IN A PHYSICS EXPERIMENT,THE DISTANCE OF X AND Y CM HAS A CONNECTION TO EACH OTHER BY 1/X + 1/Y = 1/5.IF THE DISTANCE OF Y IS INCREASING AT THE RATE 16 CMS^(-1),FIND THE RATE OF CHANGE OF THE DISTANCE OF X WHEN Y=9 CM.
So we have
1/x + 1/y = 1/5
differentiate with respect to t
(-1/x^2)dx/dt + (-1/y^2)dy/dt = 0
(-1/x2)dx/dt = (1/y^2)dy/dt
dx/dt = (-x^2/y^2)(dy/dt)
given:
when y = 9
1/x + 1/9 = 1/5
1/x = 4/45
x = 45/4
and dy/dt = ???? , not clear what you mean by
16 CM^(-1)
what ever it is, sub into
dx/dt = (-(45/4)^2 /81) (????)
= (-25/16)(????)
To find the rate of change of the distance of X when Y=9 cm, we need to differentiate the given equation with respect to time.
Given: 1/X + 1/Y = 1/5
Differentiating both sides with respect to time:
d/dt (1/X) + d/dt (1/Y) = d/dt (1/5)
Now, let's find the derivatives of each term:
d/dt (1/X) = -1/X^2 * (dX/dt)
d/dt (1/Y) = -1/Y^2 * (dY/dt) [Note: we have the value of (dY/dt)]
d/dt (1/5) = 0 [Since 1/5 is a constant]
Substituting the given values into the equation:
-1/X^2 * (dX/dt) - 1/9^2 * (16 cm/s) = 0
Simplifying the equation:
-1/X^2 * (dX/dt) - 1/81 * 16 cm/s = 0
Since we need to find the rate of change of X when Y=9 cm, we substitute Y=9 into the equation:
-1/X^2 * (dX/dt) - 1/81 * 16 cm/s = 0
Simplifying further:
-1/X^2 * (dX/dt) = 16/81 cm/s
Now, we can solve for (dX/dt):
-1/X^2 * (dX/dt) = 16/81 cm/s
Multiplying both sides by X^2:
(dX/dt) = -16X^2/81 cm/s
Therefore, the rate of change of the distance of X when Y=9 cm is given by the equation:
(dX/dt) = -16X^2/81 cm/s