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]
(-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
-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
To find the rate at which y is changing, we need to differentiate the given equation with respect to time.
Given: 1/x + 1/y = 1/9
Differentiating with respect to time (t):
(1/x)' + (1/y)' = (1/9)'
Now, let's differentiate each term using the chain rule:
(1/x)' = (-1/x^2)(dx/dt)
(1/y)' = (-1/y^2)(dy/dt)
(1/9)' = 0 (since a constant)
Substituting these derivatives into the equation:
(-1/x^2)(dx/dt) + (-1/y^2)(dy/dt) = 0
Given that dx/dt = 8 cm/s (rate of change of x) and we need to find dy/dt (rate of change of y) when x = 15 cm, we can substitute these values into the equation:
(-1/15^2)(8) + (-1/y^2)(dy/dt) = 0
Simplifying the equation:
-8/225 - (dy/dt) / y^2 = 0
Multiplying both sides by 225y^2:
-8y^2 = (dy/dt) * 225
Dividing both sides by 225:
(dy/dt) = (-8y^2) / 225
Now, we need to find the value of y when x = 15. Given the equation 1/x + 1/y = 1/9, we can substitute x = 15 to solve for y:
1/15 + 1/y = 1/9
Multiplying through by the least common denominator of 135y:
9y + 15 = 15y
Subtracting 9y and 15 from both sides:
15 - 15y = -9y
Combining like terms:
6y = 15
Simplifying:
y = 15/6
y = 2.5 cm
Now, substitute this value of y into (dy/dt) = (-8y^2) / 225:
(dy/dt) = (-8(2.5)^2) / 225
(dy/dt) = (-8(6.25)) / 225
(dy/dt) = (-50) / 225
(dy/dt) = -0.2222 cm/s
Therefore, the rate at which y is changing when x = 15 cm is approximately -0.2222 cm/s or -0.22 cm/s. Answer: -0.22 cm/s (rounded to two decimal places).