Question

The position vector of a moving point P is given by vector function r(t)= cos^2 (t) i + 2sin(t) j + 4cos2t k. Find the velocity, speed, and acceleration of the point at t=3pi/4

Answer 1

isnt the derivative of r(t)= -2costsint i + 2cost j + -8sin2t k?

Answer 2

sorry i meant r'(t)= -2costsint i + 2cost j + -8sin2t k

Answer 3

We can find the velocity by taking the derivative of the position vector:

r'(t) = -2sin(t) i + 2cos(t) j - 16cos(t)sin(t) k

At t=3pi/4, we have:

r'(3pi/4) = -2sin(3pi/4) i + 2cos(3pi/4) j - 16cos(3pi/4)sin(3pi/4) k
= -sqrt(2) i - sqrt(2) j + 8 k

To find the speed, we need to find the magnitude of the velocity vector:

|v| = sqrt((-sqrt(2))^2 + (-sqrt(2))^2 + 8^2) = sqrt(18)

Finally, we can find the acceleration by taking the derivative of the velocity vector:

r''(t) = -2cos(t) i - 2sin(t) j - 16(cos^2(t) - sin^2(t)) k
= -2cos(t) i - 2sin(t) j - 16cos(2t) k

At t=3pi/4, we have:

r''(3pi/4) = -2cos(3pi/4) i - 2sin(3pi/4) j - 16cos(3pi/2) k
= sqrt(2) i - sqrt(2) j - 16 k

So the velocity at t=3pi/4 is -sqrt(2) i - sqrt(2) j + 8 k, the speed is sqrt(18), and the acceleration is sqrt(2) i - sqrt(2) j - 16 k.

Answer 4

Yes, you're correct. My apologies for the mistake in my previous response.

The position vector of a moving point is given by:

r(t) = -2cos(t)sin(t)i + 2cos(t)j - 8sin^2(t)k

Taking the derivative of the position vector r(t), we get the velocity vector v(t):

v(t) = r'(t) = [-2sin(t)sin(t) - 2cos(t)cos(t)]i + [-2sin(t)cos(t) + 2sin(t)cos(t)]j + [-16sin(t)cos(t)]k
= -2sin(2t)i - 16sin(t)cos(t)k

Now, to calculate the velocity, speed, and acceleration of the point at t = 3π/4, we substitute t = 3π/4 in the above expression of the velocity vector:

v(3π/4) = -2sin(3π/2)i - 16sin(3π/4)cos(3π/4)k
= 2i - 8√2k

The speed |v| is the magnitude of the velocity vector:

|v| = √(|2|^2 + |-8√2|^2) = 2√33

Taking the derivative of the velocity vector v(t), we get the acceleration vector a(t):

a(t) = v'(t) = -4cos(2t)i - 16(cos^2(t) - sin^2(t))k

Substituting t = 3π/4 in the above expression of the acceleration vector:

a(3π/4) = -4cos(3π/2)i - 16(cos^2(3π/4) - sin^2(3π/4))k
= 4i - 16k

Therefore, the velocity at t = 3π/4 is 2i - 8√2k, the speed is 2√33, and the acceleration is 4i - 16k.

Answer 5

You are correct, the position vector of a moving point is given by:

r(t) = -2cos(t)sin(t)i + 2cos(t)j - 8sin²(t)k

The velocity vector is:

r'(t) = (-2cos(t)cos(t) + 2sin(t)sin(t))i + (-2sin(t)cos(t) - 2cos(t)sin(t))j + (-16sin(t)cos(t))k
-2sin(2t)i - 2cos(2t)j - 16sin(t)cos(t)k

At t = 3π/4, we have:

r'(3π/4) = (-2cos(3π/4)cos(3π/4) + 2sin(3π/4)sin(3π/4))i + (-2sin(3π/4)cos(3π/4) - 2cos(3π/4)sin(3π/4))j - 16sin(3π/4)cos(3π/4)k
-2sin(3π/2)i - 2cos(3π/2)j - 16(-√2/2)(√2/2)k
2√2i + 2j + 8k

So the velocity at t = 3π/4 is 2√2i + 2j + 8k.

The speed is the magnitude of the velocity vector, which is:

|v| = √[(2√2)² + 2² + 8²] = 2√37

The acceleration vector is the derivative of the velocity vector:

r''(t) = -2cos(2t)i + 2sin(2t)j - 16cos(t)sin(t)k

At t = 3π/4, we have:

r''(3π/4) = -2cos(3π/2)i + 2sin(3π/2)j - 16(-√2/2)(√2/2)k
= 0i - 2j + 8k

So the acceleration at t = 3π/4 is -2j + 8k.

I apologize for any confusion caused by my previous response.