calculus

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A particle is moving along the curve below.
y = sqrt(x)
As the particle passes through the point (4,2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant?

• calculus - ,

let P(x,y) be any point on the curve
y = x^(1/2)

dy/dt = (1/2)x^(-1/2) dx/dt

when x-4 and y=2 and dx/dt=4

dy/dt = (-1/2)(4^(-1/2)(4)
= ...

you finish it

• calculus - ,

A particle is moving along the curve y = 2 √{3 x + 7}. As the particle passes through the point (3, 8), 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.

• calculus - ,

Well, we know x=3 and y=8 and dx/dt=4. We need to find z (the distance from the particle to the origin) dy/dt, and finally our answer dz/dt.
First solve for dy/dt
y=2(3x+7)^(1/2)
dy/dt=[2(1/2)(3x+7)^(-1/2)](3dx/dt)
dy/dt=[(3x+7)^(-1/2)]3dx/dt
dy/dt=[(3(3)+7)^(-1/2)](3(4))
dy/dt=3
Now we can draw a right triangle, with x=3 and y=8, and using the pythagorean theorem, z=8.544
So we have z^2=x^2+y^2
(2z)dz/dt=(2x)dx/dt+(2y)dy/dt
(z)dz/dt=(x)dx/dt+(y)dy/dt
(8.544)dz/dt=(3)(4)+(8)(3)
dz/dt=4.2135