Can you please check my work?

Question

jim_thompson5910 · Answer

The good news is that you are on the right track and that you have the answer in that list of 4 points.

However, the bad news is that the other points you wrote down are incorrect.

anonymous · Answer

it looks good

anonymous · Answer

Oh okay! So I think I plugged it in wrong. Would I plug in each theta value into the x and y equations to get sets of points?

jim_thompson5910 · Answer

There is only one point that has a vertical tangent line.

There are no points that have a horizontal tangent line.

jim_thompson5910 · Answer

you plug in the values of theta into the original x,y equations, yes

anonymous · Answer

so would I only get (4,0)? I'm a bit lost.

jim_thompson5910 · Answer

yep (4,0) is the only answer

jim_thompson5910 · Answer

this is where the vertical tangent is

jim_thompson5910 · Answer

lost how?

anonymous · Answer

Ohhh I guess I was expecting more points. Thank you so much!! If you can, do you think you could take a look at a few other problems and check them as well?

jim_thompson5910 · Answer

sure

jim_thompson5910 · Answer

ok you did great until you got to the substitution part

when you went from t to theta

anonymous · Answer

the trig sub part? when t equaled sin(theta)?

jim_thompson5910 · Answer

you have 2 options to fix this

either

a) change the limits of integration on any integral with dtheta in it

or

b) keep the limits the same but remember to substitute back in for t

jim_thompson5910 · Answer

yes

jim_thompson5910 · Answer

the original integral goes from t = 0 to t = 1/2

jim_thompson5910 · Answer

when you integrate sec(theta), what did you get?

anonymous · Answer

ln|sec(theta)|

jim_thompson5910 · Answer

t = sin(theta)

so theta = arcsin(t)

jim_thompson5910 · Answer

ln| sec(theta) | = ln| sec(arcsin(t)) |

anonymous · Answer

oh I see. would I have to do the same for other parts of my answer?

jim_thompson5910 · Answer

wait, ln|sec(theta)| isn't the integral of sec(theta)

try again

anonymous · Answer

oh is it ln|sec(theta) + tan(theta)|?
(plus C, but that's not necessary for this problem, is it?)

jim_thompson5910 · Answer

the Cs will cancel when you subtract, so either way it really doesn't matter

to be really technical, yes you should have them but I don't mind

jim_thompson5910 · Answer

yes it is ln|sec(theta) + tan(theta)|

each time you see a theta, replace it with arcsin(t)

anonymous · Answer

Ohh I understand now

anonymous · Answer

and the rest is correct?

jim_thompson5910 · Answer

Yes it looks good. Tell me the updated answer you get

anonymous · Answer

ln|sec(1/2) + tan(1/2)| - ln|sec(arcsin(t)) + tan(arcsin(t))|

jim_thompson5910 · Answer

you should get this
|dw:1427782108044:dw|