# Test review for calculus, related rates & critical numbers

I got 12/60 on my latest test I want to go over it to figure out what I did wrong and why and how to improve. I was looking if there is anything specific, any pattern to this. When I review my tests I just see all sorts of problems, I forget formulae, I forget basic math, I mess up signs, I run out of time and can't finish the test. Things like that, I don't have an answer key for this test is there any way I can go over it on here and someone else can tell me why I am messing up so much?

There 10 questions and I could probably type them up in under 30 minutes, and another 30 for my work.

Anyways here is the first one. I am supposed to find the derivative of $\log 3 (x+ \sqrt {(x^2-1})$. I then set the problem up as $1/u \ln 3 (1+1/2(x^2-1)^{-1/2}2x)$; this is wrong, completely wrong.

Next questions is let $f(x) = (\cos x)^x$ and find the derivative. I couldn't even do this on the test, I forgot how to get the derivative of a variable that is an exponent, I knew what to do I just couldn't make sense of how to do and I have scrawled down about 30 attempts at nothing.

Doing this now I think what I need to do is rewrite the problem as $e^{\ln u^x}$ and then solve it from there. I could be wrong though.

3) I am supposed to solve this word problem.

A particle moves along a real axis and its location, $s$, is given by $s=t^3-12t^2+36t$ ($s$=feet)
a) find the acceleration at $t=2$ seconds

For this I know that I need to get the derivative and put in $2$ for $t$. Pretty simple, I makes the problem $3t^2-24t+36=0$; I think I forgot to do this one, not sure what happened but this is all I have written down, not sure why. I would plug in $2$ and get $0$, oh I guess I did have the answer written down. It was probably just wrong.

b) When is the particle moving in the positive direction?

I know for this I need to find the critical numbers and test points around them to see where they are increasing or decreasing. I have positive from $5$ to $\infty$.

c) What is the total distance traveled in the first $6$ seconds?

I have written down $39$ ft.

I am not sure what the rules on this are so I will not post anymore questions under this topic unless I know it is okay to. I would like to transcribe the whole test if that is possible.

• It would be much much much better for you to find some person at your institution to talk to; at the very least, hire a tutor to go over the test with you. – Gerry Myerson Oct 20 '11 at 23:41
• I can't hire a tutor, they are too much money. I can only use the free ones available at my school, and not until a week from tomorrow. – toby yeats Oct 20 '11 at 23:44
• How much money is it going to cost you to fail calculus and have to take it again? How much money is it going to cost you to fail out of college? Anyway, talk to someone in the Math Department, and/or someone in Guidance/Counseling/Careers/whatever they call it at your institution. – Gerry Myerson Oct 21 '11 at 0:15
• None of them are really helpful, none of them really seem to care as long as I pay for classes. Anyways I already failed this class so I could probably save money by just failing compared to failing and paying for a tutor. – toby yeats Oct 21 '11 at 0:19
• Then go to some other college, somewhere where they care about how students do. Meanwhile, enrol in something easier than Calculus; if your posts on m.se are anything to go by, you aren't ready for Calculus. – Gerry Myerson Oct 21 '11 at 5:25

When you have something of the form $f(x) = x^x$, you need to do something like this:

$$f(x) = x^x$$ $$\ln f(x) = \ln(x^x)$$ $$\ln f(x) = x \ln x$$

Then differentiate both sides:

$$\frac{f'(x)}{f(x)} = \ln x + 1$$ $$f'(x) = x^x(\ln x + 1)$$

It should be simple to figure out what to do when $f(x) = (\cos x)^x$. Just use the chain rule in combination with the above.

Regarding problem 3a), as it was said in another answer, the acceleration is the second derivative of the position. In this case, $a = 6t - 24$.

I agree with the commenters. I think you should get a tutor. If tutors are too expensive, maybe ask a friend for help. It sounds like you need to go over some of the basics, and just getting the answer to your problems here isn't going to be of much help.

• I have used tutors they just don't seem to really help, it is the same as this. They just help me work out problems when I can't (I never can) do it by myself. I don't have any friends either, definitely not any taking math. Just saw that you are 18...pretty embarassing that I am so stupid that I am getting lectured on how to do high school math from someone nearly 10 years younger than me. – toby yeats Oct 21 '11 at 1:54
• @Javier, Isn't using ${dy\over dx}$ when doing logarithmic differentiation a lot easier? – E.O. Dec 20 '11 at 7:23
• @Emile: What do you mean? – Javier Dec 20 '11 at 14:16
• @JavierBadia Nevermind. Its just a personal preference of mine using $dy\over dx$ when doing calculus as it more intuitive, especially when doing implicit (e.g. logarithmic) differentiation. – E.O. Dec 21 '11 at 0:41

For the 3a, the acceleration is the second derivative of the position, so you need to take one more derivative before plugging in $t=2$. It is not equal to $0$. I get $-12$ as the acceleration at $t=2$. For 3b, your derivative is the velocity, so you can find its roots. I find it positive over $(-\infty,2)\cup(6,\infty)$. For 3c, the particle moves right until $t=2$, reaching $s=32$, then returns at $t=6$ to $s=0$, so it travels a total of $64$.

• I did these problems about 6 times checking to make sure I was correct. What is wrong with me that I can't get it right? – toby yeats Oct 21 '11 at 0:22
• I don't know, but it sounds like you might have done them 6 times the same way. If the difficulty is figuring out what math to do, checking the math you did will not be productive. For 3a, it doesn't appear that you got the right relationship between position and acceleration. – Ross Millikan Oct 21 '11 at 0:29
• So the rate of change is the velocity and the rate of change of the rate of change is the acceleration right? I just got the terminology messed up. – toby yeats Oct 21 '11 at 0:40
• @Jordan: That's right. By definition, acceleration is the rate of change (i.e., derivative) of velocity. – Javier Oct 21 '11 at 1:51

For the first question: \begin{align*} f(x) &= \log_3(x + \sqrt{x^2 - 1})\\ f'(x) &= \frac{1}{\ln3}\frac{1}{x + \sqrt{x^2 - 1}}\left(1 + \frac{1}{2}\frac{1}{\sqrt{x^2 - 1}}2x\right)\\ f'(x) &= \frac{1}{\ln3}\left(\frac{1}{x + \sqrt{x^2 - 1}} + \frac{x}{(x + \sqrt{x^2 -1})\sqrt{x^2 -1}}\right)\\ f'(x) &=\frac{1}{\ln3}\left(\frac{x + \sqrt{x^2 -1}}{(x + \sqrt{x^2 -1})\sqrt{x^2 -1}}\right)\\ f'(x) &=\frac{1}{\ln3}\frac{1}{\sqrt{x^2 -1}} \end{align*}

• You're assuming that when OP writes $\log3({\rm whatever})$, OP means $\log_3({\rm whatever})$. You may be right. – Gerry Myerson Oct 21 '11 at 0:16
• @GerryMyerson That's my best guess. If he says it's $log3(...)$ I'll redo it. – Alexandar Živkovič Oct 21 '11 at 0:33