Find length of curve x = 3cos(t), y = 3sin(t), 0 <= t <= pi

[Shutterbug424]

I have some problems that are due tonight.

Find the length of the curve:

x = 3cos t, y= 3 sin t, o < t < pi

 

[galactus]

Re: Need Calculus help before it's too late!

Use the formula for parametric arc length.

\(\displaystyle \int_{0}^{\pi}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt\)

Find the respective derivatives of x and y, square them, then take the square root, integrate over the region specified.

Usually these things are made so they reduce quite nicely. This one sure does. Remember the identity, \(\displaystyle sin^{2}(t)+cos^{2}(t)=1\).

It will come in very handy.

Let us know what you get.

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

We are supposed to use the equation:

 

[galactus]

Re: Need Calculus help before it's too late!

OK, convert to rectangular form. What you have is a circle with radius 3. What is the equation of such a circle?.

Can you find it?. I have included a graph. It looks like an ellipse on the graph, but it is actually a circle.

The parametric is much easier and it is an arc length formula. Just another form. Use it.

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

Ok, this is an entirely new concept for me. I'm working on it. But, maybe you can reinforce my answers I chose for the following:

 

[galactus]

Re: Need Calculus help before it's too late!

For #9, the derivative of cot(x) is \(\displaystyle -csc^{2}(x)\). Which do you think the answer is?.

For #10, the derivative of tan(x) is \(\displaystyle sec^{2}(x)\). Go from there.

 

[Deleted member 4993]

Re: Need Calculus help before it's too late!

Ok, this is an entirely new concept for me. I'm working on it. But, maybe you can reinforce my answers I chose for the following:

What answers did you choose?

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

For 9 I chose D and for 10 I chose B.

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

This was the one I meant to post.

 

[Deleted member 4993]

Re: Need Calculus help before it's too late!

For 9 I chose D and for 10 I chose B.

Looks good to me.

 

[Deleted member 4993]

Re: Need Calculus help before it's too late!

This was the one I meant to post.

What are your thoughts?

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

I solved 16 in terms of x to be

2 pi times the integral from a to b of x*sqrt of (1+9x^(-4))

The answers are in terms of y. I can't seem to change 9x^(-4) to terms of y.

 

[Shutterbug424]

Re: Need Calculus help before it's too late!

For 17, I can set it up. But, I seem to get lost in the algebra.
I'm thinking there must be a trig identity that could help.

 

[KenR]

My answer

My answer is 6pi, is that correct?