The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).
y=1/(1+x^2)
Some call it witch of Maria Agnesi. She did not name it that, the name was a mistranslation of her work.
y'=dy/dx (1+x^2)^-1= -(1+x)^-2 * 2x
y'=-2x y^2
Put in x,y and you have it.
I am wondering..what was so difficult about this?
I just didn't have time to solve it, I had to run to class and now I only have a minute to enter all the correct answers, thank you
To find the equation of the tangent line to the curve y = 1/(1+x^2) at the point (-2, 1/5), we need to determine the slope of the tangent line at that point.
Step 1: Find the derivative of the function y = 1/(1+x^2).
To find the derivative, we can apply the quotient rule:
y' = (d/dx) [1/(1+x^2)]
= [(d/dx)(1)](1+x^2) - 1(1+x^2)' / (1+x^2)^2
= 0(1+x^2) - 2x / (1+x^2)^2
= -2x / (1+x^2)^2
Step 2: Substitute x = -2 into the derivative.
To find the slope of the tangent line at x = -2, substitute x = -2 into the derivative we found:
m = -2(-2) / (1+(-2)^2)^2
= 4 / (1+4)^2
= 4 / 25
So, the slope of the tangent line at the point (-2, 1/5) is 4/25.
Step 3: Use the point-slope form to find the equation of the tangent line.
Using the point-slope form equation, y - y1 = m(x - x1), where (x1, y1) is the given point (-2, 1/5) and m is the slope we found:
y - 1/5 = (4/25)(x - (-2))
y - 1/5 = (4/25)(x + 2)
y - 1/5 = (4/25)x + 8/25
y = (4/25)x + 8/25 + 1/5
y = (4/25)x + 8/25 + 5/25
y = (4/25)x + 13/25
Therefore, the equation of the tangent line to the curve y = 1/(1+x^2) at the point (-2, 1/5) is y = (4/25)x + 13/25.
To find the equation of the tangent line to the curve at the point (-2, 1/5), you can use the concept of differentiation. The derivative of a function represents the slope of the tangent line at any given point.
1. Start by finding the derivative of the function y = 1/(1 + x^2). To do this, use the power rule and the chain rule.
dy/dx = d/dx(1/(1+x^2))
= -2x / (1 + x^2)^2
2. Now, substitute the x-coordinate (-2) of the given point into the derivative equation to find the slope of the tangent line at that point.
dy/dx = -2 * (-2) / (1 + (-2)^2)^2
= 4 / (1 + 4)^2
= 4 / 25
3. The slope of the tangent line at the point (-2, 1/5) is 4/25. This gives us the slope-intercept form of the equation of a line, y = mx + b, where m is the slope and b is the y-intercept.
4. Substitute the coordinates of the point (-2, 1/5) into the equation to solve for b.
1/5 = (4/25)(-2) + b
1/5 = -8/25 + b
1/5 + 8/25 = b
5/25 + 8/25 = b
13/25 = b
5. The y-intercept is 13/25, so the equation of the tangent line is y = (4/25)x + 13/25.