Suppose that x=x(t) and y=y(t) are both functions of t. If
x^2+y^2=2,
and dx/dt=5 when x=1 and y=1, what is dy/dt?
2x dx/dt=-2y dy/dt
dy/dt = -x/y * dx/dt
To find dy/dt, we need to differentiate the equation x^2 + y^2 = 2 with respect to t.
Differentiating both sides of the equation, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since dx/dt = 5 when x = 1 and y = 1, we can substitute these values into the equation:
2(1)(5) + 2(1)(dy/dt) = 0
10 + 2(dy/dt) = 0
2(dy/dt) = -10
Now, we can solve for dy/dt by dividing both sides of the equation by 2:
(dy/dt) = -10/2
(dy/dt) = -5
Therefore, the value of dy/dt is -5.
To find dy/dt, we need to differentiate both sides of the equation x^2 + y^2 = 2 with respect to t.
Differentiating x^2 + y^2 = 2 with respect to t using the chain rule, we get:
2x*dx/dt + 2y*dy/dt = 0
We know dx/dt = 5 and x = 1 and y = 1, so we can substitute these values into the equation:
2(1)(5) + 2(1)(dy/dt) = 0
10 + 2(dy/dt) = 0
2(dy/dt) = -10
Dividing both sides by 2, we get:
dy/dt = -10/2
dy/dt = -5
Therefore, dy/dt = -5.