# Thread: Hey, trouble on worksheet! help plz!

1. ## Hey, trouble on worksheet! help plz!

Hi, this math sheet has been giving me a lot of problems lately. I was wondering if someone could look over it and help me out. I have done most of it, however the questions i have on the attached document i cant get a solid answer. Please give me a hand! THANK YOU VERY MUCH!

2. Hmm, the picture is very small.

3. Originally Posted by Stuck686
Hi, this math sheet has been giving me a lot of problems lately. I was wondering if someone could look over it and help me out. I have done most of it, however the questions i have on the attached document i cant get a solid answer. Please give me a hand! THANK YOU VERY MUCH!
i can't see the questions, i doubt anyone can. what format did you paste it in. i find that .gif images tend to be posted larger than .bmp images for instance

4. Originally Posted by Stuck686
Hi, this math sheet has been giving me a lot of problems lately. I was wondering if someone could look over it and help me out. I have done most of it, however the questions i have on the attached document i cant get a solid answer. Please give me a hand! THANK YOU VERY MUCH!
No one will be able to read it, it has been reduced to such a size that
of ~640 to 800 pixels.

RonL

5. Ohh thanks for the advice on the sizing. looks a lot better now, thanks.

6. Originally Posted by Stuck686
Hi, this math sheet has been giving me a lot of problems lately. I was wondering if someone could look over it and help me out. I have done most of it, however the questions i have on the attached document i cant get a solid answer. Please give me a hand! THANK YOU VERY MUCH!
Here are hints on how to do the questions:

1) (A) use the product rule:

Product Rule: (ab)' = a'b + ab'

use a = tan^-1(2x) and b = (cos(3x))^(1/2)

you know how to find the derivatives of each of those correct? you will need the chain rule for each

Chain rule: [f(g(x))]' = f'(g(x)) * g'(x)

1) (B) Use the chain rule on each. if you still have problems, say so

1) (C)
let sin^-1(2x) = theta
=> 2x = sin(theta)

draw a right-triangle with one acute angle called theta, and the side oppiste to that angle call 2x and the hypotenuse call 1, use pythagorean's theorem to find the adjacent side. then use the fatc that tan(sin^-1(2x)) = tan(theta) = opposite/adjacent. so you can change t to an algebrac function, then find the derivative normally.

2) find y' for y = xcosx, and plug in x = 2^R in the result. that will give you your m. to find b, plug in 2^R for x in the original equation and then use the coordinates (x,y) = (2^R, y(2^R)) in the point-slope form

3) the area of a trapezium is half the sum of the two parallel sides times the distance between them. write out this formula and use substitutions to get it into one variable. find the derivative and set it equal to zero, this will give you one coordinate for the max value.

6) (A) use the substitution u = 1 - cos(2x)

(B) use substitution u = tanx

(C) use the fact that sin(2x) = 2sinxcosx

so (sin(2x))/cosx = (2sinxcosx)/cosx = 2sinx

post your solutions when you're done so we can examine them. if you still have problems just say so

7. can someone like Topsquark do the answers so i can check my work? thanks

8. Hello,

to #3: As Jhevon pointed out already you have to find the equation of the trapezoid's area. I've attached a screenshot of my calculations. (Derivation is not necessarily needed)

9. Originally Posted by earboth
Hello,

to #3: As Jhevon pointed out already you have to find the equation of the trapezoid's area. I've attached a screenshot of my calculations. (Derivation is not necessarily needed)
finding the derivative isn't necessary of course, but i figured he might want to do it since this is obviously for a calculus course above precalc

10. can someone do them?