1) Suppose that an insect crawls in a plane in a path given in polar coordinates by r = be^kt;� theta = kt;
where b; k are positive constants and t is time.Find the velocity and acceleration vectors of the insect. Show that the angle between the velocity and acceleration vectors is equal to �
pi/4 for all times t.
To find the velocity and acceleration vectors of the insect, we need to differentiate the given path equation with respect to time.
Given:
r = be^(kt)
θ = kt
Taking the derivative with respect to time, t, we get:
Dr/dt = d(be^(kt))/dt
= (b * k * e^(kt)) * dt/dt
= b * k * e^(kt)
Dθ/dt = d(kt)/dt
= k * dt/dt
= k
The velocity vector, V, is defined as the derivative of the position vector, r:
V = (Dr/dt) * r-hat + (Dθ/dt) * θ-hat
Substituting the derivatives we found earlier, we have:
V = (b * k * e^(kt)) * r-hat + k * θ-hat
To find the acceleration vector, A, we take the derivative of the velocity vector with respect to time:
A = (dV/dt) = (D^2r/dt^2) * r-hat + (D^2θ/dt^2) * θ-hat
Differentiating V with respect to time, we get:
D^2r/dt^2 = d/dt(b * k * e^(kt))
= (b * k^2 * e^(kt))
D^2θ/dt^2 = d/dt(k)
= 0
Hence, the acceleration vector, A, is:
A = (b * k^2 * e^(kt)) * r-hat
Now, let's calculate the angle between the velocity and acceleration vectors:
The dot product of two vectors, V and A, can be calculated as:
V • A = |V| * |A| * cos(θ)
where θ is the angle between V and A.
Since the magnitude of both V and A is positive, we can ignore the magnitude terms:
V • A = cos(θ)
Therefore, the angle θ between V and A is given by:
θ = arccos(V • A)
Substituting the expressions for V and A, we have:
θ = arccos((b * k * e^(kt) * (b * k^2 * e^(kt)))/(b * k * e^(kt))^2)
Simplifying the expression, we get:
θ = arccos(b * k^2 * e^(kt)/(b^2 * k^2 * e^(2kt)))
= arccos(e^(-kt)/b)
Now, to prove that the angle between the velocity and acceleration vectors is equal to π/4 for all times t, we need to show that θ = π/4.
θ = π/4 if and only if arccos(e^(-kt)/b) = π/4.
To establish this, we need to solve the equation:
arccos(e^(-kt)/b) = π/4
This equation can be solved numerically for different values of t and b using numerical methods.
However, from the given information, we cannot directly prove that the angle between the velocity and acceleration vectors is always π/4 for all times t.