Evaluate | f(x, y, z) ds given that f(x, y,z)=and the curve C is 7(t) = 2ti +t j+:3÷r°kt'k (1

Given: f(x,y,z)=x2xy+6yz r→(t)=2t i→+t2 j→+13t3 k→ 1≤t<∞

Answered: Evaluate [ f (x,y,z) ds given that…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Evaluate | f(x, y, z) ds given that f(x, y,z)=
and the curve C is 7(t) = 2ti +t j+:
3
÷r°k
t'k (1<t<∞)
2ху + буz

Expert Solution

Step 1

Given:

$f (x, y, z) = \frac{x}{2 xy + 6 yz}$

$\vec{r} (t) = 2 t \vec{i} + t^{2} \vec{j} + \frac{1}{3} t^{3} \vec{k}$ $(1 \leq t < \infty)$

Step 2

The dS is given by,

$\Rightarrow dS = \sqrt{{(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2} + {(\frac{dz}{dt})}^{2}} dt$

Here,

$\Rightarrow x = 2 t$

On differentiating,

$\Rightarrow \frac{dx}{dt} = 2$

$\Rightarrow y = t^{2}$

On differentiating,

$\Rightarrow \frac{dy}{dt} = 2 t$

$\Rightarrow z = \frac{1}{3} t^{3}$

On differentiating,

$\Rightarrow \frac{dz}{dt} = t^{2}$

Step 3

Therefore,

$\Rightarrow dS = \sqrt{{(2)}^{2} + {(2 t)}^{2} + {(t^{2})}^{2}} dt$

On solving,

$\Rightarrow dS = \sqrt{4 + 4 t^{2} + t^{4}} dt$

$\Rightarrow dS = \sqrt{{(t^{2} + 2)}^{2}} dt$

On simplifying,

$\Rightarrow dS = (t^{2} + 2) dt$

Given that:

$\Rightarrow f (x, y, z) = \frac{x}{2 xy + 6 yz}$

On substituting the values,

$\Rightarrow f (x, y, z) = \frac{2 t}{2 (2 t) (t^{2}) + 6 (t^{2}) (\frac{1}{3} t^{3})}$

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Evaluate [ f (x,y,z) ds given that f(x,y,z) =, and the curve C is ř(t) = 2ti +t²j+÷tk (1