Find the point on the curve x=4y-y^2 where the tangent to the curve is a vertical line.
My work: Finding the derivative.
1=4(dy/dx)-2y(dy/dx)
1=dy/dx(4-2y)
dy/dx=1/4-2y
Therefore, y cannot equal +2 or -2
Right?
take the derivative respect to y not x
if you graph it its a sideway parabola.
x=4y-y^2
x'=4-2y
when derivative = 0 you have the min of the function.
0=4-2y
-4=-2y
2=y
plug 2 in to the equation
x=4y-y^2
x=4(2)-(2)^2=4
vertical line at point(4,2)
x=4 is vertical tangent to the curve.
Almost correct! When finding the derivative, you correctly differentiate both sides of the equation. However, in the last step, it seems like you made a mistake in simplifying the expression.
The derivative of x with respect to y (dx/dy) is simply 1 since x is not a function of y.
The derivative of 4y - y^2 with respect to y (dy/dy) is 4 - 2y.
Combining these two derivatives, we get:
1 = (4 - 2y) * (dx/dy)
To find the point on the curve where the tangent is a vertical line, we need to find the value of y that makes dx/dy undefined. In other words, the value of y that makes 4 - 2y equal to 0.
Thus, solving the equation:
4 - 2y = 0
2y = 4
y = 2
So, the point on the curve x = 4y - y^2 where the tangent to the curve is a vertical line is when y = 2.
Yes, your work is correct up to this point. You correctly found the derivative of the equation x = 4y - y^2 with respect to x. However, there is a slight mistake in your last line.
The derivative you found, dy/dx = 1/4 - 2y, represents the slope of the tangent line at any given point on the curve. In order for the tangent line to be vertical, the slope must be undefined, meaning the denominator must be zero.
Setting the denominator equal to zero:
4 - 2y = 0
2y = 4
y = 2
So, y = 2 is the critical point where the slope becomes undefined and the tangent line is vertical.
Therefore, the point on the curve x = 4y - y^2 where the tangent to the curve is a vertical line is (x, y) = (8, 2).