Asked by bre

A particle is moving along the curve y= 3sqrt3x+1. As the particle passes through the point (5,12), 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.

Answers

Answered by Reiny
Let P(x,y) be any point on the curve
then we could write P as (x,3√(3x+1))
and if D is the distance to the origin,
D^2 = x^2 + (3√(3x+1))^2
= x^2 + 27x + 9
2D(dD/dt) = 2x(dx/dt) + 27(dx/dt)

at (5,12) D^2 = 25+135+9 = 169
so D = 13

so dD/dt = (2(5)(3) + 27(3))/(2(13))
= 4.26

check my work please
Answered by sam
your answer is good

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