Question 1192714: Imagine that a particle is moving along the function (x^2) + (2*y^3) = 32 such that dx/dt is a constant rate of 6 units per second. At what rate is it moving in the y-direction when x = 4?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
If x = 4, then y is,
x^2 + 2y^3 = 32
(4)^2 + 2y^3 = 32
16 + 2y^3 = 32
2y^3 = 32-16
2y^3 = 16
y^3 = 16/2
y^3 = 8
y^3 = 2^3
y = 2 assuming we only focus on the real numbers
Therefore, the point (x,y) = (4,2) is on the curve x^2 + 2y^3 = 32
Go back to the original equation.
Apply the derivative to both sides with respect to t.
x^2 + 2y^3 = 32
d/dt[ x^2 + 2y^3 ] = d/dt[ 32 ]
d/dt[ x^2 ] + d/dt[ 2y^3 ] = d/dt[ 32 ]
2x*dx/dt + 6y^2*dy/dt = 0
Now plug in (x,y) = (4,2) and dx/dt = 6.
Then isolate the dy/dt.
Treat dy/dt as a single variable rather than two variables being divided.
2x*dx/dt + 6y^2*dy/dt = 0
2*4*6 + 6*2^2*dy/dt = 0
48 + 24*dy/dt = 0
24*dy/dt = -48
dy/dt = -48/24
dy/dt = -2
The particle is moving in the negative y direction, aka south, at a vertical velocity of 2 units per second.
This is of course in addition to the horizontal speed of dx/dt = 6 to indicate it's moving 6 units per second to the east.
This is all in the instantaneous moment the particle is at the location (4,2).
Answer:
dy/dt = -2
2 units per second to the south
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