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 .

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.