You to the two dedicated people who have been me this evening. here is my final problem for the night, and here is what i'm thinking. for a, i think i need to take the derivative of the velocity function to give me the acceleration function. i think the answer is . hopefully this first answer is correct. i get confused at b. i understand that this question is looking for a minimum, so i need to think about critical points and extrema. do you have to take the derivative of the acceleration function or just use the acceleration function as found in a? i think the former. c goes back to velocity. do you use the derivative of the velocity function to solve this (which is actually the acceleration function in a)?

Part A

Yes you find the derivative of v(t) to get the acceleration a(t). Since, v'(t) = a(t)

You are close but not quite there. 
The portion -pi*cos(pi*t) is correct, and so is the -5 at the end. The part you have incorrect is the 2pi. It should be positive instead of negative.

-pi*cos(pi*t)+2pi-5


Part B

a(t) = -pi*cos(pi*t)+2pi-5
a'(t) = -pi*pi*(-sin(pi*t))+0 ... derivative of cosine is negative sine; use chain rule
a'(t) = (pi)^2*sin(pi*t)

Find the solutions to a ' (t) = 0 on the interval [0,3]
The solutions are t = 0, 1, 2, 3

You'll find that
The local mins are at t = 0 and t = 2
The local maxes are at t = 1 and t = 3
When you use the first derivative test (I'm skipping the work to show this as it is way too lengthy)

Only focus on the local mins t = 0 and t = 2
We'll focus on the endpoints t = 0 and t = 3
It's perfectly possible to have a critical point be an endpoint as long as the function is defined on both sides of the critical point.

So we will plug t = 0, t = 2, t = 3 into the a(t) function to see which is smallest.

Plug in t = 0
a(t) = -pi*cos(pi*t)+2pi-5
a(0) = -pi*cos(pi*0)+2pi-5
a(0) = -1.8584 (approximate)

Plug in t = 2
a(t) = -pi*cos(pi*t)+2pi-5
a(2) = -pi*cos(pi*2)+2pi-5
a(2) = -1.8584 (approximate)

Plug in t = 3
a(t) = -pi*cos(pi*t)+2pi-5
a(3) = -pi*cos(pi*3)+2pi-5
a(3) = 4.4248 (approximate)

The output -1.8584 is smallest.

-1.8584


Part C

v(t) = (2pi-5)*t - sin(pi*t)
v ' (t) = a(t) = -pi*cos(pi*t)+2pi-5

Solve a(t) = 0 for t on the interval [0,2]
The solutions are these approximate values
t = 0.3660 and t = 1.6339

You'll find that a local min is at t = 0.3660 while a local max is at t = 1.6339

Focus on the local max. Ignore the local min.
We'll also focus on the endpoints as well

So we'll plug t = 0, t = 1.6339, and t = 2 into the v(t) function to see which is the largest.

Plug in t = 0
v(t) = (2pi-5)*t - sin(pi*t)
v(0) = (2pi-5)*(0) - sin(pi*(0))
v(0) = 0

Plug in t = 1.6339
v(t) = (2pi-5)*t - sin(pi*t)
v(1.6339) = (2pi-5)*(1.6339) - sin(pi*(1.6339))
v(1.6339) = 3.0094 (approximate)

Plug in t = 2
v(t) = (2pi-5)*t - sin(pi*t)
v(2) = (2pi-5)*(2) - sin(pi*(2))
v(2) = 2.5664 (approximate)

The output 3.0094 is the largest, so this is the absolute max (largest velocity) on the interval [0,2]

3.0094

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12