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2014 AP CALCULUS AB FRQs. Average rate of change = lbs/day. On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs /day. Average amount of clippings: Solve days. Tangent line at is Solve days. y = 4. f (x).

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2014 ap calculus ab frqs

On the fifteenth day the amount of grass clippings is decreasing at the rate of 0.1635 lbs/day


2014 ap calculus ab frqs

Average amount of clippings: decreasing at the rate of 0.1635

Solvedays


2014 ap calculus ab frqs

Tangent line at decreasing at the rate of 0.1635 is

Solve days


2014 ap calculus ab frqs

y decreasing at the rate of 0.1635 = 4

f(x)

Intersections of the two graphs: and


2014 ap calculus ab frqs

y decreasing at the rate of 0.1635 = 4

f(x)

Length of the rectangle:

y – f(x)


2014 ap calculus ab frqs

y decreasing at the rate of 0.1635 = 4

f(x)


2014 ap calculus ab frqs

y decreasing at the rate of 0.1635 = 4

f(x)


2014 ap calculus ab frqs

y decreasing at the rate of 0.1635 = 4

f(x)


2014 ap calculus ab frqs

, therefore if decreasing at the rate of 0.1635 g(x) increases . This happens on the intervals and (−3, 2)

If g(x) i decreases. This happens on and (0, 4)

g(x)is increasing and concave down on and (0, 2)


2014 ap calculus ab frqs

The slope of the graph of decreasing at the rate of 0.1635 f at is −2, so


2014 ap calculus ab frqs

Average acceleration of train A: decreasing at the rate of 0.1635

S meters/minute2


2014 ap calculus ab frqs

Since decreasing at the rate of 0.1635 is differentiable, it is also continuous so the Intermediate Value Theorem applies to for . Therefore, there must be at least a value such that meters/minute


2014 ap calculus ab frqs

meters decreasing at the rate of 0.1635


2014 ap calculus ab frqs

Train B decreasing at the rate of 0.1635

meters

meters

meters

z = distance between the two trains

y

Train A

x

At minutes,


2014 ap calculus ab frqs

f decreasing at the rate of 0.1635 has a local minimum when its derivative, f’, switches from negative to positive. This occurs at x = 1.


2014 ap calculus ab frqs

f decreasing at the rate of 0.1635 is twice differentiable so its derivative, f’, is both continuous and differentiable. Therefore the Mean Value Theorem can be used on f’ .

There must be a value for such that

Since


2014 ap calculus ab frqs

The problem can also be done using Rolle’s Theorem: decreasing at the rate of 0.1635 f is twice differentiable so its derivative, f’, is both continuous and differentiable.

And since Rolle’s Theorem can be used on f’.

Therefore, there must be a value for such that


2014 ap calculus ab frqs

Equation of the tangent line: decreasing at the rate of 0.1635


2014 ap calculus ab frqs

Using (0, decreasing at the rate of 0.1635 1):