Posted by **norma** on Friday, September 24, 2010 at 8:23am.

find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2 if ds/dt=5 centimeters per second.

- calculus -
**Reiny**, Friday, September 24, 2010 at 9:12am
I will assume you meant

dx/dt = 5, or else the answer to your question is given

s = √(x^2 + y^2)

= (x^2 + (x^2+2)^2)^.5

= (x^4 + 5x^2 + 4)^.5

ds/dt = (1/2)(x^4 + 5x^2 + 4)(4x^3 + 10x)(dx/dt)

so after you plug in the value, not much else can be done

Something fishy about the question, was there an x value given?

- calculus -
**Reiny**, Friday, September 24, 2010 at 9:14am
forgot the exponent ... should say'

ds/dt = (1/2)(x^4 + 5x^2 + 4)^(-1/2)(4x^3 + 10x)(dx/dt)

## Answer this Question

## Related Questions

- calculus - Find the rate of change of the distance between the origin and a ...
- calculus - find the rate of change of the distance between the origin and a ...
- Calculus - find the rate of change of the distance between the orgin and a ...
- Calculus - A particle is moving along the curve . As the particle passes through...
- Calculus - A particle is moving along the curve . As the particle passes through...
- Calculus - A particle is moving along the curve . As the particle passes through...
- calculus - A particle is moving along the curve y=4((3x+1)^.5). As the particle ...
- Calculus HELP - A particle is moving along the curve y=5 sqrt (2x+6). As the ...
- calculus - A particle is moving along the curve y=5sqrt(3x+1). As the particle ...
- Calculus - A particle is moving along the curve y = 2 √{3 x + 7}. As the ...

More Related Questions