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November 25th, 2016, 07:12 PM   #1
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Related rate

A particle is moving along the curve y = 4 (5x+6)^(1/2). As the particle passes through the point (2;16), its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
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November 25th, 2016, 07:48 PM   #2
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$\displaystyle \begin{align*} D &= \sqrt{ x^2 + y^2 } \\ &= \sqrt{ x^2 + \left[ 4\,\left( 5\,x + 6 \right) ^{\frac{1}{2}} \right] ^2 } \\ &= \sqrt{x^2 + 16\,\left( 5\,x + 6 \right) } \\ &= \sqrt{ x^2 + 80\,x + 96 } \end{align*}$

You want $\displaystyle \begin{align*} \frac{\mathrm{d}D}{\mathrm{d}t} \end{align*}$ at the point in time when the particle is at $\displaystyle \begin{align*} \left( 2, 16 \right) \end{align*}$, when $\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} = 3 \textrm{ units}/ \textrm{s} \end{align*}$. The rates are related through the chain rule as

$\displaystyle \begin{align*} \frac{\mathrm{d}D}{\mathrm{d}t} &= \frac{\mathrm{d}D}{\mathrm{d}x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t} \end{align*}$

and since $\displaystyle \begin{align*} D = \sqrt{ x^2 + 80\,x + 96 } \end{align*}$ that means

$\displaystyle \begin{align*}\frac{ \mathrm{d}D}{\mathrm{d}x} &= \frac{2\,x + 80}{2\,\sqrt{ x^2 + 80\,x + 96 }} \,\textrm{units} / \textrm{unit} \\ &= \frac{x + 40}{\sqrt{x^2 + 80\,x + 96}} \end{align*}$

so when $\displaystyle \begin{align*} x = 2 \end{align*}$ that means

$\displaystyle \begin{align*} \frac{\mathrm{d}D}{\mathrm{d}x} &= \frac{2 + 40}{\sqrt{2^2 + 80 \left( 2 \right) + 96}} \\ &= \frac{42}{\sqrt{ 4 + 160 + 96 }} \\ &= \frac{42}{\sqrt{260}} \\ &= \frac{42}{2\,\sqrt{ 65 }} \\ &= \frac{21}{\sqrt{65}} \\ &= \frac{21\,\sqrt{65}}{65} \end{align*}$

and thus at the point $\displaystyle \begin{align*} \left( 2, 16 \right) \end{align*}$

$\displaystyle \begin{align*} \frac{\mathrm{d}D}{\mathrm{d}t} &= \frac{\mathrm{d}D}{\mathrm{d}x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t} \\ &= \frac{21\,\sqrt{65}}{65} \cdot 3 \,\textrm{units}/ \textrm{s} \\ &= \frac{63\,\sqrt{65}}{65} \,\textrm{units} / \textrm{s} \end{align*}$
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November 25th, 2016, 07:58 PM   #3
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Whoever you are, you enabled me to solve this problem 1 minute before due time. One of my best clapbacks ever. Thanks ^^
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November 28th, 2016, 04:29 AM   #4
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"Solve" it or "copy" the solution given?
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