Asked by mat
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]
Answers
Answered by
Reiny
(-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
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
Answered by
Damon
-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
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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