Posted by **mat** on Friday, July 22, 2011 at 8:28pm.

the distance x cm and y cm are related by the eqt. 1/x+1/y=1/9, if x is increasing at a rate of 8 cm/s, calculate the rate at which y is changing when x=15. [ans:-18]

- math -
**Reiny**, Friday, July 22, 2011 at 8:35pm
(-1/x^2)dx/dt - (1/y^2)dy/dt = 0

when x=15

1/15 + 1/y = 1/9

1/y = 2/45

y = 45/2

(-1/225)(8) - (4/2025)(dy/dt) = 0

I will let you do the arithmetic

- math -
**Damon**, Friday, July 22, 2011 at 8:37pm
-dx/x^2 - dy/y^2 = 0

so

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

when x = 15, find y

1/15 + 1/y = 1/9

1/y = 5/45 - 3/45 = 2/45

so y = 45/2

dy/dy = - (45^2/4/225)(8) = -18

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