Question

Find the equation of the line tangent to the path of a particle traveling in the xy-plane with position vector r(t)=<t^3 +2t, t^2> at the point (3, 1).

Answer 1

x = t^3 + 2t

y = t^2

dx/dt = x velocity = 3 t^2 + 2
dy/dt = y velocity = 2 t
tangent slope = dy/dt / dx/dt
= 2t/(3t^2+2)

now what is t when x = 3 and y = 1?
t^2 = 1 = 1
so t = 1 (or -1 but we will see that won't work:)
1^3 + 2*1 = 3, sure enough
so slope = 1/4
y = (1/4) x + b
1 = (1/4)(3) + b
4/4 = 3/4 + b
b = 1/4
so
4y = x + 1

Answer 2

But the answer choices are

y=2/5x + 11/5
y=6/11x - 7/11
y=y=2/5x + 13/5
y=2/5x - 1/5
y=6/11x - 18/11

Answer 3

Why didn't you post the answer choices before? Damon went to all of this work without having all of the necessary information.

Answer 4

Well shouldn't Damon have known what to do and not need answer choices?

Answer 5

Oh, the slope is 2/4 = 1/2

You can do it from there I think

Answer 6

tangent slope = dy/dt / dx/dt

= 2t/(3t^2+2)

when t = 1
that is 2/(3+2) = 2/5

Answer 7

[AI-generated answer — Explain Bot]

To find the equation of the line tangent to the path of the particle at the given point, we need to find the derivative of the position vector.

The position vector of the particle is given by r(t) = <t^3 + 2t, t^2>.

To find the derivative, we differentiate each component of the vector with respect to t.

Differentiating the x-component of the position vector:
dx/dt = d/dt (t^3 + 2t)
= 3t^2 + 2

Differentiating the y-component of the position vector:
dy/dt = d/dt (t^2)
= 2t

So, the derivative of the position vector is r'(t) = <3t^2 + 2, 2t>.

To find the slope of the tangent line at the point (3, 1), we substitute t = 3 into the derivative:

m = (dy/dt)/(dx/dt)
= (2(3))/(3(3)^2 + 2)
= 6/(27 + 2)
= 6/29

Therefore, the slope of the tangent line at the point (3, 1) is m = 6/29.

To find the equation of the tangent line, we use the point-slope form of a line:

y - y1 = m(x - x1)

Substituting the values of (x1, y1) = (3, 1) and m = 6/29, we get:

y - 1 = (6/29)(x - 3)

Simplifying the equation further, we can write it in the slope-intercept form:

y = (6/29)x - 18/29 + 29/29
y = (6/29)x + 11/29

Therefore, the equation of the tangent line to the path of the particle at the point (3, 1) is y = (6/29)x + 11/29.