Chapter 12: Q. 35 (page 961)

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.
$\frac{d ρ}{d t} when ρ = \sqrt{x^{2} + y^{2} + z^{2}}, x = \sqrt{t}, y = t^{2}, z = t^{3}$

Short Answer

Expert verified

The value of $\frac{d ρ}{d t} = \frac{1}{2} (1 + 4 t^{3} + 3 t^{5}) {(t + t^{4} + t^{6})}^{\frac{- 1}{2}}$ .

Step by step solution

Step 1. Given Information.

$ρ = \sqrt{x^{2} + y^{2} + z^{2}} x = \sqrt{t} y = t^{2} z = t^{3}$

Step 2. Calculation.

By Theorem 12.34, we have

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

So first we find

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

So we have

$\frac{\partial ρ}{\partial x} = \frac{1 \cdot 2 x}{2 \sqrt{x^{2} + y^{2} + z^{2}}} = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} \frac{\partial ρ}{\partial y} = \frac{1 \cdot 2 y}{2 \sqrt{x^{2} + y^{2} + z^{2}}} = \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} \frac{\partial ρ}{\partial z} = \frac{1 \cdot 2 z}{2 \sqrt{x^{2} + y^{2} + z^{2}}} = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \frac{d x}{d t} = \frac{1}{2 t^{\frac{1}{2}}} \frac{d y}{d t} = 2 t \frac{d z}{d t} = 3 t^{2}$

Step 3. Calculation.

Use these above values in (1) we get,

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

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

$\frac{d ρ}{d t} = \frac{x}{2 t^{\frac{1}{2}} {(x^{2} + y^{2} + z^{2})}^{\frac{1}{2}}} + \frac{2 y t}{{(x^{2} + y^{2} + z^{2})}^{\frac{1}{2}}} + \frac{3 t^{2} z}{{(x^{2} + y^{2} + z^{2})}^{\frac{1}{2}}} \frac{d ρ}{d t} = \frac{1}{\sqrt{t + t^{4} + t^{6}}} (\frac{1}{2} + 2 t^{3} + 3 t^{5}) \frac{d ρ}{d t} = \frac{1}{2} (1 + 4 t^{3} + 3 t^{5}) {(t + t^{4} + t^{6})}^{\frac{- 1}{2}}$

Step 4. Conclusion.

The value of $\frac{d ρ}{d t} = \frac{1}{2} (1 + 4 t^{3} + 3 t^{5}) {(t + t^{4} + t^{6})}^{\frac{- 1}{2}}$ .

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Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.
$\frac{d ρ}{d t} when ρ = \sqrt{x^{2} + y^{2} + z^{2}}, x = \sqrt{t}, y = t^{2}, z = t^{3}$

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.dρdtwhenρ=x2+y2+z2,x=t,y=t2,z=t3

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

Decision Maths

Applied Mathematics

Pure Maths

Statistics

Discrete Mathematics

Calculus

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.
$\frac{d ρ}{d t} when ρ = \sqrt{x^{2} + y^{2} + z^{2}}, x = \sqrt{t}, y = t^{2}, z = t^{3}$