#6 t goes from 0 to 1 5. Evaluate the line integral yz dy + ry dz, where C is the curve 7 (t) = (Vi, t, t) and0

Consider the given integral as ∫Cyxds, where C is the curve r⇀t=t4,t3 and 0≤t≤1. Note that, the line…

Answered: 6. Set up but do NOT evaluate Evaluate…

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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#6 t goes from 0 to 1

5. Evaluate the line integral yz dy + ry dz, where C is the curve 7 (t) = (Vi, t, t) and
0<t<1.
6. Set up but do NOT evaluate Evaluate the line integral .
curve ř (t) = (t", t"), where 0 <t <0.
ds, where C is the

Expert Solution

Step 1

Consider the given integral as $\int_{C} \frac{y}{x} d s$ , where C is the curve $\overset{⇀}{r} (t) = (t^{4}, t^{3})$ and $0 \leq t \leq 1$ .

Note that, the line integral of f(x,y) over the curve C is given by $\int_{C} f (x, y) d s = \int_{a}^{b} f (x (t), y (t)) ||\vec{r (t)}|| d t$ , where $a \leq t \leq b$ .

Here, given that $f (x, y) = \frac{y}{x}$ and $x (t) = t^{4} a n d y (t) = t^{3}$ .

Now find the derivative of

$||\vec{r (t)}|| = \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} = \sqrt{{(4 t^{3})}^{2} + {(3 t^{2})}^{2}} = \sqrt{16 t^{6} + 9 t^{4}} = t \sqrt{16 t^{2} + 9}$

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