Let y = f(x) be the continuous function that satisfies the equation x^4 - (5x^2)(y^2) + 4y^4 = 0 and whose graph contains the points (2,1) and (2,2). Let L be the line tangent to the graph of f at x = 2.
(a) Find and expression for y'.
(b) Write an equation for the line L.
(c) Give the coordinates of a point that is on the graph of f but is not on line L.
(d) Give the coordinates of a point that is on the line L but is not on the graph of f.
I think a0 is (-4x^3+10xy^2)/(-10x^2y+16y^3)
and that makes b y=1/2x. However, I can't seem to get anything for c nor d. How do I start?
To find the expression for y', the derivative of y with respect to x, we can differentiate the given equation implicitly.
(a) First, differentiate both sides of the equation with respect to x:
d/dx(x^4) - d/dx((5x^2)(y^2)) + d/dx(4y^4) = d/dx(0)
The derivative of x^4 with respect to x is 4x^3.
For the middle term, we need to apply the chain rule. Let's consider (5x^2)(y^2) as (f(x)g(x)), where f(x) = 5x^2 and g(x) = y^2.
Using the chain rule, the derivative of (f(x)g(x)) with respect to x is:
d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
Differentiating f(x) = 5x^2 with respect to x, we get f'(x) = 10x.
Differentiating g(x) = y^2 with respect to x gives g'(x) = 2y * dy/dx.
Now let's apply these results back into the equation:
4x^3 - (5x^2)(2y * dy/dx) + 4(4y^3 * dy/dx) = 0
Now, we can solve for dy/dx:
dy/dx = (5x^2 * 2y - 16y^3) / (4x^3 - 8y^2)
(b) To find the equation for the line L, we need a point and the slope. We are given that the line L is tangent to the graph of f at x = 2. Therefore, the point (2, f(2)) lies on the line.
Using the equation x^4 - 5x^2y^2 + 4y^4 = 0 and substituting x = 2, we can solve for y:
(2^4) - 5(2^2)(y^2) + 4y^4 = 0
16 - 20y^2 + 4y^4 = 0
Factoring out a common factor, we get:
4y^2(1 - y^2) = 0
This gives us two possible values for y: y = 0 or y = ±1.
For y = 0, the point is (2, 0), which is not on line L since L is tangent to the graph at x = 2.
For y = 1, the point is (2, 1), which is on line L.
For y = -1, the point is (2, -1), which is on line L.
Therefore, we can choose either (2, 1) or (2, -1) as the point on line L.
Now, we can find the slope of the tangent line by substituting x = 2 and y = 1 (or y = -1) into the expression for y':
dy/dx = (5x^2 * 2y - 16y^3) / (4x^3 - 8y^2)
Using y = 1 and x = 2:
dy/dx = (5(2^2) * 2(1) - 16(1^3)) / (4(2^3) - 8(1^2))
= (20 - 16) / (32 - 8)
= 4 / 24
= 1/6
So, the slope of line L is 1/6.
By using the point-slope form of a line, we can write the equation for line L:
y - y1 = m(x - x1)
Using the point (2, 1) or (2, -1) and the slope m = 1/6, we get:
For the point (2, 1):
y - 1 = (1/6)(x - 2)
For the point (2, -1):
y - (-1) = (1/6)(x - 2)
Either equation can be used as the equation for line L.
(c) To find a point that is on the graph of f but not on line L, we need to find a point that satisfies the equation x^4 - 5x^2y^2 + 4y^4 = 0 and is not on the line L.
One such point can be found by substituting different values of x and y in the equation and solving for y. For example, let's choose x = 0 and solve for y:
(0^4) - 5(0^2)(y^2) + 4y^4 = 0
0 - 0 + 4y^4 = 0
4y^4 = 0
This equation has only one solution, which is y = 0. Therefore, the point (0, 0) is on the graph of f but not on line L.
(d) To find a point that is on line L but not on the graph of f, we can choose any point on line L that is not on the graph of f.
Using the equation for line L (either y - 1 = (1/6)(x - 2) or y - (-1) = (1/6)(x - 2)), we can substitute different values of x and solve for y.
For example, let's choose x = 4 and use the equation y - 1 = (1/6)(x - 2):
y - 1 = (1/6)(4 - 2)
y - 1 = (1/6)(2)
y - 1 = 1/3
y = 1/3 + 1
y = 4/3
Therefore, the point (4, 4/3) is on line L but not on the graph of f.