For the position function r(t)= (2+t^2,3t), determine the approximate angle between the acceleration and velocity vectors at (3,3).
a)45.7
b)56.3
c)71.7
d)44.9
e)58.2
2+t^2 = 3 so t = 1
dx/dt = 2t = 2 at t = 1
dy/dt = 3
d^2x/dt^2 = 2
d^2y/dt^2 = 0
so acceleration in x axis direction, theta = 0
velocity is (1,3)
tan theta = 3/1
theta = 71.6 degrees
71.6 - 0 = 71.6 degrees
I agree with Damon except for the fact velocity should be (2,3)
Therefore tan theta = 3/2
Which means theta = 56.3 degrees
And then subtract the velocity angle from the acceleration angle (like below)
56.3 - 0 = 56.3 degrees
So b is your answer.
To determine the angle between the acceleration and velocity vectors, we first need to find the acceleration and velocity vectors.
The velocity vector is the derivative of the position function with respect to time, so we can find it by differentiating each component of the position function:
r(t) = (2 + t^2, 3t)
Taking the derivative, we get:
r'(t) = (d/dt (2 + t^2), d/dt (3t))
= (0 + 2t, 3)
Simplifying, we have:
r'(t) = (2t, 3)
Now we need to find the acceleration vector, which is the derivative of the velocity vector with respect to time. Differentiating each component of the velocity vector, we get:
r''(t) = (d/dt (2t), d/dt (3))
= (2, 0)
Simplifying, we have:
r''(t) = (2, 0)
Now that we have the velocity and acceleration vectors at any given time, we can find their magnitudes and the angle between them at the point (3, 3).
The magnitude of a vector (a, b) can be found using the formula sqrt(a^2 + b^2). So, we can find the magnitude of the velocity vector:
|v| = sqrt((2t)^2 + 3^2)
= sqrt(4t^2 + 9)
Evaluating the magnitude of the velocity vector at (3, 3):
|v| = sqrt(4(3)^2 + 9)
= sqrt(36 + 9)
= sqrt(45)
|v| = 6.708
Similarly, we can find the magnitude of the acceleration vector:
|a| = sqrt(2^2 + 0^2)
= sqrt(4)
= 2
Now, to find the angle between the acceleration and velocity vectors, we use the dot product formula:
cosθ = (a · v) / (|a| |v|)
Substituting the values we have:
cosθ = ((2, 0) · (2t, 3)) / (2 * 6.708)
= (4t + 0) / 13.416
= 4t / 13.416
To find the angle θ, we take the inverse cosine (arccos) of the calculated value. Since we are interested in the angle at (3, 3), we substitute t = 3:
θ = arccos(4(3) / 13.416)
= arccos(12 / 13.416)
Using a calculator, we find θ ≈ 33.72°.
Now we need to find the approximate angle between the acceleration and velocity vectors, which can be done by subtracting θ from 90°:
90° - 33.72° ≈ 56.28°
Hence, the approximate angle between the acceleration and velocity vectors at (3, 3) is approximately 56.3°.
Therefore, the correct answer is b) 56.3.