This week in calculus, we took the AP test Monday and Tuesday and reviewed them the rest of the week.
Some problems that I learned how to do this week were
If a particle moves on a line according to the law s=t^5 + 2t^3, then the number of times it reverses direction is
A. 4 B. 3 C. 2 D. 1 E. 0
The key words in this problem is reverses direction. Reverse direction only means how many maxs or mins there are the the problem.
To find the max and min, take the derivative and set it equal to zero to find the zeros. Then set up intervals and test the intervals with the zeros found. In order to have a max or min or a change in direction in this case, the interval must go from negative to positive or positive to negative.
Derivative: 5t^4 + 6t^2=0
t^2(5t^2+6) = 0
t=0, +- the square root of -6/5
Since you can't have an imaginary number, the second two zeros do not count so the only one you have to test is
(-infinity, 0) and (0, infinity)
To test the intervals, plug in a number between them.
-1: 5(-1)^4 + 6(-1)^2 = positive
1: 5(1)^4 + 6(1)^2 = positive
Since both intervals are positive, there are no maxs or mins and therefore no reverse direction so the answer is E. 0.
Suppose f(x) - x^2 +x / x if x is not equal to 0 and f(0) = 1. Which of the following statements is (are) true of f?
I. f is defined at x=0
II. lim as x->0 f(x) exists.
III. f is continuous at x=0
A. I only B. II only C. I and II only D. None of the statements are true. E. All are true.
For this problem, the key words are the roman numerals. You would have to check each to see if it is true.
I. Defined means that f is there at x=0. To test this, take the find the limit at 0. First you will have to factor the top equation.
x(x+1) / x The x's cancel and you are left with x+1. If you plug in 0, you get 1. Therefore, f is defined at x=0 so I is TRUE.
II. This is basically stating the same thing as I. Factor, and plug in 0. You get 1, so the limit of f(x) exists at 0. TRUE.
III. Since you canceled the x's and found a removable at x=0, this function would not be continuous BUT since they state in the question, f(0)=1, it closes in the removable, making this a TRUE statement.
The answer is E. All are true.
f(x) = the integral of [0,x^2+2] the square root of 1+cos t dt. Then f'(x) =
Since the derivative and the integral cancel each other out, for this problem all you have to do is plug in b for and multiply that by the derivative of b. So the square root of 1+cos (x^2+2) X 2x or 2x the square root of 1+cos(x^2+2).
I am having a harder time with the calculator portion of the AP than I am the non-calculator portion. I also need help with related rates and area under a curve.
Don't forget to look over the short answer for tomorrow.
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For the calculator portion, you (and I both) need to focus more on using our calculator...like for instance if it gives you velocity and you need the acceleration at a certain time... don't take the derivative by hand. Simply plug in your calculator...it's much faster than you.
ReplyDeleteIn general, almost every question on the calculator portion should (as mrs. robinson says) require something in your calculator.
Basically what I'm trying to say is I know you are looking at it as if it wasn't calculator allowed :o. Use your calculator. Use the derivative, the integral, the e^x function. Use all what you can :-)
And don't forget how to find zeros, intersections, and max's, and min's with a calculator :o.
ok so related rates. first you have to figure out everything that is already given to you.
ReplyDeleteonce you have all that sorted out nicely, figure out what they are ASKING for you to FIND.
after that, i always look for the equation i will be using (for example, if they say, the volume of water in a conical tank is ...blah blah blah, then you would use the equation for volume of a cone)
then, plug everything in! also, remember if it says the rate at which the radius is changing, the would be dr/dt = whatever. if it says rate @ which volume is changing, dv/dt, rate @ which area is changing, da/dt, etc.