Chapter 12: Q. 36 (page 961)

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.
$\frac{d θ}{d t} when θ = \tan^{- 1} (\frac{yz}{x}), x = e^{t}, y = e^{2 t}, z = e^{3 t}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

$θ = \tan^{- 1} (\frac{yz}{x}) x = e^{t} y = e^{2 t} z = e^{3 t}$

Step 2. Calculation.

By Theorem 12.34, we have

$\frac{d θ}{d t} = \frac{\partial θ}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial θ}{\partial y} \cdot \frac{d y}{d t} + \frac{\partial θ}{\partial z} \cdot \frac{d z}{d t} - - - - - - - (1)$

So first we find

$\frac{\partial θ}{\partial x}, \frac{d x}{d t}, \frac{\partial θ}{\partial y}, \frac{d y}{d t}, \frac{\partial θ}{\partial z}, \frac{d z}{d t}$

So we have

$\frac{\partial θ}{\partial x} = \frac{1}{1 + {(\frac{y z}{x})}^{2}} \cdot (- \frac{y z}{x^{2}}) = - \frac{y z}{x^{2} + y^{2} z^{2}} \frac{d x}{d t} = e^{t} \frac{\partial θ}{\partial y} = \frac{1}{1 + {(\frac{y z}{x})}^{2}} \cdot (1 \cdot \frac{z}{x}) = \frac{z x}{x^{2} + y^{2} z^{2}} \frac{d y}{d t} = 2 e^{2 t} \frac{\partial θ}{\partial z} = \frac{y x}{x^{2} + y^{2} z^{2}} \frac{d z}{d t} = 3 e^{3 t}$

Step 3. Calculation

Use these above values in (1) we get,

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

$\frac{d θ}{d t} = \frac{1}{x^{2} + y^{2} z^{2}} \{- y z e^{t} + 2 z x e^{2 t} + 3 y x e^{3 t}\} \frac{d θ}{d t} = \frac{1}{e^{2 t} + e^{10 t}} \{- e^{5 t} \cdot e^{t} + 2 e^{4 t} \cdot e^{2 t} + 3 e^{3 t} \cdot e^{3 t}\} \frac{d θ}{d t} = \frac{4 e^{4 t}}{1 + e^{8 t}}$

Step 4. Conclusion.

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

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Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.
$\frac{d θ}{d t} when θ = \tan^{- 1} (\frac{yz}{x}), x = e^{t}, y = e^{2 t}, z = e^{3 t}$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Calculation

Step 4. Conclusion.

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Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.dθdtwhenθ=tan-1yzx,x=et,y=e2t,z=e3t

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Calculation

Step 4. Conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Probability and Statistics

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.
$\frac{d θ}{d t} when θ = \tan^{- 1} (\frac{yz}{x}), x = e^{t}, y = e^{2 t}, z = e^{3 t}$