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
a. Differentiate implicitly.
4x^3 -10x*y^2 -5x^2*2y*dy/dx +16y^3*dy/dx = 0
4x^3 -10x*y^2 -10 x^2*y*dy/dx +16y^3*dy/dx = 0
Plug in x = 2 and y = 1 to get dy/dx.
4 -20 -40 dy/dx +16 dy/dx = 0
-24 dy/dx = 16
dy/dx = -2/3
b. (y - 1)/(x-2) = -2/3
Check my work.
a. Oh, boy! We've got ourselves a math problem. To find the expression for y', we need to use the chain rule. Are you ready for some wild derivatives? Here we go!
First, let's rearrange the given equation:
x^4 - 5x^2y^2 + 4y^4 = 0
Differentiating implicitly with respect to x:
4x^3 - 10xy^2 - 10x^2yy' + 16y^3y' = 0
Now, let's plug in the point (2,1):
4(2)^3 - 10(2)(1)^2 - 10(2)^2(1)y' + 16(1)^3y' = 0
After some calculations:
32 - 40 - 40y' + 16y' = 0
-8y' = 8
y' = -1
So the expression for y' is y' = -1. Easy peasy!
b. Now let's find the equation for the line tangent to the graph at x = 2. Let's call this line "l". To find the slope of the tangent line, we can use the value of y' we just found.
We know that the slope of a line is given by y' at any given point. So the slope of line l is -1.
Using the point (2,1) on the line, we can use the point-slope form to find the equation:
y - y1 = m(x - x1)
y - 1 = -1(x - 2)
y - 1 = -x + 2
y = -x + 3
Voila! The equation for line l is y = -x + 3. Good luck with all these math shenanigans!
To find the expression for y', we need to differentiate the given equation with respect to x. Let's differentiate both sides of the equation:
d/dx (x^4 - 5x^2y^2 + 4y^4) = d/dx (0)
Using the chain rule and product rule, we can differentiate each term:
4x^3 - 10x^2y^2 - 10x^2yy' + 16y^3y' = 0
Rearranging the terms, we get:
y'(10x^2y + 16y^3) = 4x^3 - 10x^2y^2
Dividing both sides by (10x^2y + 16y^3), we obtain:
y' = (4x^3 - 10x^2y^2)/(10x^2y + 16y^3)
To find the equation for line l, a tangent line to the graph at x = 2, we need to find both the slope and a point on the line.
a. To find the slope, substitute x = 2 into the expression we found for y':
y' = (4(2)^3 - 10(2)^2y^2)/(10(2)^2y + 16y^3)
Simplifying,
y' = (32 - 40y^2)/(40y + 16y^3)
b. To find a point on the line, substitute x = 2 into the original equation, y = f(x):
(2)^4 - 5(2)^2y^2 + 4y^4 = 0
16 - 20y^2 + 4y^4 = 0
Dividing by 4, we get:
4 - 5y^2 + y^4 = 0
Rearranging, we have:
y^4 - 5y^2 + 4 = 0
To solve this equation, we can factor:
(y^2 - 4)(y^2 - 1) = 0
This gives us two possible solutions: y = Β±2 and y = Β±1.
Using the point (2, 1), we can substitute these values into the expression for y' to find the slope of the tangent line:
y' = (4(2)^3 - 10(2)^2(1)^2)/(10(2)^2(1) + 16(1)^3)
= (32 - 40)/(40 + 16)
= -8/56
= -1/7
Therefore, the slope of the tangent line at x = 2 is -1/7.
Using the point-slope form of a line, where the slope is -1/7 and the point (2, 1) lies on the line, we can write the equation:
y - 1 = (-1/7)(x - 2)
Simplifying,
y - 1 = (-1/7)x + 2/7
Adding 1 to both sides,
y = (-1/7)x + 2/7 + 7/7
= (-1/7)x + 9/7
Thus, the equation for line l is y = (-1/7)x + 9/7.
To find an expression for y', we need to take the derivative of the given equation with respect to x.
1. Taking the derivative of the equation x^4 - 5x^2y^2 + 4y^4 = 0 with respect to x, we get:
4x^3 - 10xy^2 - 10x^2yy' + 16y^3y' = 0.
2. Now, let's substitute the given point (2, 1) into the equation to find the value of y' at x = 2:
4(2)^3 - 10(2)(1)^2 - 10(2)^2(1)y' + 16(1)^3y' = 0.
Simplifying this expression gives:
32 - 20 - 40y' + 16y' = 0.
12 - 24y' = 0.
-24y' = -12.
y' = -12/(-24) = 1/2.
Hence, the expression for y' is y' = 1/2.
To write the equation for the line l, we need to find the slope and the y-intercept.
1. The slope of the line tangent to the graph of f at x = 2 is given by the value of y' at x = 2, which we found to be 1/2.
2. Now, let's substitute the point (2, 1) and the slope 1/2 into the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Using (2, 1) and m = 1/2, the equation becomes:
y - 1 = 1/2(x - 2).
3. Simplifying this equation gives:
2y - 2 = x - 2.
2y = x.
y = (1/2)x.
Hence, the equation for line l is y = (1/2)x.