Given that y=x^(5/2)- x^(-1/2) .
a) Find the coordinates where the tangent line is parallel to the line y-4x-4=0.
b) Find the coordinates where the tangent line is perpendicular to the line 2y-3x-12=0.
First find the slope of the tangent line = y'
y' = 5/2 * x^(3/2) + 1/2 * x^(-3/2)
The line in question has slope m=4
So, you want y'=4
5/2 * x^(3/2) + 1/2 * x^(-3/2) = 4
let u = x^(3/2) We can rewrite that as
5u/2 + 1/2u = 4
Multiply by 2u
5u^2 + 1 = 8u
5u^2 - 8u + 1 = 0
u = 1.46332 or 0.13667
so, x = u^(2/3) = 1.2889 or .26532
Evaluate y for those values of x to get the actual points.
for the perpendicular line, the slope m = -2/3. Do the same steps.
To find the coordinates where the tangent line of the function y=x^(5/2)- x^(-1/2) is parallel or perpendicular to a given line, we need to find the derivative of the function and find the slope of the given line. Then we can use the relationship between slopes of parallel or perpendicular lines to find the points of tangency.
a) To find the coordinates where the tangent line is parallel to the line y-4x-4=0:
1. Find the derivative of the function y=x^(5/2)- x^(-1/2). Using the power rule, we differentiate each term separately:
- The derivative of x^(5/2) is (5/2)x^(5/2-1) = (5/2)x^(3/2).
- The derivative of x^(-1/2) is (-1/2)x^(-1/2-1) = (-1/2)x^(-3/2).
Therefore, the derivative of y is dy/dx = (5/2)x^(3/2) - (1/2)x^(-3/2).
2. Set dy/dx equal to the slope of the given line to find the x-coordinate(s) where the tangent line is parallel.
The slope of the line y-4x-4=0 is the coefficient of x, which is 4.
We set dy/dx = 4 and solve for x:
(5/2)x^(3/2) - (1/2)x^(-3/2) = 4.
We can simplify this equation by multiplying through by 2 to eliminate the fractions:
5x^(3/2) - x^(-3/2) = 8.
Now, let's solve for x. We can rearrange the equation as follows:
5x^(3/2) - 8 = x^(-3/2).
To simplify further, let's introduce a substitution. Let u = x^(1/2).
This substitution simplifies the equation to:
5u^3 - 8 = 1/u^3.
Multiply through by u^3 to eliminate the fraction:
5u^6 - 8u^3 - 1 = 0.
Now we have a cubic equation in terms of u. We can use numerical methods or algebraic techniques to solve for u and then back substitute to find the values of x.
b) To find the coordinates where the tangent line is perpendicular to the line 2y-3x-12=0:
1. Find the derivative of the function y=x^(5/2)- x^(-1/2):
dy/dx = (5/2)x^(3/2) - (1/2)x^(-3/2).
2. Find the slope of the given line by rearranging the equation in slope-intercept form (y=mx+b):
2y - 3x - 12 = 0
2y = 3x + 12
y = (3/2)x + 6.
The slope of this line is the coefficient of x, which is 3/2.
3. Determine the slope of the perpendicular line by taking the negative reciprocal of the slope of the given line.
The negative reciprocal of 3/2 is -2/3.
4. Set dy/dx equal to the negative reciprocal of the slope of the given line (-2/3) to find the x-coordinate(s) where the tangent line is perpendicular.
(5/2)x^(3/2) - (1/2)x^(-3/2) = -2/3.
Similarly to part a), you would need to manipulate the equation and solve for x, which may involve substitutions and solving a cubic equation.
Once you find the x-coordinates for both cases (parallel and perpendicular), you can substitute them back into the original equation (y=x^(5/2)- x^(-1/2)) to find the corresponding y-coordinates for the points of tangency.