This week in calculus we took our two APs on tuesday and wednesday because the juniors were not at school on Monday. We went over corrections on thursday and friday we gave our sub a going away party. So lets go over some material:
First derivative test:
You are given a function and are either asked to find increasing or decreasing or max and mins. You take the derivative of the function and set equal to zero. You then solve for the x values also known as critical points. You then set those points up into intervals between negative infinity and infinity.
Second derivative test:
You are given a function and are asked to find whether the graph is concave up or down or where there is a point of inflection. You take the derivative of the function twice and solve for the critical points once more. You set the x values up into intervals between negative infinity and infinity.
Tangent lines:
You are given a function and an x value. Sometimes they may give you a y value but if not you can find it by plugging the x value into the original function and solving for y. Then you take the derivative of the function and plug in the x value to get the slope. Then you set everything up into point slope form y-y1=slope(x-x1).
Limits:
If the degree on top is bigger than the degree on the bottom, the limit is infinity
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
If they give you a limit problem where there is any letter going to 0 and they have a huge problem with parenthesis in it, you take the derivative of what is behind the parenthesis and plug in for x if needed.
Things i need help with:
integrals with trig functions in them
three question problems
normal lines
integrating problems with e and ln in the same problem
Have a great weekend :)
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A normal line is the same as a tangent line EXCEPT you use the negative reciprocal of the slope instead of the slope. The one on the most recent test was...
ReplyDeleteGive the equation of the normal line to the graph of
y= 3x (x^2 +6)^1/2 + 4 at the point (0,4).
You already have a x and a y value so all you have to do is find the slope, which you do by taking the derivative and plugging in the x-value. We will have to do product rule.
3x (1/2 (x^2 +6 )^-1/2 (2x) + (x^2 +6)^1/2 (3)
OR 3x/ 2(x^2 + 6)^1/2 + 3(x^2+6)^1/2
Now this looks ugly, but the x-value is zero so anything with an x in it is 0.
Plugging in 0: 0/anything =0 + 3 (6)^1/2
The slope is 3 (the square root of 6)
BUT you are finding normal line, so take the negative reciprocal.
Slope of the normal line is -1/ 3 (the square root of 6)
Point slope: y-4 = -1/3 (the square root of 6) (x-0)
To simplify, multiply each side by 3 (square root of 6) to get rid of it: 3(square root of 6) y - 12 (Square root of 6) = -x
x+ 3(square root of 6) y = 12 (Square root of 6)
(Answer choice D)
something that can help you integrate trig functions is to use your trig identities to simplify your original funtion
ReplyDeleteIntegrating Trig Functions:
ReplyDelete1. Opposite of derivatives
2. Use your Trig Identities that we learned in Advanced Math/Calculus
all it is memorization really.