I need help solving these CALCULUS QUESTIONS

So I'm really just stuck on solving any of these. Any help would be great!

1. A particle moves along the x-axis so that a time t>0 its position is given by x(t) = ln(l+t^2). At the instant when the acceleration becomes zero, what is the value of the velocity?

2. An ordinary pop can has a volume of 355 cm^3. Find the dimensions(radius r and height h) that minimize the total surface area(including top and bottom) of the can.

3. Suppose f(x)=x^5-3x^2+4. Find all numbers c that satisfy the conclusion of the Mean Value Theorem on the closed interval [1,3]. Approximate your answers to three decimal places.

4. A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank? Justify your answer.

1.

x(t) = ln(1 + t^2)
v(t) = 2t / (1 + t^2)
a(t) = (2 - 2t^2)/(1 + t^2)^2

Set the acceleration equal to 0.

2 - 2t^2 = 0
2 = 2t^2
1 = t^2
+/- 1 = t

We only care for positive values in this case. Disregard the negative value.

v(1) = 2(1) / (1 + 1^2) = 2/2 = 1.


2.

V = π * r^2 * h
V = 355
355 = π * r^2 * h
355/πr^2 = h

SA = 2πr^2 + 2πrh
SA = 2πr^2 + 2πr(355/πr^2)
SA = 2πr^2 + 710/r

Differentiating yields:

SA ' = 4πr - 710/r^2
SA ' = (4πr^3 - 710) / r^2

Setting the numerator equal to 0, gives:

4πr^3 - 710 = 0
4πr^3 = 710
r^3 = 710/4π
r = (710/4π)^(1/3)
r = 18.84

355 = π * 18.84^2 * h
355/π18.84^2 = h
.3183 = h


3.

f(x) = x^5 - 3x^2 + 4

(f(3) - f(1)) / (3 - 1) = f '(c)
(220 - 2) / 2 = f '(c)
218/2 = 109 = f '(c)

f '(c) = 5c^4 - 6c
109 = 5c^4 - 6c
109 = c(5c^3 - 6)