Find the equation of the tangent to the curve at x = 3 for the parametric equations below:
x= t+1/t
y= t^2+1/t^2, with t>0
a) y= 2x+1
b) y= 6x-11
c) y= 6x-39
d) y= 2x-1
A seldom used trick:
let's look at the t + 1/t and the t^2 + 1/t^2
t^2 + 1/t^2 = (t + 1/t)^2 - 2 , (expand the right side to see that is true)
so if x = t + 1/t, then
y = (t+1/t)^2 - 2 = x^2 - 2
so you simply have the parabola y = x^2 - 2
when x = 3, y = 7
and dy/dx = 2x , which at the point (3,7) is 6
so we have a slope of 6 and the point (3,7)
tangent equation is y-7 = 6(x - 3)
I see that equation in the y = mx + b form
To find the equation of the tangent to the curve at x = 3, we need to find the derivative of y with respect to x and evaluate it at x = 3, and then use this to find the equation of the tangent line.
1. First, let's find dy/dt, the derivative of y with respect to t.
Given: x = t + 1/t, y = t^2 + 1/t^2
To find dy/dt, we will differentiate y with respect to t.
dy/dt = d/dt(t^2 + 1/t^2)
= d/dt(t^2) + d/dt(1/t^2)
= 2t - 2/t^3
2. Now, let's find dx/dt, the derivative of x with respect to t.
Given: x = t + 1/t, y = t^2 + 1/t^2
To find dx/dt, we will differentiate x with respect to t.
dx/dt = d/dt(t + 1/t)
= d/dt(t) + d/dt(1/t)
= 1 - 1/t^2
3. Next, we need to find dy/dx, the derivative of y with respect to x.
To find dy/dx, we will divide dy/dt by dx/dt.
dy/dx = (dy/dt) / (dx/dt)
= (2t - 2/t^3) / (1 - 1/t^2)
= (2t^3 - 2) / (t^3 - t^2)
4. Now, let's evaluate dy/dx at x = 3 to find the slope of the tangent line.
Since x = t + 1/t, when x = 3, we have:
3 = t + 1/t
3t = t^2 + 1
t^2 - 3t + 1 = 0
Using the quadratic formula, we can solve for t:
t = (3 ± √(3^2 - 4(1)(1))) / (2(1))
t = (3 ± √5) / 2
Since t > 0, we choose t = (3 + √5) / 2.
Substituting t = (3 + √5) / 2 into dx/dt, we have:
dx/dt = 1 - 1/((3 + √5) / 2)^2
= 1 - 4 / (9 + 6√5 + 5)
= 1 - 4 / (14 + 6√5)
= (1(14 + 6√5) - 4) / (14 + 6√5)
= (14 + 6√5 - 4) / (14 + 6√5)
= (10 + 6√5) / (14 + 6√5)
= (5 + 3√5) / (7 + 3√5)
Substituting t = (3 + √5) / 2 into dy/dt, we have:
dy/dt = 2((3 + √5) / 2)^3 - 2
= (3 + √5)^3 - 2
= 27 + 27√5 + 15 + 3√5 + 5√5 + √5 - 2
= 40 + 35√5
Substituting dx/dt and dy/dt into dy/dx, we have:
dy/dx = (40 + 35√5) / (10 + 6√5)
= (8 + 7√5) / (2 + √5)
= (8 + 7√5)(2 - √5) / (2 + √5)(2 - √5)
= (16 - 3√5 + 14√5 - 35) / (4 - 5)
= (-19 + 11√5) / -1
= 19 - 11√5
So, the slope of the tangent line at x = 3 is 19 - 11√5.
5. Finally, let's find the equation of the tangent line using point-slope form.
Using point-slope form, we have:
y - y1 = m(x - x1)
where (x1, y1) is a point on the curve (x = 3) and m is the slope of the tangent line.
Substituting x1 = 3, y1 = (3^2 + 1/3^2) = 10/3 into the equation, we have:
y - 10/3 = (19 - 11√5)(x - 3)
Simplifying further, we have:
y - 10/3 = (19 - 11√5)x - (57 - 33√5)
y = (19 - 11√5)x - 57 + 33√5 + 10/3
y = (19 - 11√5)x - (171 - 99√5) / 3
Therefore, the equation of the tangent to the curve at x = 3 is y = (19 - 11√5)x - (171 - 99√5) / 3.
To find the equation of the tangent to the curve at a specific point, you need to follow these steps:
1. Differentiate the parametric equations with respect to the parameter (t) to find the derivatives: dx/dt and dy/dt.
dx/dt = 1 - 1/t^2
dy/dt = 2t - 2/t^3
2. Find the value of t that corresponds to the given x-coordinate (x = 3).
Set x = 3 and solve for t:
3 = t + 1/t
Multiply both sides by t to remove the fraction:
3t = t^2 + 1
Rearrange the equation:
t^2 - 3t + 1 = 0
3. Use the quadratic formula to solve for t:
t = (3 ± √(3^2 - 4*1*1)) / (2*1)
t = (3 ± √(9 - 4)) / 2
t = (3 ± √5) / 2
Since t must be greater than 0 (as specified in the parametric equations), discard the negative value of t.
4. Substitute the value of t into the parametric equations to find the corresponding y-coordinate:
For t = (3 + √5) / 2,
y = ((3 + √5) / 2)^2 + 1 / ((3 + √5) / 2)^2
5. Calculate the slope of the tangent line by evaluating dy/dt at the specific value of t found above:
For t = (3 + √5) / 2,
slope, m = dy/dt = 2t - 2/t^3 = 2(3 + √5) / 2 - 2 / ((3 + √5) / 2)^3
6. Write the equation of the tangent line using the point-slope form (y - y₁ = m(x - x₁)) and the point (x₁, y₁) found above. Since x = 3, y = y₁, and the slope is m:
The equation of the tangent to the curve at x = 3 is y - y₁ = m(x - 3).
By substituting the value of y₁ and m into the equation, you can determine which option (a), (b), (c), or (d) matches the equation of the tangent.