Aparticle moves along the plane curve c described by r(t)=ti+t2j. find the curvature of the plane curve at t=2. round your answer to two decimal places.

The curvature of the plan curve at t=2 is 0.

Step-by-step explanation:

First, we find take a derivative of r(t) = ti + t^2j

r'(t) = i + 2tj

we can also write it like r'(t) = < 1, 2t>

Now, we take a magnitude of it

|r'(t)| = √(1)^2 + (2t)^2

|r'(t)| = √17

Now, we assume a variable T(t)

T(t) = r'(t)/|r'(t)|

T(t) = (i + 2tj) / √17

T(t) = 1/√17 (< 1, 2t>)

Take derivative of T(t)

T'(t) = 1/√17 (< 0, 2>)

|T'(t)| = √(1/√17)^2 + (0)^2 + (2)^2

|T'(t)| = √69/17

Therefore,

Curvature = |T'(t)| / |r'(t)|

Curvature = (√69/17) /  √17

Curvature = √69 .

x+y = 29, 5x + 2y = 100

step-by-step explanation:

i'm guessing it was a multiple first option was the correct answer

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.