Science:Math Exam Resources/Courses/MATH110/December 2012/Question 03 (d)

MATH110 December 2012

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Question 03 (d)
Calculate the following derivative. You may leave your answer unsimplified. $\displaystyle f'(\theta )$, where $f(\theta )$ is the area of the triangle with vertices at the origin, $P$ and $Q$, shown below. Note that the circle is of radius $1$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Identify the base and height of the triangle in order to arrive at the area formula. Finding the height will involve a trig ratio.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we need to find a formula for $f(\theta )$, the area of the triangle. The area of a triangle is $(b)(h)/2$. See the labeled picture below: The base of the triangle is a radius of the circle, which is equal to 1. We can find the height by a trig ratio. Using the blue right triangle, we can see that $\sin \theta =h/1$. So $h=\sin \theta$. Plugging these into our area formula, we get: $\displaystyle f(\theta )=(1)(\sin \theta )/2=(\sin \theta )/2$ It is now easy to find the derivative, $\displaystyle f'(\theta )=(\cos \theta )/2$ completing the question.