1 / 32

2014 AP CALCULUS AB FRQs - PowerPoint PPT Presentation

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).

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

PowerPoint Slideshow about '2014 AP CALCULUS AB FRQs' - torie

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

2014 AP CALCULUS AB FRQs

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

Average amount of clippings: decreasing at the rate of 0.1635

Solvedays

Tangent line at decreasing at the rate of 0.1635 is

Solve days

y decreasing at the rate of 0.1635 = 4

f(x)

Intersections of the two graphs: and

y decreasing at the rate of 0.1635 = 4

f(x)

Length of the rectangle:

y – f(x)

y decreasing at the rate of 0.1635 = 4

f(x)

y decreasing at the rate of 0.1635 = 4

f(x)

y decreasing at the rate of 0.1635 = 4

f(x)

, 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)

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

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

S meters/minute2

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

meters decreasing at the rate of 0.1635

Train B decreasing at the rate of 0.1635

meters

meters

meters

z = distance between the two trains

y

Train A

x

At minutes,

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.

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

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

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

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