# Homework Help: Line integral problem

1. Apr 16, 2014

### BiGyElLoWhAt

ok, my turn to ask a question.

Problem: evaluate $∫_{C}xyds$ for $x=t^2 and y = 2t from 0\leq t \leq 5$

not sure what I did wrong, but here it goes:

solve for ds:
$ds =\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} = \sqrt{4t^2+4}=2\sqrt{t^2+1}$

substitute:
$∫_{0}^5 4t^3\sqrt{t^2+1}dt$

substitute again:
$t=tan(\theta) → dt = sec(\theta)d\theta → ∫_{L}4tan^3(\theta)sec^2(\theta)d\theta$

$4∫_{L}tan^3(\theta)(tan^2(\theta)+1)d\theta =4 ∫_{0}^5t^3(t^2+1) =4∫_{0}^5t^5 +t^3 =4\frac{5^6}{6} + 5^4$
...
But it's wrong

2. Apr 16, 2014

### jbunniii

Surely the fact that you transformed $\int_{0}^{5}4t^3 \sqrt{t^2 + 1} dt$ into $\int_{0}^{5}4t^3 (t^2 + 1) dt$ is a red flag. Where did the square root go?

Double check your derivative of $\tan(\theta)$.

3. Apr 16, 2014

### SammyS

Staff Emeritus
What is the derivative of tan(Θ) ?

4. Apr 16, 2014

### BiGyElLoWhAt

ok, its sec*tan, stupid mistake, but i'm still not getting my root back. substituting my tan in under the root gimves me sqrt(sec^2)=sec, so i now have 1 exrtra power of tan, but dt is sec*tan, and that gives me 2 secants, one from the root, 1 from my substituted differential.... I do agree with you, I always remember the root coming back in calc 2... I'll check it again when i get off work

5. Apr 16, 2014

### BiGyElLoWhAt

Alright I got it, I was making the wrong substitution. I tried u=t^2 +1 and that got me the right answer. I just saw the sqrt(t^2 +1) and assumed a trig substitution, but once I actually wrote it down and looked at it, it was a useless substitution, as I either had all secants or all tangents, and no good differential to sub in.

Thanks for catching my d/dtheta (tan(theta)) error guys =]

6. Apr 16, 2014

### jbunniii

The derivative of $\tan(\theta)$ is not $\sec(\theta) \tan(\theta)$.

If you are not sure, you can always calculate it by applying the quotient rule to $\tan(\theta) = \sin(\theta)/\cos(\theta)$.

7. Apr 16, 2014