Given the relationship 2x^2 + y^3 =10, with y > 0 and dy, dt = 3 units/min., find the value of dx, dt at the instant x = 1 unit.
Given the relationship 2x^2 + y^3 =10, with y > 0 and dy/dt = 3 units/min., find the value of dx/ dt at the instant x = 1 unit.
To find the value of dx, dt at the instant x = 1 unit, we need to differentiate the given relationship with respect to time t and then solve for dx, dt.
First, differentiate both sides of the equation 2x^2 + y^3 = 10 with respect to t using the chain rule. The derivative of 2x^2 with respect to t is 4x(dx, dt), and the derivative of y^3 with respect to t is 3y^2(dy, dt).
So, the differentiated equation becomes:
4x(dx, dt) + 3y^2(dy, dt) = 0
We are given that dy, dt = 3 units/min, so we can substitute this value into the equation:
4x(dx, dt) + 3y^2(3) = 0
Since the equation does not provide the value of y directly, we need to find it. We can do this by substituting the x value of 1 unit into the original relationship.
2(1)^2 + y^3 = 10
2 + y^3 = 10
y^3 = 8
y = 2
Now we have the value of y, so we can substitute it back into the differentiated equation:
4x(dx, dt) + 3(2)^2(3) = 0
Simplifying the equation further:
4x(dx, dt) + 3(4)(3) = 0
4x(dx, dt) + 36 = 0
To find the value of dx, dt, we need to solve for it:
4x(dx, dt) = -36
(dx, dt) = -36/4
(dx, dt) = -9
Therefore, at the instant x = 1 unit, the value of dx, dt is -9 units/min.
4x dx/dt + 3y^2 dy/dt = 0
Now just plug in your values and solve for dx/dt.