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