Question

Find the length of the arc of the curve from point P to point Q.

y=1/2(x^2)
P(-3,9/2)
Q(3,9/2)

I'm just confused as to how to find the endpoints. Would it simply be 0 to 6 or would it be from -3 to 3?

Answer 1

well your function is in term of x
and your x's go from x=-3 to x=3

Answer 2

\[\int\limits_{-3}^{3}\sqrt{1+(y')^2} dx \]

Answer 3

I will show you a trick to finish this without tricky calculus
once you think you got the integral

Answer 4

or i mean the correct integrand

Answer 5

nope wait trick calculus might be needed
sorry

Answer 6

Haha! So would I just put everything in terms of y and integrate it from -3 to 3?

Answer 7

you would differentiate y=1/2(x^2) w.r.t x

Answer 8

And then square it. I misunderstood your previous equation!

Answer 9

\[\int\limits_{-3}^{3}\sqrt{1+x^2} dx \]

Answer 10

yep

Answer 11

and here I would do a trig sub
tan(theta)=x

Answer 12

dx=sec^2(theta) d theta

Answer 13

we also need to write the x limits in terms of theta limits

Answer 14

So, I got to the point where it's sec^3(theta).

Answer 15

\[\int\limits_{-3}^{3}\sqrt{1+x^2} dx=2 \int\limits_{0}^{3}\sqrt{1+x^2} dx \text{ since } \\ g(x)=\sqrt{1+x^2} \text{ is an even function } \\ \text{ This will make life easier \in the future for this one } \\ x=\tan(\theta) \\ dx=\sec^2(\theta) d \theta \\ 2 \int\limits_{0}^{\arctan(3)} \sqrt{1+\tan^2(\theta)} \sec^2(\theta) d \theta \\ \text{ Since } \sec(\theta) >0 \text{ on } [0,\arctan(3)] \text{ we have } \\ 2\int\limits_{0}^{\arctan(3)}\sec^3(\theta) d \theta \text{ now we do integration by parts } \]

Answer 16

\[\int\limits_{}^{}\sec^3(\theta) d \theta= \int\limits_{}^{}\sec^2(\theta ) \sec(\theta) d \theta =\tan(\theta) \sec(\theta)-\int\limits_{}^{}\tan(\theta) \sec(\theta) \tan(\theta) d \theta \]
That should help you proceed
(you will end up getting 2 int (sec^3(theta), d theta) on the left hand side

Answer 17

Thank you so much once again!