Post #31

More APs this week.
I'll review some from the calculator portion because I have more trouble with that one.

9. What is the approximate instantaneous rate of change of the function
f(t) = the interval of cos (x) dx on [0,8t] at t= pi/3?
A. -1.333 B. -6.928 C. -.8660 D. -4 E. -.5000

Instantaneous rate of change means to take the derivative, and since there is an integral, this problem wants you to take the derivative of a integral which you do by plugging in b and multiplying that by the derivative of b.
cos (8t) (8) = 8 cos(8t)
Now plug in pi/3.
8 cos (8 pi/3)
Plug into your calculator and you get -4 D.

10. What is the error when the integral sin (pi x) dx on [0,1] is approximated by the trapezoidal rule with n=3.
A. 0.059 B. 0.051 C. 0.032 D. 0.109 E. 0.011

The formula for trapezoidal rule is delta x/ 2 [f(a) + 2f (a + delta x) + 2f(a + 2 delta x)+....... + f(b)]
delta x is found by doing b-a/n
Delta x = 1-0/3 = 1/3
Divide by 2 and you get 1/6
Plugging into the formula: 1/6[f(0) + 2f(1/3) + 2f (2/3) + f(1)] which equals .5773333333
Now you have to find the actual value of the integral. You can do this by entering into your calculator.
The exact value is .63661977
To find the error, subtract the two values: .63661977-.5773333333= .059
The answer is A. 0.059.

Jumping to the non calculator portion. This question has been on a few APs and it is really simple so I'm going to explain it.

28. Determine dx/x ln (ln (2-cos (x)))
The derivative of a natural long is just 1/whatever the natural log is of times the derivative of the inside
1/ ln (2-cos (x)) takes care of the first natural log
times that by the derivative of the inside which is 1/ (2- cos (x)
Now times that natural log by its inside (sin (x)).
Everything together is 1/ ln (2-cos (x)) (1/ (2-cos (x) (sin (x)) OR
sin (x) / ln (2-cos (x)) (2- cos (x))
That is answer choice E.

I have questions on a few problems on both portions.
Non calculator- #'s 13, 14, 18, 23, and 24.
Calculator- #'s 2 (invertible again) , 8, and 11.