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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-




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
Apointmovesonthecurveofthefunctionf(x)=x2+5suchthatitsxcoordinateincreasesatarateof310cm/s.Findtherateofchangeofitsdistancefromthepoint(1,0)whenx=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=x2+5s2=(x1)2+(y0)2s2=2x22x+6atx=2:s2=2×222×2+6=10s=102sdsdt=4xdxdt2×dxdtsdsdt=(2x1)dxdt10dsdt=(2×21)310dsdt=9m/s

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