1. A baby is being bounced approximately according to the position function 10*sin(2*pi*t/2) (distance in centimeters, t in seconds). Compute the acceleration function and report the maximum acceleration experienced.






2. What x value minimizes x/5 + 3/x , when x>0 ? Express it precisely, not just its decimal version (so, just asking your calculator to minimize it won’t be sufficient)

#1. If y = a sin(bx), then

y' = ab cos(bx)
y" = -ab^2 sin(bx)
and what's with the 2*pi/2 ? Why not just say pi?
and, of course, max acceleration is where -ab^3 cos(bx) = 0
That will be at the max or min of the graph for y(x), since that is where the curvature is greatest.

#2. y = x/5 + 3/x
y' = 1/5 - 3/x^2 = (x^2-15)/(5x^2)
so, where is that zero?
Be sure to check that it's the minimum, not the maximum.
check your work at some handy online graphing site.

so every even t is a full period?

1. To compute the acceleration function, we need to differentiate the position function twice. Let's break down the process step by step:

Step 1: Differentiate the position function to find the velocity function.
The derivative of the position function 10*sin(2*pi*t/2) with respect to time (t) gives us the velocity function.

To differentiate 10*sin(2*pi*t/2), we apply the chain rule:
- The derivative of sin(u) with respect to u is cos(u).
- The derivative of u = 2*pi*t/2 with respect to t is (d/dt) 2*pi*t/2 = pi.

Therefore, the velocity function v(t) is obtained by multiplying the derivative of the position function by the derivative of the inside function:
v(t) = 10*cos(2*pi*t/2) * pi = 10*pi*cos(pi*t).

Step 2: Differentiate the velocity function to find the acceleration function.
Similarly, we differentiate the velocity function to find the acceleration function.

To differentiate 10*pi*cos(pi*t), we apply the chain rule again:
- The derivative of cos(u) with respect to u is -sin(u).

The derivative of u = pi*t with respect to t is (d/dt) pi*t = pi.

Therefore, the acceleration function a(t) is obtained by multiplying the derivative of the velocity function by the derivative of the inside function:
a(t) = -10*pi^2*sin(pi*t).

To find the maximum acceleration experienced, we will determine the critical points of the acceleration function.

Step 3: Find the critical points of the acceleration function.
To find the critical points, we set the acceleration function equal to zero and solve for t:

-10*pi^2*sin(pi*t) = 0.

This equation is satisfied when sin(pi*t) = 0. Sin(pi*t) equals 0 at t = 0, t = 1, t = 2, t = 3, and so on.

Step 4: Evaluate the second derivative test.
We need to analyze the sign changes in the acceleration function to determine the maximum acceleration experienced.

The second derivative of the acceleration function is:

(d^2/dt^2) a(t) = -10*pi^3*cos(pi*t).

Evaluating the second derivative at the critical points:

(d^2/dt^2) a(0) = -10*pi^3*cos(pi*0) = -10*pi^3*cos(0) = -10*pi^3.

(d^2/dt^2) a(1) = -10*pi^3*cos(pi*1) = -10*pi^3*cos(pi) = 10*pi^3.

(d^2/dt^2) a(2) = -10*pi^3*cos(pi*2) = -10*pi^3*cos(2*pi) = -10*pi^3.

We observe that (d^2/dt^2) a(t) changes sign at t = 0 and t = 2. These are the intervals where the maximum acceleration could be located.

Step 5: Determine the maximum acceleration.
To determine if the maximum acceleration is positive or negative, we evaluate the acceleration function at the end points of the intervals (t = 0, t = 1) and (t = 1, t = 2):

a(0) = -10*pi^2*sin(pi*0) = -10*pi^2*sin(0) = 0.

a(1) = -10*pi^2*sin(pi*1) = -10*pi^2*sin(pi) = 0.

a(2) = -10*pi^2*sin(pi*2) = -10*pi^2*sin(2*pi) = 0.

Since the acceleration is zero at both t = 0 and t = 2, there is no maximum acceleration experienced.

Therefore, the maximum acceleration experienced in this case is 0 cm/s^2.

2. To find the value of x that minimizes the expression (x/5) + (3/x), we need to find the critical points where the derivative is equal to zero.

Step 1: Differentiate the expression (x/5) + (3/x) with respect to x.
To differentiate the expression (x/5) + (3/x), we apply the quotient rule:
- The derivative of x/5 is 1/5.
- The derivative of 3/x is -3/x^2.

Therefore, the derivative of the expression is obtained by subtracting the derivative of the denominator from the derivative of the numerator:
f'(x) = (1/5) - (3/x^2) = 1/5 - 3/x^2.

Step 2: Set the derivative equal to zero and solve for x:
1/5 - 3/x^2 = 0.

Multiply through by x^2 to eliminate the fraction:
x^2/5 - 3 = 0.

Multiply through by 5 to remove the fraction:
x^2 - 15 = 0.

Step 3: Solve the quadratic equation for x:
x^2 - 15 = 0.

Rearrange the equation:
x^2 = 15.

Take the square root of both sides:
x = ±√15.

Since the question specifies that x > 0, we need to take the positive square root:
x = √15.

Therefore, the precise value of x that minimizes the expression (x/5) + (3/x) when x > 0 is √15.