1. Let y = f(x) be the continuous function that satisfies the equation x^4-5x^2y^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 an expression for yβ
b. Write an equation for line l
differentiate implicitly:
4x^3 - 10xy^2 - 10x^2yy' + 16y^3y' = 0
y'(16y^3 - 10x^2y) = 10xy^2 - 4x^3
y' = x(5y^2 - 2x^2) / y(8y^2 - 5x^2)
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y'(2) = 2(5-8) / (8-20) = -6/-12 = 1/2
Line l is (y-1)/(x-2) = 1/2
a. To find an expression for y', we need to differentiate the given equation with respect to x using implicit differentiation.
The given equation is x^4 - 5x^2y^2 + 4y^4 = 0.
Differentiating both sides with respect to x, we get:
4x^3 - 10xy^2 - 10x^2yy' + 16y^3y' = 0.
Rearranging the equation, we have:
10x^2yy' + 16y^3y' = 4x^3 - 10xy^2.
Factoring out y' from the left side, we have:
y'(10x^2 + 16y^3) = 4x^3 - 10xy^2.
Dividing both sides by (10x^2 + 16y^3), we get:
y' = (4x^3 - 10xy^2) / (10x^2 + 16y^3).
So, the expression for y' is y' = (4x^3 - 10xy^2) / (10x^2 + 16y^3).
b. To write the equation for the line l tangent to the graph of f at x = 2, we need to find the slope of the tangent line and a point it passes through.
Substituting x = 2 into the given equation, we get:
(2)^4 - 5(2)^2y^2 + 4y^4 = 0.
16 - 20y^2 + 4y^4 = 0.
Factoring the equation, we get:
4(y^2 - 2)(y^2 - 2) = 0.
(y^2 - 2)^2 = 0.
Taking the square root of both sides, we get:
y^2 - 2 = 0.
y^2 = 2.
y = Β±β2.
The point (2, 1) lies on the graph of f, so we can use it to determine the appropriate value of y.
Since the line l is tangent to the graph at x = 2, it will have the same slope as the derivative at x = 2.
Substituting x = 2 and y = 1 into the expression for y', we get:
y' = (4(2)^3 - 10(2)(1)^2) / (10(2)^2 + 16(1)^3).
y' = (32 - 20) / (40 + 16).
y' = 12 / 56.
y' = 3 / 14.
So, the slope of the tangent line l is 3/14.
Using the point-slope form of a linear equation, we can write the equation for line l as:
y - 1 = (3/14)(x - 2).
Rearranging the equation, we have:
y = (3/14)x + (2/7) - (3/14).
Simplifying, we get:
y = (3/14)x + (4/7).
Therefore, the equation for line l is y = (3/14)x + (4/7).
To find the expression for y', we need to find the derivative of y with respect to x. Let's differentiate the given equation implicitly:
x^4 - 5x^2y^2 + 4y^4 = 0
Differentiating with respect to x, we get:
4x^3 - 10x^2y^2 - 10x^2yy' + 16y^3y' = 0
Rearranging the terms, we have:
4x^3 - 10x^2yy' + 16y^3y' = 10x^2y^2
Factoring out y' from the second and third terms, we get:
y'(4x^3 - 10x^2y + 16y^3) = 10x^2y^2
Dividing both sides by (4x^3 - 10x^2y + 16y^3) gives us the expression for y':
y' = (10x^2y^2) / (4x^3 - 10x^2y + 16y^3)
Now let's move on to part b: finding the equation for the line l tangent to the graph of f at x = 2.
Since the line is tangent to the graph of f at x = 2, it will have the same slope as the derivative of f at x = 2, which is y'(2).
Using the expression we found for y', we can substitute x = 2 to find y'(2):
y'(2) = (10(2)^2(1)^2) / (4(2)^3 - 10(2)^2(1) + 16(1)^3)
y'(2) = 40 / (32 - 40 + 16)
y'(2) = 40 / 8
y'(2) = 5
So, the slope of the tangent line l is 5.
Now, we can use the point-slope form of a line to write the equation for line l using the slope and the point (2, 1):
y - 1 = 5(x - 2)
Expanding and simplifying the equation gives us the final answer:
y = 5x - 9
Therefore, the equation for the line l tangent to the graph of f at x = 2 is y = 5x - 9.