Post #29

We had another AP tests this week, so I am going to review some of those questions.

The problems I always forget how to do are the ones like 5 and 10.
5 says the graph of a piecewise linear function f, for 0
A. 1 B. 4 C. 8 D. 10 E. 13

I can't show the graph, but all you have to do is divide the graph into rectangles and triangles then find the area of each shape. After that, you add the areas together. Finally you subtract the area of the top of the graph and the area of the bottom of the graph.
The first shape is a triangle so (1/2) (2) (2) = 2
Next is a rectangle (2)(2) = 4
Then a triangle (1/2) (1)(2) = 1
4+2+1 = 7
Bottom of the graph is two triangles.
(1/2)(1)(2) = 1
(1/2) (2)(2) = 2
1+2 = 3
Now subtract: 7-3 =4
The answer is B. 4

And number 10 says A car's velocity is shown on the graph above. Which of the following gives the total distance traveled from t=0 to t=16 (in kilometers)?

A. 360 B. 390 C. 780 D. 1000 E. 1360
Its the same steps as number 5, except there is no graph below the axis so you do not have to subtract.
The first shape is a triangle (1/2) (4) (60) = 120
Rectangle (4)(30) = 120
Rectangle (4)(90) = 360
Triangle (1/2)(4)(90)= 180
120+120+360+ 180 = 780
The answer is C. 780

Another one I got wrong was number 24. the lim as x -> 0 tan(3x) + 3x / sin (5x) =

A. 0 B. 3/5 C. 1 D. 6/5 E. Nonexistent

At first glance, I thought this problem was the shortcut so the answer would be 3/5, but that does not work for this problem.
To find the limit as x-> 0, you usually plug in 0, but in this case, you get 0/0, you have to us L'hopitals rule. L'hopitals rule states take the derivative of the top and the derivative of the bottom until you no longer have 0/0.

tan (3x) + 3x = sec^2 (3x) (3) + 3 = 3 sec^2 (3x) + 3
sin (5x) = cos (5x) (5) = 5 cos (5x)
Now plug in 0: 3 sec^2 (0) +3 / 5 cos (0) = 6/5
The answer is D. 6/5.

I need help on the volume and area problems. I never know if I need to solve the equation for x or y or when I have to subtract the equations by a number so pretty much everything to do with those problems.