alaina's blog, 10 oct. 2010

Alright, this week is exam week and I’m nervous about the exam for this class. Actually, I'm FREAKING OUT. I’m hoping to make a good grade, but I guess we will have to see. Anyway…

I’ll explain finding tangent lines when it gives you an equation or formula, and a point. When this happens you’re going to do what you would usually do which is figure out what kind of problem you’re dealing with. Once that’s done you’re going to set up your problem and take the derivative, then solve. After getting an answer, which will be what dy/dx is equal to, you’re going to plug in the point they give you. Then you will have to plug into point slope form.

Here’s an example:Find the tangent line to the graph given by x^2(x^2+y^2) at the point (square root of 2/2, square root of 2/2).
*First thing you do is distributeX^4+x^2y^2=y^2
**Notice it is now product rule, so now you can set up & take the derivative4x^3+x^2(2ydy/dx)+y^2(2x)=2ydy/dx
*Solve and set dy/dx on one sideDy/dx=4x^3+2xy^2/2y-2x^2y
*Now plug inDy/dx=4(square root 2/2)^3+2(square root 2/2)(square root 2/2)^2/2(square root 2/2)-2(square root 2/2)^2(square root 2/2)*Solve to get slopeM=3
*Put in point slope
ANSWER: y-square root of 2/2=3(x-square root of 2/2)

One of the things that I’m still not too comfortable with is the word problems. When it doesn’t tell me what the problem is, or what formula to use, I get confused and can’t figure out how to work it. I can set it up, but I get a little confused from there. I guess I just need more practice; it also will help me once I get all the formulas.