The curve y = x/(1 + x^2) is called a serpentine. Find an equation of the tangent line to this curve at the point(2, 0.40).(Round the slope and y-intercept to two decimal places.)
use the quotient rule to get
dy/dx = [(1+x^2)(1) - x(2x)]/(1+x^2)^2
when x = 2,
dy/dx = [ 5 - 8]/25 = -3/25
equation of tangent:
y - .40 = (-3/25)(x-2)
25y - 10 = -3x + 6
25y = -3x + 16
y = (-3/25)x + 16/25
or
y = -.12x + .64 or 3x + 25y = 16
(why would you round anything? The numbers are exact)
I guess whoever wrote this problem didn't test it out first.
Well, well, a serpentine curve, huh? Sounds like a slippery situation! Let's see if we can untangle it.
To find the equation of the tangent line, we'll need to determine the slope and the y-intercept. Let's start with the slope.
The slope of the tangent line can be found by taking the derivative of the curve. So, let's do some math magic and differentiate the function y = x/(1 + x^2).
Using the quotient rule, we get:
y' = [(1 + x^2)(1) - (x)(2x)] / (1 + x^2)^2
Simplifying that, we have:
y' = (1 + x^2 - 2x^2) / (1 + x^2)^2
y' = (-x^2 + 1) / (1 + x^2)^2
Now, let's substitute x = 2 into this lovely equation:
y' = (-(2)^2 + 1) / (1 + (2)^2)^2
= (-4 + 1) / (1 + 4)^2
= -3 / 25
Hence, the slope is -3/25. Now, let's find the y-intercept.
We know that the point (2, 0.40) lies on the tangent line. So, we substitute x = 2 and y = 0.40 into the original equation to find the corresponding y-intercept:
0.40 = 2/(1 + 2^2)
0.40 = 2/5
Oh, look at that! The point (2, 0.40) was actually on the curve itself! Therefore, the y-intercept of the tangent line is also 0.40.
Putting it all together, the equation of the tangent line is:
y = mx + b
y = (-3/25)x + 0.40
Voila! We've cracked the joke... I mean, the problem! I hope this tangentially amusing answer put a smile on your face.
To find the equation of the tangent line to the curve at the given point, we need to find the slope of the tangent line and then use the point-slope form of a line to write the equation.
Step 1: Find the derivative of the curve.
The slope of the tangent line at any point on the curve y = x/(1 + x^2) can be found by taking the derivative of the function with respect to x.
Let's find the derivative of y = x/(1 + x^2):
y = x/(1 + x^2)
To simplify, let's rewrite the equation as y = x * (1 + x^2)^(-1).
Using the chain rule, the derivative dy/dx can be found as follows:
dy/dx = (1 + x^2)^(-1) * 1 - x * d/dx(1 + x^2)^(-1)
dy/dx = (1 + x^2)^(-1) + x * (1 + x^2)^(-2) * 2x
Simplifying further, we get:
dy/dx = (1 + x^2)^(-1) + 2x^2 / (1 + x^2)^2
Step 2: Find the slope at the given point.
Substitute x = 2 into the derivative to find the slope at the given point (2, 0.40):
m = dy/dx = (1 + 2^2)^(-1) + 2(2)^2 / (1 + 2^2)^2
m = 1/5 + 2(4) / 25
m = 1/5 + 8/25
m = 5/25 + 8/25
m = 13/25
The slope of the tangent line at the point (2, 0.40) is 13/25.
Step 3: Write the equation of the tangent line.
Using the point-slope form of a line, we have:
y - y1 = m(x - x1)
Substituting the values: (x1, y1) = (2, 0.40) and m = 13/25, we get:
y - 0.40 = (13/25) * (x - 2)
Simplifying further, we get:
y - 0.40 = (13/25)x - (13/25) * 2
y - 0.40 = (13/25)x - 26/25
y = (13/25)x - 26/25 + 0.40
y = (13/25)x - 26/25 + 10/25
y = (13/25)x - 16/25
Therefore, the equation of the tangent line to the curve at the point (2, 0.40) is y = (13/25)x - 16/25.