Area in Polar

Today in class, we learned how to find the area of a function in polar. In the rectangular coordinate system, we can find the area of a function by breaking the area into many small rectangles that can then be added together to find the total area. Instead, in the polar format we can find the area of a function by breaking up the area into many sectors that all come out of the origin. The upper and lower bound of the integral are the radial lines between theta = a and theta = b that bind the area you are trying to find.

Think about a radar on a weather map as the line travels in a counter-clockwise motion. This motion is similar to how we find the area.

When solving, think about certain identities that may be useful to help solve the integral - as we did in today's example.

Slopes of Tangent Lines in Polar

Today in class, we learned about how to find the slope of a tangent line to a curve in Polar. Since slope is defined as change in y over change in x, we need to know r and theta in terms of x and y. So we use x = rcos(theta) and y = rsin(theta). We differentiate each with respect to theta and then divide dy/d(theta) by dx/d(theta), similar to what we did in parametrics, because that will give us dy/dx so that we can plug in theta and find the slope of the line tangent at that point. We can also find the points where there are horizontal tangent lines and the slope of the tangent line at the pole for that graph. It can be quite tedious with a LOT of trig. Hope this helps!

Polar Coordinates

Today in class we learned about using polar coordinates. The points for polar coordinates are (r,θ), where r is the directed distance from the origin (otherwise known as a pole), and where θ is the angle measured in standard position. We also learned how to switch from polar to rectangular equations and how to switch from rectangular to polar equations. There are also some examples we did in class to help.

Parametric Equations in a Plane

Today in class we learned about particle motion in a plane using position, velocity, speed and displacement. We learned that the distance traveled by the particle is equal to the arc length. Using the integral of speed, you can find the arc length, which is synonymous to the distance traveled by the particle.

Arc Length

Today in class, we learned about Arc Length. Since we know the length of a line segment is equal to the square root of (change in x squared plus change in y squared), we can multiply that formula by 1 in the form of change in t over change in t to get the formula for arc length. This formula turns out to be the integral from t1 to t2 of the square root of (dv/dt)^2 + (dy/dt)^2. We can now use this formula to find the arc length of an integral for parametric equations.

Second Derivative with Parametric Equations

Today in class we built upon the knowledge of last weeks lesson (finding the slope of tangent lines to find the slope of lines tangent to parametric equations) by learning an additional step: finding the second derivative of a parametric equation.

Differentiating Parametric equations

Today in class we used our prior knowledge in finding the slope of tangent lines to find the slope of lines tangent to parametric equations. By using the Chain rule, we found that the derivative of y with respect to x equals the derivative of y with respect to t divided by the derivative of x with respect to t (as shown in notes). Using prior knowledge from the fall semester, it is easy to apply the same rules to this concept with a new variable t. Answers may be left in the y-intercept form when solving for the equation of the line tangent to the curve.

Parametric Equations

Today in class we briefly reviewed the fundamentals of parametric equations. Parametric equations are three-dimensional graphs that relate a point in a plane to a corresponding value of time, t. Every function has an infinite number of parametric equations that can be associated with it. Using the example illustrated in my notes, it can be seen that the graph y=x^2 can be represented parametrically with both x=t, y=t^2, and x=sint, y=(sint)^2. The only difference between the two graphs is that x=sint, y=(sint)^2 is bounded by [-1,1] and oscillates because of the behavior of the sin function.

Improper Integrals- 2/8/2010

Today in class we went over improper integrals (integrals that approach infiniti). To write the improper integral properly, we need to set up the integral as a limit (refer to attachment!). Another important idea was divergence vs. convergence; this will be important after spring break.

Enjoy!

Integration by Partial Fractions

So today in class we started Integration by partial fractions. To integrate using partial fractions, in the rational function, the degree of the top always has to be less than the degree of the bottom. If it is not, then you have to divide the two. To use integration by partial fractions, the bottom function should be factorable. Once you factor it, put A and B over the two factors. Cross multiply, so you end up with the top of the rational function on the left side, and A times one factor plus B times the other factor on the other side. Set one factor equal to zero and solve A or B of the other factor using the x you used to make the first factor zero. Then replace A and B with their values, and integrate!!

Integration by Parts (cont.)

Today in class, we continued discussing integration by parts. The main concept we learned was that we could make sense of confusing or looping integrals by combining like terms and dividing by the coefficient on each side. Homework was pg. 531, # 36-38, 49, 73-78, 84

Integration by Parts

Today in Calculus we learned a new method for integration known as integration by parts. This method is useful for solving antiderivatives of functions that are combinations (two functions multiplied together), which cannot be solved by regular u substitution.

if uv = something then
d(uv) = udv + vdu (product rule)
uv = fint (udv) + fint(vdu)
uv - fint(vdu) = fint(udv)

So the idea is to take your function and find the most complicated part of the function (usually this is the highest order function in the function hierarchy, but it can also be the most complex sometimes like a trig function). This complicated part must have a known antiderivative (that you can do by integration rules). The "left over" part is the v. Solve for dv and u. You then set up a table:

v
dv
u
du

Finally, when you have identified the elements/ finished the table you plug in the appropriate expressions in the correct place in the equation above. Sometimes when evaluating a 'simplified' expression, you may come across an antiderivative of another combination function (aka the fint(vdu) or fint(udv) cannot be automatically solved). This could mean that you may have set up the table wrong but it can also mean that further integration by parts is necessary and that your work is at an intermediate step to the ultimate simplified form.

Today we talked about the logistic population growth model, which is just another kind of differential equation. The one we learned before this, the exponential growth model is in the form dy/dt = kt, while the logistic model is in the form dy/dt = ky(1-(y/L)), where L is the carrying capacity of the environment where the population lives (determined through research).

Given this growth model, we can determine both the long term trends of this population and when the rate of growth of the population is the largest. Although we do not know how to solve the logistic differential equation yet, we know that its solution is y = L / (1 + Ce ^ -kt).

Slope Fields: On January 27th, we learned about Slope Fields. To find the graph of a differentiable equation, one can find the slope fields of each point by plugging in the x & y coordinates of each point. The number found by plugging into the equation is the slope of the slope field.

Many times, the graph may look the same, such as graph 1, where even though the second point is not the same as the first, it is the same shape.

To check if a line drawn on a slopefield is correct, plug in the given points into point-slope formula, and substitute into the given dy/dx formula.

You can find this in the calculator by following the instructions on page (2).