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 November 25th, 2016, 07:12 PM #1 Newbie   Joined: Nov 2016 From: Montréal Posts: 9 Thanks: 0 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.
 November 25th, 2016, 07:48 PM #2 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 \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*} Thanks from Simongekyo
 November 25th, 2016, 07:58 PM #3 Newbie   Joined: Nov 2016 From: Montréal Posts: 9 Thanks: 0 Whoever you are, you enabled me to solve this problem 1 minute before due time. One of my best clapbacks ever. Thanks ^^
 November 28th, 2016, 04:29 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 "Solve" it or "copy" the solution given?

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