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A-point-moves-on-the-curve-of-the-function-f-x-x-2-5-such-that-it-s-x-coordinate-increases-at-a-rate-of-3-10-cm-s-Find-the-rate-of-change-of-it-s-distance-from-the-point-1-0-when-x-

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Question Number 199662 by cortano12 last updated on 07/Nov/23
A point moves on the curve of the function f(x)=(√(x^2 +5)) such that it′s x−coordinate increases at a rate of 3(√(10)) cm/s. Find the rate of change of it′s distance from the point (1,0) when x = 2
$$ \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{moves}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{5}}\:\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{it}'\mathrm{s}\:\mathrm{x}−\mathrm{coordinate}\:\mathrm{increases}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{3}\sqrt{\mathrm{10}}\:\:\mathrm{cm}/\mathrm{s}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{change}\:\mathrm{of}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{0}\right) \\ $$$$\mathrm{when}\:\mathrm{x}\:=\:\mathrm{2} \\ $$
Answered by mr W last updated on 07/Nov/23
y=(√(x^2 +5)) s^2 =(x−1)^2 +(y−0)^2 s^2 =2x^2 −2x+6 at x=2: s^2 =2×2^2 −2×2+6=10 ⇒s=(√(10)) 2s(ds/dt)=4x(dx/dt)−2×(dx/dt) s(ds/dt)=(2x−1)(dx/dt) (√(10))(ds/dt)=(2×2−1)3(√(10)) ⇒(ds/dt)=9 m/s
$${y}=\sqrt{{x}^{\mathrm{2}} +\mathrm{5}} \\ $$$${s}^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{0}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6} \\ $$$${at}\:{x}=\mathrm{2}: \\ $$$${s}^{\mathrm{2}} =\mathrm{2}×\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}+\mathrm{6}=\mathrm{10}\:\Rightarrow{s}=\sqrt{\mathrm{10}} \\ $$$$\mathrm{2}{s}\frac{{ds}}{{dt}}=\mathrm{4}{x}\frac{{dx}}{{dt}}−\mathrm{2}×\frac{{dx}}{{dt}} \\ $$$${s}\frac{{ds}}{{dt}}=\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{{dx}}{{dt}} \\ $$$$\sqrt{\mathrm{10}}\frac{{ds}}{{dt}}=\left(\mathrm{2}×\mathrm{2}−\mathrm{1}\right)\mathrm{3}\sqrt{\mathrm{10}} \\ $$$$\Rightarrow\frac{{ds}}{{dt}}=\mathrm{9}\:{m}/{s} \\ $$

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