Post #33

HAPPEE EASTER!

Tangent lines:
To find the equation of a tangent line. you are usually given an equation and a x-value. To find the equation, you need a slope, and a point. To find the point, plug in the x-value into the original equation to find the y. To find the slope, take the derivative of the equation and plug in the x- value into the derivative. The final step is to plug the point and the slope into point slope form which is y-y1 = m(x-x1).

Examples:
Find the equation of the line tangent to y=4x^3 - 7x^2 at x=3.
1. Find y by plugging x into original: 4(3)^3 - 7(3)^2 = 45
The point is (3, 45)
2. Find slope by taking derivative and plugging in x: 12x^2 - 14x
12(3)^3 - 14(3) = 66
3. Point slope form; y-45= 66 (x-3)

The answer may be in other forms, so you might have to solve the equation in order to get an answer given.
Other possible answers are: y-45 = 66x - 198 = y= 66x - 153
OR 66x - y = 153

Normal lines:
To find the equation of a normal line, follow the same steps as you would finding the equation of a tangent line EXCEPT when you find slope, you have to take the negative reciprocal of it.
Example:
An equation of the line normal to the graph of y = (3x^2 +2x)^1/2 at (2,4) is
1. Since you already have a point, you do not need to plug in the x to find the y.
2. Derivative: 1/2 (3x^2 + 2x) ^ -1/2 (6x +2)
Plug in x to find slope: 6(2) +2 / 2 (3(2)^2 + 2(2)) ^1/2
14/8 = 7/4
Since this is normal, we need the negative reciprocal of the slope, which is -4/7
3. Now we can plug in: y-4 = -4/7 (x-2)
Lastly, manipulate the equation to get an answer choice.
In this case, it helps to get rid of the fraction.
7y - 28 = -4 (x-2)
7y - 28 = -4x + 8
7y = -4x + 36
4x + 7y = 36

Something to remember: Position, velocity, acceleration
It helps when given a problem such as
A particle's position is given by s=t^3 - 6t^2 + 9t. What is its acceleration at time t=4?
Since we are given position and are looking for acceleration, we know we have to take the second derivative and then plug in 4.
3t^2 - 12t + 9
6t - 12
6(4) - 12 = 12

Another example: A particle moves along the x-axis so that its position at time t, in seconds, is given by x(t) = t^2 -7t +6. For what value(s) of t is the velocity of the particle zero?
Given position and looking for velocity means take derivative.
2t - 7
We are looking for where the velocity equals zero, so set the derivative equal to zero and solve for t.
2t-7 = 0
t = 7/2 OR 3.5

I can use a review on problems such as number 2 on the calculator portion of the last AP (the one with invertible and give the derivative of f-1). I know we went over this in class, but I didn't quite catch on to what we have to do.
I can also use a review on optimization and related rates.