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.
SUPER CONFUSED DONT GET THE EQUATION!!!!!!!
This is a typical exercise using the chain rule.
2x^2 + y^3 =10
4x dx/dt + 3y^2 dy/dt = 0
Now, using your values, when x=1, y=2. So
4(1) dx/dt + 3(2^2)(3) = 0
4 dx/dt + 36 = 0
dx/dt = -9
To find the value of dx/dt at the instant x = 1 unit, we need to differentiate the given equation implicitly with respect to time (t).
Step 1: Differentiate both sides of the equation with respect to time (t).
d/dt(2x^2 + y^3) = d/dt(10)
Step 2: Apply the chain rule. For the left side of the equation, we need to consider the derivatives of x and y with respect to t.
2(d/dt)(x^2) + 3y^2(dy/dt) = 0
Step 3: Substitute the given value of dy/dt into the equation.
2(d/dt)(x^2) + 3y^2(3) = 0
Step 4: Simplify the equation.
2(d/dt)(x^2) + 9y^2 = 0
Step 5: Plug in the value x = 1 into the equation.
2(d/dt)(1^2) + 9y^2 = 0
2(d/dt)(1) + 9y^2 = 0
2(dx/dt) + 9y^2 = 0
Step 6: Solve for dx/dt.
2(dx/dt) = -9y^2
dx/dt = -9y^2 / 2
Since y > 0, we know that dy/dt = 3 units/min. Thus, dy/dt = 3.
Step 7: Substitute dy/dt = 3 into the equation.
dx/dt = -9y^2 / 2
dx/dt = -9(3^2) / 2
dx/dt = -9(9) / 2
dx/dt = -81 / 2
Therefore, the value of dx/dt at the instant x = 1 unit is -81 / 2 units/min.
To find the value of dx/dt at the instant x = 1 unit, we need to use implicit differentiation.
First, let's differentiate both sides of the given equation with respect to t:
d/dt(2x^2) + d/dt(y^3) = d/dt(10)
Since we are looking for dx/dt at the instant x = 1 unit, we can substitute x = 1 into the equation.
d/dt(2(1)^2) + d/dt(y^3) = d/dt(10)
Simplifying, we can remove the terms with no t dependence:
0 + d/dt(y^3) = 0
Now, we can find dy/dt:
d/dt(y^3) = d/dt(y) * d/dy(y^3)
Since dy/dt = 3 units/min, we can substitute dy/dt into the equation:
d/dt(y^3) = 3 * d/dy(y^3)
Simplifying, we have:
d/dt(y^3) = 3 * 3y^2 * dy/dt
Since y > 0, we know that dy/dt is positive. Thus, we have:
d/dt(y^3) = 9y^2 * dy/dt
Now, substitute this equation back into our original equation:
9y^2 * dy/dt = 0
Since y > 0 and dy/dt = 3 units/min, we have:
9y^2 * 3 = 0
Simplifying further, we have:
27y^2 = 0
This equation has no solution. Therefore, at the instant x = 1 unit, dx/dt does not exist.