Given the parametric equations: x=t - 1/t, y=t + 1/t
Find the slope of the tangent to the curve at (8/3,10/3)
Find the area of the region bounded by the given curve and the line y=5/2
1. Find the slope of the tangent to the curve at (8/3, 10/3).
First, we will need to find the derivative of y with respect to x (dy/dx). To do this, we will first find dx/dt and dy/dt.
Given the parametric equation x = t - 1/t and y = t + 1/t, we can differentiate both equations with respect to t.
(dx/dt) = d(t - 1/t)/dt = 1 + 1/t^2
(dy/dt) = d(t + 1/t)/dt = 1 - 1/t^2
Now, we can find dy/dx by dividing (dy/dt) by (dx/dt):
dy/dx = (dy/dt) / (dx/dt) = (1 - 1/t^2) / (1 + 1/t^2)
At the point (8/3, 10/3), we need to find the value of t. Using the x-parametric equation, we can solve for t when x = 8/3:
8/3 = t - 1/t
t^2 - 8/3*t + 1 = 0
Multiplying by 3 to clear the denominator:
3*t^2 - 8t + 3 = 0
Factoring:
(3t - 1)(t - 3) = 0
So t = 1/3 or t = 3. Using the y-parametric equation, we can find the corresponding y-coordinate for each t-value:
t = 1/3: y = (1/3) + 3 = 10/3
t = 3: y = 3 + 1/3 = 10/3
Both values of t give the same point (8/3, 10/3). We can use either t-value to find the slope of the tangent. We will choose t = 1/3 to find the slope:
dy/dx = (1 - 1/t^2) / (1 + 1/t^2)
dy/dx = (1 - 1/(1/3)^2) / (1 + 1/(1/3)^2)
dy/dx = (1 - 9) / (1 + 9)
dy/dx = -8/10
The slope of the tangent to the curve at (8/3, 10/3) is -8/10, or -4/5.
2. Find the area of the region bounded by the given curve and the line y = 5/2.
To find the area of the region, we can use the calculus technique of finding the integral of y*dx/dt from the points of intersection.
First, we need to find the points of intersection between the curve and the line y = 5/2. Using the y-parametric equation, we can find the values of t when y = 5/2:
t + 1/t = 5/2
t^2 + 1 = 5t/2
4t^2 - 10t + 4 = 0
2t^2 - 5t + 2 = 0
The quadratic equation has two solutions for t:
t1 = 1/2 and t2 = 2
Now, we can find the integral of y*dx/dt from t = t1 to t = t2:
Area = ∫[y*dx/dt] dt from 1/2 to 2
Area = ∫[(t + 1/t)((1 - 1/t^2)/(1 + 1/t^2))] dt from 1/2 to 2
This integral is difficult to calculate by hand, but can be found using a calculator or online tool. The area of the region bounded by the curve and the line y = 5/2 is approximately 1.528 square units.