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.)
To find the equation of the tangent line to the curve at a given point, we need to find the slope of the curve at that point and the coordinates of the point.
First, let's find the derivative of the curve to find the slope at any point (x, y). The derivative of y with respect to x, denoted as dy/dx, can be found using the quotient rule.
The quotient rule states that if we have a function of the form y = f(x) / g(x), then the derivative dy/dx is given by:
dy/dx = (g(x) * f'(x) - f(x) * g'(x)) / [g(x)]^2
In this case, our function is y = x / (1 + x^2), so f(x) = x and g(x) = 1 + x^2.
Let's find the derivatives of f(x) and g(x):
f'(x) = 1 (since the derivative of x with respect to x is 1)
g'(x) = 2x (since the derivative of 1 + x^2 with respect to x is 2x)
Now, substituting these values into the quotient rule formula, we have:
dy/dx = [(1 + x^2) * 1 - (x) * (2x)] / [1 + x^2]^2
= (1 + x^2 - 2x^2) / [1 + x^2]^2
= (1 - x^2) / [1 + x^2]^2
Now, let's find the slope at the point (2, 0.40) by substituting x = 2 into the derivative:
dy/dx = (1 - 2^2) / [1 + 2^2]^2
= (1 - 4) / (1 + 4)^2
= -3 / 25
So, the slope of the curve at the point (2, 0.40) is -3/25.
To find the equation of the tangent line, we'll use the point-slope form of a line, which is:
y - y1 = m(x - x1)
Where (x1, y1) is the given point on the line, and m is the slope.
Plugging in the values from the point (2, 0.40) and the slope -3/25, we have:
y - 0.40 = (-3/25)(x - 2)
Expanding the equation:
y - 0.40 = (-3/25)x + (6/25)
Rearranging the equation to the slope-intercept form, we get:
y = (-3/25)x + (6/25) + 0.40
y = (-3/25)x + (6/25) + (10/25)
y = (-3/25)x + (16/25)
So, the equation of the tangent line to the curve at the point (2, 0.40) is y = (-3/25)x + (16/25).