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Question Number 199662 by cortano12 last updated on 07/Nov/23

$$$$$$\mathrm{A}\mathrm{point}\mathrm{moves}\mathrm{on}\mathrm{the}\mathrm{curve}\mathrm{of}$$$$\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2+5}\mathrm{such}$$$$\mathrm{that}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{x}-\mathrm{coordinate}\mathrm{increases}$$$$\mathrm{at}\mathrm{a}\mathrm{rate}\mathrm{of}3\sqrt{10}\mathrm{cm}/\mathrm{s}.\mathrm{Find}$$$$\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{point}(1,0)$$$$\mathrm{when}\mathrm{x}=2$$

Answered by mr W last updated on 07/Nov/23

$$y=\sqrt{x^2+5}$$$$s^2={(x-1)}^2+{(y-0)}^2$$$$s^2=2x^2-2x+6$$$$atx=2:$$$$s^2=2\times 2^2-2\times 2+6=10\Rightarrow s=\sqrt{10}$$$$2s\frac{ds}{dt}=4x\frac{dx}{dt}-2\times \frac{dx}{dt}$$$$s\frac{ds}{dt}=(2x-1)\frac{dx}{dt}$$$$\sqrt{10}\frac{ds}{dt}=(2\times 2-1)3\sqrt{10}$$$$\Rightarrow \frac{ds}{dt}=9m/s$$

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