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November 20, 2014

November 20, 2014

Posted by **sanjay** on Tuesday, October 26, 2010 at 6:52am.

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 -
**Reiny**, Tuesday, October 26, 2010 at 8:29amlet 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 -
**Akash**, Tuesday, October 26, 2010 at 2:41pmA 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 -
**Stephanie**, Saturday, February 11, 2012 at 5:27pmWell, 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

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