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
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.
isnt the derivative of r(t)= -2costsint i + 2cost j + -8sin2t k?
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.
sorry i meant r'(t)= -2costsint i + 2cost j + -8sin2t k
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.
To find the velocity, speed, and acceleration of the point at t=3π/4, we need to find the derivatives of the position vector function with respect to time (t).
1. Velocity:
To find the velocity, we need to take the derivative of the position vector function, r(t), with respect to time (t). The derivative of r(t) gives us the velocity vector, v(t).
So, let's find the derivative of r(t):
r(t) = cos^2(t) i + 2sin(t) j + 4cos(2t) k
Differentiating each component of r(t) with respect to t, we get:
r'(t) = -2sin(t) cos(t) i + 2cos(t) j - 8sin(2t) k
Therefore, the velocity vector at t=3π/4 is given by:
v(t) = -2sin(3π/4) cos(3π/4) i + 2cos(3π/4) j - 8sin(2(3π/4)) k
= -√2/2 * (-√2/2) i + √2/2 j - 8 * (-√2/2) k
= 1 i + √2/2 j + 4√2 k
So, the velocity vector at t=3π/4 is 1 i + √2/2 j + 4√2 k.
2. Speed:
The speed of a moving point is the magnitude of its velocity vector. To find the speed at t=3π/4, we need to find the magnitude of the velocity vector.
The magnitude (speed) of the velocity vector can be found using the formula:
Speed = √[ (v_x)^2 + (v_y)^2 + (v_z)^2 ]
Substituting the respective values, we get:
Speed = √[ (1)^2 + (√2/2)^2 + (4√2)^2 ]
= √[ 1 + 1/2 + 32 ]
= √[ 33/2 ]
= √(33)/√2
= √(33)/√2 * √2/√2
= √(66)/2
Therefore, the speed at t=3π/4 is √(66)/2.
3. Acceleration:
To find the acceleration, we need to take the derivative of the velocity vector function (v(t)) with respect to time (t).
Differentiating each component of v(t) with respect to t, we get:
v'(t) = 0 i - 2sin(t) i - 8cos(2t) k
Therefore, the acceleration vector at t=3π/4 is given by:
a(t) = 0 i - 2sin(3π/4) i - 8cos(2(3π/4)) k
= 0 i + 2/√2 i - 8cos(3π/2) k
= 0 i + 2/√2 i + 8k
= 0 i + √2 i + 8k
So, the acceleration vector at t=3π/4 is √2 i + 8k.