivative expression: \(\frac{dy}{dt} = 2(2)\cos(2^2) = 4\cos(4)\). The rate of change at \(t = 2\) is \(4\cos(4)\). (c) \(y=\cos \left(t^{3}+1\right), t=1\)
05

Find the derivative of y with respect to t

Applying the chain rule with \(u = t^3 + 1\). We have \(\frac{\mathrm{d} y}{\mathrm{d} u} = -\sin(u)\) and \(\frac{\mathrm{d} u}{\mathrm{d} t} = 3t^2\). Multiply the two derivatives: \(\frac{dy}{dt} = -\sin(u) * 3t^2 = -3t^2\sin(t^3 + 1)\)
06

Evaluate the derivative at the given point

At \(t = 1\), substitute into the derivative expression: \(\frac{dy}{dt} = -3(1)^2\sin(1^3 + 1) = -3\sin(2)\). The rate of change at \(t = 1\) is \(-3\sin(2)\). (d) \(y=\left(t^{3}-1\right)^{2 / 3}, t=2\)
07

Find the derivative of y with respect to t

We will apply the chain rule, with \(u = t^3 - 1\). We have \(\frac{\mathrm{d} y}{\mathrm{d} u} = \frac{2}{3}u^{-1/3}\) and \(\frac{\mathrm{d} u}{\mathrm{d} t} = 3t^2\). Multiply the two derivatives: \(\frac{dy}{dt} = \frac{2}{3}u^{-1/3} * 3t^2 = 2t^2(u)^{-1/3}\)
08

Evaluate the derivative at the given point

At \(t = 2\), substitute into the derivative expression: \(\frac{dy}{dt} = 2(2^2)((2^3 - 1)^{-1/3}) = \frac{8}{(7)^{1/3}}\). The rate of change at \(t = 2\) is \(\frac{8}{\sqrt[3]{7}}\). (e) \(y=4 \mathrm{e}^{\cos t}, t=\frac{\pi}{2}\)
09

Find the derivative of y with respect to t

Apply the chain rule with \(u = \cos(t)\). We have \(\frac{\mathrm{d} y}{\mathrm{d} u} = 4\mathrm{e}^{\cos(t)}\) and \(\frac{\mathrm{d} u}{\mathrm{d} t} = -\sin(t)\). Multiply the two derivatives: \(\frac{dy}{dt} = -4\mathrm{e}^{\cos(t)}\sin(t)\)
10

Evaluate the derivative at the given point

At \(t = \frac{\pi}{2}\), substitute into the derivative expression: \(\frac{dy}{dt} = -4\mathrm{e}^{\cos(\pi/2)}\sin(\pi/2) = -4\mathrm{e}^0 \cdot 1 = -4\). The rate of change at \(t = \frac{\pi}{2}\) is \(-4\).

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