Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 08

MATH102 December 2016

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Question B 08

Jenna is riding off a snowboarding ramp while her father takes a video of her flight. The ramp tilts up with a slope of 1/2. Her father stands 3 meters in front of the ramp with his camera pointed at the ramp and held at the same level as the top of the ramp. The angular elevation of the camera, $\theta (t)$, starts at 0 and increases with time so that the camera follows Jenna. When Jenna is a horizontal distance x from the ramp, she is a vertical distance $f(x)$ above the top end of the ramp. She is moving horizontally at a speed of 2 m/s as she leaves the ramp. How rapidly must Jenna’s father be rotating the camera so as to follow her flight path just as she leaves the ramp ($x=0$)? Note that $f(0)=0$ and $f'(0)=1/2$.

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
Find the relation between $\theta$ and $f$. (we can consider $x$ as Jenna's horizontal distance from the ramp in $t$, i.e., $x=x(t)$) Then, find ${\frac {d\theta }{dt}}$ at $t=0$.

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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. The first step is to find an equation for $\theta (t)$ in terms of $x$ and $f(x)$ . As we see in the diagram we have $\tan(\theta )={\frac {f(x)}{3-x}}={\frac {f(x(t))}{3-x(t)}}$, now we find th derivative of both sides with respect to $t$ using the chain rule: ${\frac {d(\tan(\theta ))}{dt}}={\frac {d}{dt}}\left({\dfrac {f(x(t))}{3-x(t)}}\right)={\frac {d}{dx}}\left({\dfrac {f(x)}{3-x}}\right)\cdot {\frac {dx}{dt}}$ $(1+\tan ^{2}\theta )\cdot {\frac {d\theta }{dt}}={\frac {f'(x)(3-x)+f(x)}{(3-x)^{2}}}\cdot {\frac {dx}{dt}}$ , so we have ${\frac {d\theta }{dt}}={\frac {f'(x)(3-x)+f(x)}{(3-x)^{2}}}\cdot {\frac {dx}{dt}}\cdot {\frac {1}{1+\tan ^{2}\theta }}$ , now we need to evaluate this expression at $t=0$ . Note that $x(0)=0$ and also at $t=0$ we have $\tan(\theta (0))={\frac {f(x(0))}{3-x(0)}}={\frac {f(0)}{3-0}}=0$ . $\color {blue}{\frac {d\theta }{dt}}{\Big |}_{t=0}={\frac {3f'(0)+f(0)}{(3)^{2}}}\cdot 2\cdot 1={\frac {1}{3}}$ .