Think about a linear function and its first and second derivative

The second derivative of a linear function always equals 0

Anytime you are trying to find the limit of a piecewise function, find the limit approaching from the left and right ­ if they are equal then the limit exists

Remember you can't differentiate at a vertical tangent line Differentiabilty implies continuity ­ the graph isn't continuous at x = 0

1 To find the velocity, take the first derivative of the position function. To find where the velocity equals zero, set the first derivative equal to zero

2 From the graph, we know that f (1) = 0; Since the graph is increasing, we know that f '(1) is positive; since the graph is concave down, we know that f ''(1) is negative

Possible points of inflection ­ need to set up a table to see where concavity changes (that verifies points of inflection)

(­∞, ­1) (­1, 0) (0, 2) (2, ∞) + ­ + + concave up concave down concave up concave up

Since the concavity changes when x = ­1 and when x = 0, they are points of inflection. x = 2 is not a point of inflection

3 We need to find the derivative, find the critical numbers and use them in the first derivative test to find where f (x) is increasing and decreasing

(­∞, 0) (0, ∞) ­ + Decreasing Increasing

4 This is the graph of y = cos x, shifted right

(A) and (C) are the only graphs of sin x, and (A) is the graph of ­sin x, which reflects across the x­axis

5 Take the derivative of the velocity function to find the acceleration function

There are no critical numbers, because v '(t) doesn't ever equal zero (you can verify this by graphing v '(t) ­ there are no x­intercepts). That means the maximum must occur at the endpoints of the interval (either x = 0 or x = 3)

This is finding the area between two curves. The rule for this is:

6 two solutions = 2 points of intersection

1 point of No points of No points of No points of intersection intersection intersection intersection

7 There is a hole at x = a

Parallel Lines ­ Same slope Slope of tangent line ­ find the derivative

Set the derivatives equal to each other and solve the equation

Solve by graphing

8 (+) (+) (+) (+) Max Min (+) (­) (­) (­) (­) (­) No Critical Numbers No Max/Min

MVT doesn't apply ­ there isn't a closed interval Max: Goes from +, to 0, to ­ Min: Goes from ­, to 0, to +

above the x­axis ­ positive; below the x­axis ­ negative

9 How to set the window Set f ' (x) = 0

Graph and solve y­min = ­1, y­max = 1

can't take the derivative at x = 0 (corner)

F '(x) = f (x)

New Interval: x = 1, u = 2(1) = 2 x = 3, u = 2(3) = 6

10 Original population doubles when t = 11

11 To store zero in calculator: Find the zero, press 2nd & QUIT STO ⇒ A (or B, C, D whatever) Can use to substitute more accurately (once you find the zero & quit; ALPHA X pulls up the zero; Y pulls up the y­coordinate)

Graph and solve

We need the slope and a point to write the equation, we know that m = 1 and x = 0.2367

12 negative

negative

Let g (x) = ­x (a negative function) x = ­2 is a min & x = 2 is a max

13 continuous & smooth

IVT

y = 3

Integrate

14 (a).

(b). Initial Amount + Water Pumped ­ Leaked Water

(c).

(d).

(0,63) (63,120) + ­ Increasing Decreasing

A relative maximum occurs at t = 63 since A '(t) changes from positive (A(t) is increasing) to negative (A(t) decreasing) on (63, 120)

15 (a).

(b).

(c). To find a vertical tangent line, find what would make dy/dx undefined (set the denominator = 0)

When x = 0, ∴ There isn't a vertical tangent line at x = 0

16 (a).

Change to a log (LN) by using the circle rule

Find c using the given info

Change to an exponential using the circle rule

(b). Domain: (can't take the LN of a negative number)

Range: (­∞, ∞)

17 (+) 1

2 (­) r = 2

(a). (­3, ­2) f' > 0 when ­3 ≤ x ≤ ­2 (b). x = 0, x = 2 f' changes from decreasing to increasing at x = 0; f' changes from increasing to decreasing at x = 2 (c). f '(0) = ­2

(d).

18 2011 AP Calculus AB Free­Response Questions (Form B)

1. A cylinder can of radius 10 millimeters is used to measure rainfall in Stormville. The can is initially empty, and rain enters the can during a 60­day period. The height of water in the can is modeled by the function S, where S (t) is measured in millimeters and t is measured in days for 0 ≤ t ≤ 60. The rate at which the height of the water is rising in the can is given by S '(t) = 2sin(0.03t) + 1.5 (a). According to the model, what is the height of the water in the can at the end of the 60­day period?

(b). According to the model, what is the average rate of change in the height of water in the can over the 60­day period? Show the computations that lead to your answer. Indicate units of measure.

Average rate of change

­­OR­­

(c). Assuming no evaporation occurs, at what rate is the volume of water in the can changing at time t = 7? Indicated units of measure. Volume of a cylinder:

(d). During the same 60­day period, rain on Monsoon Mountain accumulates in a can identical to the one in Stormville. The height of the water in the can on Monsoon Mountain is modeled by the function M, where

The height M(t) is measured in millimeters, and t is measured in days for 0 ≤ t ≤ 60. Let D (t) = M '(t) ­ S '(t). Apply the Immediate Value Theorem to the function D on the interval 0 ≤ t ≤ 60 to justify that there exists a time t, 0 ≤ t ≤ 60, at which the heights of water in the two cans are changing at the same time.

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