Consider curve given by
x^2+ 4y^2= 7 + 3xy..
Three part question
A) show that dy/dx = 3y-2x/8y-3x
B) Show that there is a point P with
x-coordinate 3 at whiich the line tangent to the curve at P is horizontal. Find the y-coordinate of P.
C) find the value of d^2y/ dx^2 at the point p found in part (B)
2x dx+8ydy=3ydx + 3xdy
solve for dy/dx
Then, take the next derivative.
for B), put x=3 into the original equation, find y. Then find the slope (dy/dx) at that point.
To solve this problem, we will need to find the derivative of the given equation, use that to find the slope of the tangent line, and then find the second derivative to determine the concavity at a specific point.
A) To find dy/dx, we need to implicitly differentiate the equation x^2 + 4y^2 = 7 + 3xy with respect to x.
Differentiating both sides with respect to x, we get:
2x + 8y(dy/dx) = 3x(dy/dx) + 3y
Rearranging the terms, we get:
(8y - 3x) * (dy/dx) = 3y - 2x
Finally, dividing both sides by 8y - 3x, we get the derivative:
dy/dx = (3y - 2x) / (8y - 3x)
B) To find the x-coordinate of the point P where the tangent line is horizontal, we set dy/dx = 0 and solve for x.
Setting (3y - 2x) / (8y - 3x) = 0, we get:
3y - 2x = 0
We substitute x = 3 into the equation and solve for y:
3y - 2(3) = 0
3y - 6 = 0
3y = 6
y = 2
So the point P has x-coordinate 3 and y-coordinate 2.
C) To find d^2y/dx^2 at the point P, we need to find the second derivative of the given equation, and then evaluate it at the point P.
Taking the derivative of dy/dx = (3y - 2x) / (8y - 3x) with respect to x, we get:
d^2y/dx^2 = (9(dy/dx) - 2) / (8y - 3x)
Now, substitute x = 3 and y = 2 into this equation to find the value at point P:
d^2y/dx^2 = (9((3(2) - 2(3)) / (8(2) - 3(3)) - 2) / (8(2) - 3(3))
Simplifying this expression will give the value of d^2y/dx^2 at the point P.