Both x and y denote functions of t that are related by the given equation. Use this equation and the given derivative information to find the specified derivative.
Equation:35x^2+20y^2=1
Given that dx/dt=4, find dy/dt
when (x,y)=(1/7sqrt2, 1/5sqrt2)
To find dy/dt, we need to differentiate both sides of the equation with respect to t.
Differentiating the equation 35x^2 + 20y^2 = 1 with respect to t, we get:
70x(dx/dt) + 40y(dy/dt) = 0
Plugging in the values dx/dt = 4, and the given coordinates (x,y) = (1/7√2, 1/5√2), we can solve for dy/dt.
70(1/7√2)(4) + 40(1/5√2)(dy/dt) = 0
Simplifying the equation we get:
40(1/5√2)(dy/dt) = - 40(1/7√2)(4)
Canceling out the common terms:
dy/dt = - (4/5)(1/7)
Simplifying further:
dy/dt = -(4/35)
Hence, dy/dt = -4/35.
To find dy/dt using the given equation and derivative information, we need to take the derivative of both sides of the equation and then solve for dy/dt.
First, let's take the derivative of both sides of the equation 35x^2 + 20y^2 = 1 with respect to t. Remember that we need to apply the chain rule when differentiating y since it is a function of t.
d/dt (35x^2 + 20y^2) = d/dt (1)
70x(dx/dt) + 40y(dy/dt) = 0 (since d/dt (1) is 0)
Now, substitute the given value of dx/dt = 4 and the given point (x, y) = (1/7√2, 1/5√2) into the equation to solve for dy/dt.
70(1/7√2)(4) + 40(1/5√2)(dy/dt) = 0
(4/√2) + (8/√2)(dy/dt) = 0
Simplifying this equation further:
4 + 8(dy/dt) = 0
8(dy/dt) = -4
dy/dt = -4/8
dy/dt = -1/2
So, the value of dy/dt is -1/2.
70 x dx/dt + 40 y dy/dt = 0
280 x + 40 y dy/dy = 0
280/7sqrt2 + 40/5sqrt2 dy/dt = 0