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Posted by 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 - , 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 - , 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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