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Maria and Richard are professional figure skaters training for the olympics. Maria's skating path is defined by the polar curve r=4. Richard's path is defined by the polar curve r=3+2cosX.
At x = pi/3 and x = 5pi/3, Maria and Richard's paths meet and together they do a jump. In the olympics, scores are given based on level of complexity. The closer they are, the higher scores they get. Maria and Richard must calculate the area between their two paths from the first jump to the second.
I know how to find the area! You have to take 1/2 of the integral of each polar function squared from the first jump to the second jump . Then you subtract my path from Maria's!
Thankfully, Richard remembers taking Mr. Lucid's Calc BC class in high school.
However, there is a problem: the hockey team is practicing at the same time as the figure skaters! At x = pi/2, a hockey player chased a stray puck and crashed into Richard!!!
In order to prevent this from happening again, we must calculate the slope of the stray hockey puck, AKA the line tangent to Richard’s curve at x = pi/2.
We can do this by using the formula where the slope = (r’sinx+rcosx)/(r’cosx-rsinx) where r is specifically Richard’s path and x = pi/2! The derivative of Richard’s path is r’ = -2sinx.
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