@jim_thompson5910

http://cyh.leeschools.net/UserContent/Documents/AP%20CalcBC%20SumAssign%2014-15.pdf

Question

@jim_thompson5910 

http://cyh.leeschools.net/UserContent/Documents/AP%20CalcBC%20SumAssign%2014-15.pdf

marissalovescats · Answer

Okay so I'm pretty sure 3c is doing the integral 0 to 3 - the integral 3 to 4 right?

akashdeepdeb · Answer

Which one do you need help with? 

Glad to see someone giving AP Calc BC here! ^^

oleg3321 · Answer

does anyone know if ganeshie is online right now?

oleg3321 · Answer

can someone please answer  my questoin? is ganeshie online. please answer

akashdeepdeb · Answer

Are you talking about Part 2? Part 2 - Question 3 Okay, so the particle moves from t = 0 to t = 16. Its velocity at any point is given by $v(t) = e^{2 \sin t} - 1$ At time t = 0, it is at the origin. a) You have to sketch the graph. Input it into the calculator, easy work. b) The particle moves to the left, whenever the velocity is negative. So after plotting the graph check at what intervals of (t) the particle reverses its motion. c) You can use the calculator here too. We know that, $$v = \frac{dx}{dt}$$ $$v.dt = dx$$ $$\int v.dt = \int dx$$ $$x = \int v(t) ~.~dt$$ Now, you are given v as a function of time so, find out it's integral with variable 't'. And because they have mentioned the time intervals (0-4), obviously you have to use those limits for you integral. Hope this helps. :) Here is a video, that helped me get a 5 on Calc BC. Check it out @ http://www.youtube.com/watch?v=Ct_IMcvNyIM

oleg3321 · Answer

can someone please answer my questoin? is ganeshie online. please answer

oleg3321 · Answer

ok thanks for answering. you get a medal for that.

akashdeepdeb · Answer

Exploit your calculator.

No, the integral is not this:

$$\int_{0}^{3} e^{2 \sin t} - 1~.~dt - \int_{3}^{4} e^{2 \sin t} - 1~.~dt$$
Because, '3' is not where the graph becomes negative! It is at $\pi$ when the graph is zero and makes its transition to the negative y axis.

Check it on your calculator.

This should be the integral then:

$$\int_{0}^{\pi} e^{2 \sin t} - 1~.~dt - \int_{\pi}^{4} e^{2 \sin t} - 1~.~dt$$

Getting this? :)

nincompoop · Answer

someone has been going through the problems one by one with her

anonymous · Answer

yeah ganeshie is on

marissalovescats · Answer

Thanks @akashdeepdeb for answering anyways. We worked on a different place because last night when I needed help OS like wouldn't load. 

And yes BC :) Along with 3 other people in my whole school.