Q. 36 - Calculus

Question

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.

$\frac{d\theta }{dt} \mathrm{when} \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{yz}}{\mathrm{x}}\right), \mathrm{x}={\mathrm{e}}^{\mathrm{t}}, \mathrm{y}={\mathrm{e}}^{2\mathrm{t}}, \mathrm{z}={\mathrm{e}}^{3\mathrm{t}}$

Step-by-Step Solution

Answer

The value of $\frac{d\theta }{dt}=\frac{4e^{4t}}{1+e^{8t}}$

Step 1. Given Information.

$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{yz}}{\mathrm{x}}\right) \\ \mathrm{x}={\mathrm{e}}^{\mathrm{t}} \\ \mathrm{y}={\mathrm{e}}^{2\mathrm{t}} \\ \mathrm{z}={\mathrm{e}}^{3\mathrm{t}} \end{gathered}$$

Step 2. Calculation.

By Theorem 12.34, we have

$\frac{d\theta }{dt}=\frac{\partial \theta }{\partial x}·\frac{dx}{dt}+\frac{\partial \theta }{\partial y}·\frac{dy}{dt}+\frac{\partial \theta }{\partial z}·\frac{dz}{dt}-------\left(1\right)$

So first we find 

$\frac{\partial \theta }{\partial x},\frac{dx}{dt},\frac{\partial \theta }{\partial y},\frac{dy}{dt},\frac{\partial \theta }{\partial z},\frac{dz}{dt}$

So we have 

$$\begin{gathered} \frac{\partial \theta }{\partial x}=\frac{1}{1+{\left({\displaystyle \frac{yz}{x}}\right)}^2}·\left(-\frac{yz}{x^2}\right)=-\frac{yz}{x^2+y^2{z}^2} \\ \frac{dx}{dt}=e^t \\ \frac{\partial \theta }{\partial y}=\frac{1}{1+{\left({\displaystyle \frac{yz}{x}}\right)}^2}·\left(1·\frac{z}{x}\right)=\frac{zx}{x^2+y^2{z}^2} \\ \frac{dy}{dt}=2e^{2t} \\ \frac{\partial \theta }{\partial z}=\frac{yx}{x^2+y^2{z}^2} \\ \frac{dz}{dt}=3e^{3t} \end{gathered}$$

Step 3. Calculation

Use these above values in (1) we get,

$$\begin{gathered} \frac{d\theta }{dt}=\frac{\partial \theta }{\partial x}·\frac{dx}{dt}+\frac{\partial \theta }{\partial y}·\frac{dy}{dt}+\frac{\partial \theta }{\partial z}·\frac{dz}{dt} \\ \frac{d\theta }{dt}=\frac{1}{x^2+y^2{z}^2}\left\{-yz{e}^t+2zx{e}^{2t}+3yx{e}^{3t}\right\} \end{gathered}$$

So from here, putting the value of $x,y,z$in term of $t$ we get

$$\begin{gathered} \frac{d\theta }{dt}=\frac{1}{x^2+y^2{z}^2}\left\{-yz{e}^t+2zx{e}^{2t}+3yx{e}^{3t}\right\} \\ \frac{d\theta }{dt}=\frac{1}{e^{2t}+e^{10t}}\left\{-e^{5t}·e^t+2e^{4t}·e^{2t}+3e^{3t}·e^{3t}\right\} \\ \frac{d\theta }{dt}=\frac{4e^{4t}}{1+e^{8t}} \end{gathered}$$

Step 4. Conclusion.

The value of $\frac{d\theta }{dt}=\frac{4e^{4t}}{1+e^{8t}}$.