Science:Math Exam Resources/Courses/MATH110/April 2018/Question 06 (b)

MATH110 April 2018

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Question 06 (b)
Read the problem below and answer the questions in part (a) and (b). Makes sure your solution includes a sketch labeled consistently with variables in your calculations. Your asnwers should be a numerical value, but you do not need to simplify it. A small spider is crawling along the graph of a parabola $y=x^{2}$ in the first quadrant (where $x$ and $y$ are measured in cm) in such a way that its $x$ -coordinate increases at a constant rate of $1{\text{cm}}/{\text{s}}$ . The spider is pulling a thin thread of silk with it that is fixed at the origin. (b) As the spider crawls, the angle between the silk thread and the $x$ -axis is changing. How fast is that angle changing when the spider is at the point $(3,9)$ ?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How is the angle between the silk thread and the $x$ -axis related with the $x$ -coordinate of the location of the spider?

Hint 2
Use the implicit differentiation.

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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. Let $x$ be the $x$ -coordinate of the location of the spider in time $t$ . Since the spider moves along the graph $y=x^{2}$ , the location of the spider in time $t$ is $(x,x^{2})$ . The angle $\theta$ between the silk thread and the $x$ -axis can be expressed as $\tan \theta ={\frac {x^{2}}{x}}=x.$ Now, we use the implicit differentiation to get ${\frac {d\theta }{dt}}$ ; considering ${\frac {d}{dt}}\tan \theta ={\frac {d}{d\theta }}(\tan \theta )\cdot {\frac {d\theta }{dt}}=\sec ^{2}\theta \cdot {\frac {d\theta }{dt}},$ we have ${\frac {d}{dt}}\tan \theta ={\frac {d}{dt}}x\implies \sec ^{2}\theta \cdot {\frac {d\theta }{dt}}={\frac {dx}{dt}}.$ At the location $(3,9)$ , we have $\tan \theta =3$ and hence $\sec ^{2}\theta =1+\tan ^{2}\theta =10$ . Combining this with ${\frac {dx}{dt}}=1$ , we finally obtain the derivative of the angle at $(3,9)$ , ${\frac {d\theta }{dt}}{\bigg \|}_{(x,y)=(3,9)}={\frac {1}{\sec ^{2}\theta }}\cdot {\frac {dx}{dt}}{\bigg \|}_{(x,y)=(3,9)}={\frac {1}{10}}.$ Answer: $\color {blue}{\frac {1}{10}}({\mbox{rad}}/{\mbox{s}})$