10th Post

Calculus Week #10

Okay so this week we learned a few different things. We learned Implicit derivatives, related rates, and angle of elevation problems.

Implicit derivatives are terribly too easy. The only ones that even get remotely hard are the ones that have about 15 product rules, and even those are all about taking it slow and making sure to write the right thing down when you do them. The basic steps for implicit are just to take the derivative; however, whenever you take a derivative of a y term, make sure to write either y' (y prime which stands for derivative) or dy/dx (which stands for the derivative of y as well). Writing dy/dx is a better practice because similar notation comes in on related rates. Anyway, so you take the derivative, write dy/dx by the y terms you took a derivative and then you solve for dy/dx. So let's do an example.

x^2 + y^2 = 4

Derivative would be:

2x + 2y(dy/dx) = 0

Now, solve for dy/dx to get:

dy/dx = -2x/2y

That would be your answer. :-) Now, second derivatives with implicit are the same exact thing. Except in your final answer, when you solve for d^2y/dx^2, you will have a dy/dx in your answer still. This is easily fixed, however, by plugging in the first derivative and then algebraically simplifying your answer.

So implicits are a joke. Anyway, related rates are fun too. You know you have a related rates problem when the problem seems to be like a broken record telling you "rates". Anyway, it's kind of like implicit, but with a picture. When you are doing related rates...you basically get a whole bunch of things...like the rate of this is blah, this is an equation that models another rate, at x=blah, what is the rate of blah2? Basically all you have to do, though, is take the derivative of the equation and plug in. Whenever you take the derivative of x or of y, since it is a related rates problem and the rates are all in reference to time, you do dx/dt, dy/dt. That basically says derivative of x in reference to time, or of y in reference to time. Anyway, then you solve for that it wants and plug in. If in the problem it gives you dx/dt already, just solve for dy/dt and then plug in to find your dy/dt. It's pretty simple once you get a hang of it. Sometimes the problems get a little more complex, but if you take your time and look at what's given and what it's asking for, it's really easy.

The last thing we got, which I'm slightly getting lost on, is the same thing as related rates really...sort of. It's angle of elevation. It always uses a triangle. And there is always one side that is staying constant. So, you take the derivative tan(theta)=y/x and plug in for y or x whichever one is constant. The only thing I get confused about here when they change up the problem too much. Like for instance, the one in homework about the shadow...I was a bit unsure about how that all worked out.

Can someone maybe explain to me how the shadow problem works out? (Yes, I did my homework for once lol. I'm glad I did too. :-D)

Anyway, overall good week. As soon as I get that problem I can go back to saying that this week was a breeze. :-) (so help me do that haha)

-John