# Math Help - NEED HELP! AP Exam tomorrow! Variety of topics

1. ## NEED HELP! AP Exam tomorrow! Variety of topics

Since I don't feel like forcing you guys to analyze my unique typing methods, I photoshopped my questions to make them easier to analyze. Thanks a million!

Problem: Let the velocity at a time t of a point moving on a line be defined by V(t). How many centimeters did the point travel from t = 0s to t = 2.5s.

Answer: 9.882, but how? I tried integrating and got this

Problem 2:

Find the value of b such that:

Answer: Don't have it with me, but all of the answers are written in terms of a, for example, here is a (possibly) incorrect answer of "a squared over quantity a squared minus one". Oh, and I have NO idea how to do this one. What does D sub x even mean?

More problems to come, Thanks all! And no, I don't expect everyone to photoshop me an answer

2. your solution is correct for the given velocity function.

3. Originally Posted by skeeter
your solution is correct for the given velocity function.
So could my answer key be wrong or am I misinterpreting the problem?

4. Originally Posted by Thikr
So could my answer key be wrong or am I misinterpreting the problem?

5. Originally Posted by Thikr
So could my answer key be wrong or am I misinterpreting the problem?
have you posted the complete problem ?

6. Originally Posted by skeeter
have you posted the complete problem ?
Word for word.

7. $D_x$ means the derivative w/r to x.

$\frac{d}{dx} \left(\frac{e^{bx}}{e^{ax}}\right) =$

$\frac{e^{ax} \cdot be^{bx} - e^{bx} \cdot ae^{ax}}{e^{2ax}}=$

$\frac{e^{ax+bx}(b - a)}{e^{2ax}} = e^{bx-ax}(b-a)$

$\frac{f'(x)}{g'(x)} = \frac{be^{bx}}{ae^{ax}} = \frac{b}{a}e^{bx-ax}$

$b-a = \frac{b}{a}$

$ab-a^2 = b
$

$ab-b = a^2$

$b(a-1) = a^2$

$b = \frac{a^2}{a-1}$

btw, you will not see a question like this on the AP exam ... too much time-consuming algebraic manipulation.

8. Originally Posted by skeeter
$D_x$ means the derivative w/r to x.

$\frac{d}{dx} \left(\frac{e^{bx}}{e^{ax}}\right) =$

$\frac{e^{ax} \cdot be^{bx} - e^{bx} \cdot ae^{ax}}{e^{2ax}}=$

$\frac{e^{ax+bx}(b - a)}{e^{2ax}} = e^{bx-ax}(b-a)$

$\frac{f'(x)}{g'(x)} = \frac{be^{bx}}{ae^{ax}} = \frac{b}{a}e^{bx-ax}$

$b-a = \frac{b}{a}$

$ab-a^2 = b
$

$ab-b = a^2$

$b(a-1) = a^2$

$b = \frac{a^2}{a-1}$

btw, you will not see a question like this on the AP exam ... too much time-consuming algebraic manipulation.
Damn you properties of exponentials... that makes total sense, thanks so much man, more coming later if you are up for a late night tutoring session. What program/site do you use to post your work?