# calculus

posted by on .

A particle is moving along the curve y=5sqrt(3x+1). As the particle passes through the point (5,20) 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.

• calculus - ,

dx/dt = 3 units/s at the time of interest
dy/dt = dy/dx * dx/dt = 3 dy/dx

dy/dx = (5/2)/sqrt(3x+1) = (5/2)/4 = 5/8

Compute dy/dt and then use
dR/dt = sqrt[(dy/dt)^2 + (dx/dt)^2] ]
sqrt[3^2 + (15/8)^2] = 3.538

• calculus - ,

I just plugged it in and it says it isn't right. =/

• calculus - ,

drwls left out the factor of 3 in the derivative, it should have been

dy/dx = (5/2)(3) / √(3x+1)
so when x=5
dy/dx = (15/2) / √16 = 15/8

take it from there

• calculus - ,

Thanks for the correction, Reiny :-)
I am getting sloppier in my old age

• calculus - ,

at the point .