The curve y =|x|/(sqrt(5−x^2))
is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (2,2).
see the post below for the way to get this done.
To find the equation of the tangent line to the curve at the point (2, 2), we need to find the derivative of the curve first.
Differentiating the function y = |x|/(sqrt(5−x^2)) with respect to x using the quotient rule:
Let f(x) = |x| and g(x) = sqrt(5 - x^2).
f'(x) = (1/|x|) * sign(x), where sign(x) is the sign function (returns -1 if x < 0, 1 if x > 0, and 0 if x = 0).
g'(x) = (-2x)/[2 * sqrt(5 - x^2)] = -x/sqrt(5 - x^2).
Using the quotient rule, we can find the derivative of y as:
dy/dx = [f'(x) * g(x) - f(x) * g'(x)] / [g(x)]^2
= [(1/|x|) * sign(x) * sqrt(5 - x^2) - |x| * (-x/sqrt(5 - x^2))] / [sqrt(5 - x^2)]^2
= [(sqrt(5 - x^2) + x^2) / |x|] / (5 - x^2)
= (sqrt(5 - x^2) + x^2) / (|x|(5 - x^2))
Now, let's evaluate the derivative dy/dx at x = 2, since we want the equation of the tangent line at the point (2, 2):
dy/dx (at x = 2) = (sqrt(5 - 2^2) + 2^2) / (|2|(5 - 2^2))
= (sqrt(1) + 4) / (2(5 - 4))
= (1 + 4) / 2
= 5/2
So, the slope of the tangent line at the point (2, 2) is 5/2.
Now, we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1), where (x1, y1) is the point (2, 2) and m is the slope (5/2).
Plugging in the values, we get:
y - 2 = (5/2)(x - 2)
Simplifying, we get:
y - 2 = (5/2)x - 5
Rewriting in slope-intercept form, we get:
y = (5/2)x - 3/2
Therefore, the equation of the tangent line to the curve y = |x|/(sqrt(5 - x^2)) at the point (2, 2) is y = (5/2)x - 3/2.
To find the equation of the tangent line to the bullet-nose curve at the point (2,2), we need to find its slope and then use the point-slope form of a line equation.
1. Start by finding the derivative of the bullet-nose curve function, which will give us the slope of the tangent line at any point.
- The bullet-nose curve is given by: y = |x| / (sqrt(5 - x^2))
- To find the derivative, we can split the function into two cases based on the sign of x:
- For x > 0, y = x / (sqrt(5 - x^2))
- For x < 0, y = -x / (sqrt(5 - x^2))
- In both cases, the derivative is given by the quotient rule:
- dy/dx = (1 * (sqrt(5 - x^2)) - x * (1 / 2) * (-2x)) / (sqrt(5 - x^2))^2
- Simplifying further, we get: dy/dx = (sqrt(5 - x^2) + x^2) / (sqrt(5 - x^2))^2
2. Evaluate the derivative at the point (2,2) to find the slope of the tangent line at that point.
- Substituting x = 2 into the derivative expression, we get: dy/dx = (sqrt(5 - 2^2) + 2^2) / (sqrt(5 - 2^2))^2
= (sqrt(5 - 4) + 4) / (sqrt(5 - 4))^2
= (sqrt(1) + 4) / (sqrt(1))^2
= (1 + 4) / 1
= 5
3. Now that we have the slope of the tangent line, we can use the point-slope form of a line equation and plug in the point (2,2) to find the equation of the tangent line.
- The point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
- Plugging in the values, we get: y - 2 = 5(x - 2)
= y - 2 = 5x - 10
= y = 5x - 8
So, the equation of the tangent line to the bullet-nose curve at the point (2,2) is y = 5x - 8.