y = 3sqrt(5x+1)
dy/dt=7.5/sqrt(5x+1) * dx/dt
so dx/dt=5 at x=3, find dy/dt from last line above, then
you have dx/dt, dy/dt
d distance/dt= sqrt (dx/dt ^2 + dy/dt ^2)
A particle is moving along the curve y = 3sqrt5x+1. As the particle passes through the point (3,12), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
I've tried several times to calculate the answers but I still can't find it. Could anyone help me with this question?
2 answers
let the distance be z
z^2 = x^2 + y^2
z dz/dt = x dx/dt + y dy/dt
dx/dt = 4
dy/dt = 15/(2√(5x+1)) dx/dt
Now just plug in your numbers to find dz/dt
z^2 = x^2 + y^2
z dz/dt = x dx/dt + y dy/dt
dx/dt = 4
dy/dt = 15/(2√(5x+1)) dx/dt
Now just plug in your numbers to find dz/dt
3 answers