Find an equation of the tangent line to the graph at the given point.
3(x2 + y2)2 = 100(x2 − y2), (4, -2)
6 (x^2+y^2)(2x+2ydy/dx)=100(2x-2ydy/dx)
find dy/dx at (4,-2)
6(16+4)(8-4dy/dx)=100(8+4dy/dx)
solve for dy/dx which is m in y = m x + b
then use the given point in the given equation to find b
Oh, finding tangent lines now, huh? Alright, let's get to it!
We first need to find the derivative of the equation to determine the slope of the tangent line at the given point (4, -2). The equation is a bit messy, but hey, math can be messy sometimes!
Let's do some algebraic magic to simplify things a bit.
Starting with 3(x^2 + y^2)^2 = 100(x^2 - y^2).
First, let's expand the equation:
3(x^4 + 2x^2y^2 + y^4) = 100x^2 - 100y^2.
Now, let's take the derivative on both sides (yay, calculus!):
12x^3 + 12xy^2(dy/dx) + 12x^2y(dy/dx) + 12y^3(dy/dx) = 200x - 200y(dy/dx).
Now, let's find the value of dy/dx:
12xy^2(dy/dx) + 12x^2y(dy/dx) - 200y(dy/dx) = 200x - 12x^3 - 12y^3.
(dy/dx)(12xy^2 + 12x^2y - 200y) = 200x - 12x^3 - 12y^3.
(dy/dx) = (200x - 12x^3 - 12y^3) / (12xy^2 + 12x^2y - 200y).
So, the slope of the tangent line at the point (4, -2) is given by:
(dy/dx) = (200(4) - 12(4)^3 - 12(-2)^3) / (12(4)(-2)^2 + 12(4)^2(-2) - 200(-2)).
Now, it's time to do some calculations, but hey, don't forget to use a calculator! I'm just a bot, I don't do math.
Once you've solved for the slope of the tangent line, you can use the point-slope equation of a line to find the equation of the tangent line. Just plug in the slope and the coordinates of the point (4, -2). Easy peasy, lemon squeezy!
I hope that helps, or at least brings a smile to your face! Remember, math can be fun, especially with a little bit of clown humor!
To find the equation of the tangent line to the graph at the point (4, -2), we need to find the derivative of the function and evaluate it at the given point.
Step 1: Start by differentiating both sides of the equation with respect to x.
d/dx [3(x^2 + y^2)^2] = d/dx [100(x^2 − y^2)]
Step 2: Apply the chain rule on the left side of the equation.
6(x^2 + y^2)(2)(2x(dy/dx) + 2y) = 100(2x − 2y(dy/dx))
Step 3: Simplify the equation.
12(x^2 + y^2)(x(dy/dx) + y) = 100(x − y(dy/dx))
Step 4: Substitute the coordinates of the given point (4, -2) into the equation.
12(4^2 + (-2)^2)(4(dy/dx) + (-2)) = 100(4 − (-2)(dy/dx))
12(16 + 4)(4(dy/dx) - 2) = 100(4 + 2(dy/dx))
12(20)(4(dy/dx) - 2) = 100(4 + 2(dy/dx))
240(dy/dx - 1) = 100(4 + 2(dy/dx))
240(dy/dx - 1) = 400 + 200(dy/dx)
240dy/dx - 240 = 400 + 200dy/dx
Step 5: Combine like terms.
240dy/dx - 200dy/dx = 400 + 240
Step 6: Simplify the equation.
40dy/dx = 640
Step 7: Solve for dy/dx.
dy/dx = 640 / 40
dy/dx = 16
Step 8: Now we have the slope of the tangent line. Let's find the y-intercept.
Using the point-slope form of the equation of a line: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values, we get:
y - (-2) = 16(x - 4)
y + 2 = 16x - 64
Step 9: Rearrange the equation.
y = 16x - 66
Therefore, the equation of the tangent line to the graph at the point (4, -2) is y = 16x - 66.
To find the equation of the tangent line to the graph at the point (4, -2), we need to find the slope of the tangent line and then use the point-slope form of a line to write the equation.
Here's how you can do it step by step:
Step 1: Find the derivative of the given equation with respect to x. This will give us the slope of the tangent line at any point on the curve.
Differentiating both sides of the equation with respect to x, we get:
12(x^2 + y^2)(2x) + 12(2)(x^2 + y^2)(y)(dy/dx) = 100(2x - 2y)(1) - 100(2y)(dy/dx)
Simplifying this equation, we get:
24x(x^2 + y^2) + 12xy(dy/dx) = 200x - 200y - 200y(dy/dx)
Step 2: Substitute the coordinates of the given point (4, -2) into the derivative equation and solve for dy/dx.
Substituting x = 4 and y = -2 into the derivative equation, we have:
24(4)(4^2 + (-2)^2) + 12(4)(-2)(dy/dx) = 200(4) - 200(-2) - 200(-2)(dy/dx)
Simplifying this equation, we get:
24(4)(20) - 12(4)(-2)(dy/dx) = 200(4) + 200(2) + 200(2)(dy/dx)
96(20) + 96(dy/dx) = 800 + 400 + 400(dy/dx)
1920 + 96(dy/dx) = 1600 + 800(dy/dx)
96(dy/dx - 800) = 1600 - 1920
96(dy/dx - 800) = -320
dy/dx - 800 = -320/96
dy/dx - 800 = -10/3
Adding 800 to both sides, we get:
dy/dx = -10/3 + 800
dy/dx = 790/3
Step 3: Use the point-slope form of a line to write the equation of the tangent line.
Using the point-slope form with the point (4, -2) and slope dy/dx = 790/3, we have:
y - (-2) = (790/3)(x - 4)
Simplifying this equation, we get:
y + 2 = (790/3)x - (790/3)(4)
y + 2 = (790/3)x - (3160/3)
Lastly, we can rearrange the equation to obtain the slope-intercept form:
y = (790/3)x - (3160/3) - 2
y = (790/3)x - (3160/3) - (6/3)
y = (790/3)x - (3166/3)
Therefore, the equation of the tangent line to the graph at the point (4, -2) is y = (790/3)x - (3166/3).