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

MATH110 April 2018

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Question 06 (a)
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. (a) When the spider is at the point $(3,9)$ , how fast is the silk thread lengthening?

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
Find the relation between the length of the silk thread and the $x$ -coordinate of the location of the spider.

Hint 2
Denoting the $x$ -coordinate of the location of the spider by $x$ , we have ${\frac {dx}{dt}}=1({\text{cm}}/{\text{s}})$ .

Hint 3
Use either the chain rule (Solution 1) or the implicit differentiation (Solution 2).

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Solution 1
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 $(x,x^{2})$ . Note that the length $l$ of the silk thread is the distance between two points $(x,x^{2})$ and $(0,0)$ . Using the distance formula, we get $l(x)={\sqrt {(x-0)^{2}+(x^{2}-0)^{2}}}={\sqrt {x^{2}+x^{4}}}$ Since $l$ is a function of $x$ and $x$ is the function of $t$ , by the chain rule, we have ${\frac {dl}{dt}}={\frac {dl}{dx}}\cdot {\frac {dx}{dt}}.$ By the Hint 2, we have ${\frac {dx}{dt}}=1$ . Since $l$ can be written as the composite of two functions $l(x)=f(g(x))$ , where $f(u)={\sqrt {u}}$ and $g(x)=x^{2}+x^{4}$ , we apply the chain rule with $f'(u)={\frac {1}{2{\sqrt {u}}}}$ and $g'(x)=2x+4x^{3}$ to get ${\frac {dl}{dx}}=(f(g(x)))'=f'(g(x))\cdot g'(x)=f'(x^{2}+x^{4})\cdot g'(x)={\frac {1}{2{\sqrt {x^{2}+x^{4}}}}}\cdot (2x+4x^{3}).$ Plugging ${\frac {dl}{dx}}$ and ${\frac {dx}{dt}}$ into the formula for ${\frac {dl}{dt}}$ , we obtain ${\frac {dl}{dt}}={\frac {1}{2{\sqrt {x^{2}+x^{4}}}}}\cdot (2x+4x^{3}).$ At the location of the spider $(3,9)$ , i.e., $x=3$ , we therefore have ${\frac {dl}{dt}}{\bigg \|}_{x=3}={\frac {1}{2{\sqrt {3^{2}+3^{4}}}}}\cdot (2\cdot 3+4\cdot 3^{3})={\frac {1}{2{\sqrt {90}}}}\cdot (114)={\frac {19}{\sqrt {10}}}{\text{cm}}/{\text{s}}.$ Answer: $\color {blue}{\frac {19}{\sqrt {10}}}{\text{cm}}/{\text{s}}$

Solution 2
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 $(x,x^{2})$ . Note that the length $l$ of the silk thread is the distance between two points $(x,x^{2})$ and $(0,0)$ . Using the distance formula, we get $l(x)={\sqrt {(x-0)^{2}+(x^{2}-0)^{2}}}={\sqrt {x^{2}+x^{4}}}.$ Now, we find ${\frac {dl}{dt}}$ using the implicit differentiation. Since $l(x)^{2}=x^{2}+x^{4},$ we take the derivative with respect to ${\frac {d}{dt}}$ to get ${\frac {d}{dt}}(l^{2})=2l\cdot {\frac {dl}{dt}}.$ On the other hand, we compute the derivative of the function $u(x)=x^{2}+x^{4}$ with respect to $t$ using chain rule (note that x depends on time t): ${\frac {d}{dt}}(l(x))={\frac {d}{dt}}(x^{2}+x^{4})=(2x+4x^{3}){\frac {dx}{dt}}.$ Therefore ${\frac {d}{dt}}(l^{2})={\frac {d}{dt}}(x^{2}+x^{4})\implies 2l\cdot {\frac {dl}{dt}}=(2x+4x^{3}){\frac {dx}{dt}}.$ Noting that ${\frac {dx}{dt}}=1$ and $l(3)={\sqrt {3^{2}+9^{2}}}={\sqrt {90}}=3{\sqrt {10}}$ , we have $2\cdot 3{\sqrt {10}}\cdot {\frac {dl}{dt}}{\bigg \|}_{(x,y)=(3,9)}=(2\cdot 3+4\cdot 3^{3})\cdot 1=114.$ Finally we isolate the derivative of $l$ with respect to $t$ to get ${\frac {dl}{dt}}{\bigg \|}_{(x,y)=(3,9)}={\frac {(2\cdot 3+4\cdot 3^{3})}{2\cdot 3{\sqrt {10}}}}={\frac {19}{\sqrt {10}}}.$ Answer: $\color {blue}{\frac {19}{\sqrt {10}}}{\text{cm}}/{\text{s}}$