Derivatives with Polar Coordinates

The basic idea of finding the tangent line of a graph that is using the polar coordinate system as opposed to the Cartesian system is basically by finding a point and a slope like you would with normal equations. However instead of simply taking the derivative of an equation you have to do a little bit more work.




This is the most important equation when trying to find the tangent line. This equation is basically using the derivatives of the equations for crossing between the polar and cartesian planes, the equations being

and



so in order to find an equation at a point you will be given an equation in polar form and a point that you want to find it at.
at



All you have to do is take the derivative of that equation with respect to theta and use that to plug it into the master formula. Once you simplify it down to a more manageable equation you then plug in
and solve to get your slope. Then to find a point you will need to plug in

into the original equation that you had. This will give you an "r" value that then finally you just need to take the crossing over equations and put in the r value and the initial condition in and solve which will give you an x and y point.

and



this will give you a Cartesian set of points and a slope and all you have to do from there is plug it into the equation


That is all you have to do.
For more help on this subject you can go to this site.
For more help on polar equations in general you can go to here.

Hyperbolic Functions!

A quick review -- Hyperbolic functions are analogs of trigonometric functions. Cos(x) and sin(x) relate to points corresponding with unit circle, while cosh(x) and sinh(x) relate to points corresponding with the right half of the hyperbola.

Three main functions:

And their graphs:

There are certain hyperbolic identities that seem fairly similar to trigonometric identities...

To solve hyperbolic function can be solved my just plugging x into the main functions mentioned earlier. You can also solve hyperbolic integrals using non-hyperbolic integral rules. Check out this example we used on our powerpoint:

Integration by Partial Fractions

QUIZ

BLOG POST

Surface area of a solid of revolution

Surface area of a solid of revolution blog post

Integration by Parts

Integration by parts is a method of solving complicated integrals when the integrand is too tough to be solved with substitution or the reverse chain rule. Integration by parts can be expressed with the following equation:



This equation originates from the product rule. We can find the integration by parts formula by writing out the product rule:


When you integrate both sides of the equation, you are left with:



Rearranging the terms by subtracting gives you the integration by parts formula that we showed above:



Now that we know where it comes from, we can use integration by parts to solve tricky integration problems. Here's the trickier example that we used in class:



Using substitution doesn't help here, because neither function is really a form of an antiderivative of the other - and trying to do a reverse chain rule with a combination of product rule would probably give you a headache. So, let's use integration by parts.

1. Decide which function in the integrand will be f(x) and which will be g'(x). g'(x) should be one that's easy to antiderive. In this case, let's make represent g'(x) and represent f(x). This means that f'(x)=2x and g(x)=.
2. Then, use the integration by parts formula to solve for

= -

Next, simplify.



We'll have to use the integration by parts rule again to solve the last part. Now, f(x)=x but g'(x) is the same as before. When we do that, our equation looks like this:



Now, we can simplify and integrate to find our answer... and don't forget the +C!



It's not pretty, but it's an answer!

Our other example was much simpler - .

For this one we said that and .

We then plug into the formula to get this equation:


This simplifies to and gives you an answer of

Mr. Khan has a great video to explain integration by parts using a verrrrrry similar example. You can see it by clicking this link:
http://www.khanacademy.org/video/integration-by-parts--part-6-of-indefinite-integration?playlist=Calculus

In our presentation, we broke into groups and used these example problems to help everyone understand the concept of integration by parts. The problems can be found here: http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartsdirectory/IntByParts.html For each problem, a detailed answer key is given (look to the right of the problem).

We finished our presentation by giving a short quiz.

Student Feedback

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Final Project

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