Answer:
Explanation:
a. To sketch the plane curve defined by the vector function r(t) = ti + ln(t)j when t = 2, we can first evaluate the position vector at t = 2:
r(2) = 2i + ln(2)j
Next, we plot the point (2, ln(2)) on the x-y plane. Since the curve is two-dimensional, we don't need to determine the vectors T(t) and N(t) for this sketch.
b. To find the scalar tangential component of acceleration aT for the particle with the position vector r(t) = e^t⟨cos(4t), sin(4t), 0⟩, we need to differentiate the position vector twice with respect to time to obtain the acceleration vector a(t).
First, let's differentiate r(t) with respect to t:
r'(t) = (d/dt) (e^t⟨cos(4t), sin(4t), 0⟩)
= e^t⟨-sin(4t), cos(4t), 0⟩ + e^t⟨-4cos(4t), 4sin(4t), 0⟩
= e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩
Now, we differentiate r'(t) with respect to t to find the acceleration vector:
a(t) = (d/dt) (e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩)
= e^t⟨-4cos(4t) + 16sin(4t), -4sin(4t) - 16cos(4t), 0⟩
To find the scalar tangential component of acceleration aT, we project the acceleration vector a(t) onto the tangent vector T(t). Since the tangent vector is the derivative of the position vector with respect to t, we have T(t) = r'(t):
T(t) = e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩
Finally, we can find the scalar tangential component of acceleration aT by taking the dot product of a(t) and T(t):
aT = a(t) · T(t)
= e^t⟨-4cos(4t) + 16sin(4t), -4sin(4t) - 16cos(4t), 0⟩ · e^t⟨-sin(4t) - 4cos(4t), cos(4t) + 4sin(4t), 0⟩
= (-4cos(4t) + 16sin(4t))(-sin(4t) - 4cos(4t)) + (-4sin(4t) - 16cos(4t))(cos(4t) + 4sin(4t))
The resulting expression for aT is quite complex and difficult to simplify without a specific value of t.