How do you find the distance travelled from #0

Question

How do you find the distance travelled from #0<=t<=1# by an object whose motion is #x=e^tcost, y=e^tsint#?

William Abdalla · Accepted Answer

#sqrt 2 \ (e - 1)# Let #z(t) = x(t) + i \ y(t) = e^((1+i)t)# #dot z = (1+i) e^((1+i)t) # Speed (not velocity) is needed to calculate distance #s#: #s = int_0^1 sqrt (abs(dotz)^2) \ dt# #abs(dotz)^2 = dot z bar (dot z)# # = (1+i) e^((1+i)t) * (1-i) e^((1-i)t) = 2 e^(2t)# #implies s =sqrt 2 int_0^1 e^(t) \ dt = sqrt 2 \ (e - 1)#

Luke Alvarez · Answer

undefined To find the distance traveled by the object from $0 \leq t \leq 1$, you need to integrate the magnitude of its velocity vector with respect to time over the given interval. The velocity vector of the object is given by:

$$ \mathbf{v}(t) = \frac{d}{dt}(x(t), y(t)) = (e^t \cos t - e^t \sin t, e^t \sin t + e^t \cos t) $$

To compute the magnitude of this velocity vector, use the formula for the magnitude of a vector:

$$ |\mathbf{v}(t)| = \sqrt{(e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2} $$

This simplifies to:

$$ |\mathbf{v}(t)| = \sqrt{2e^{2t}} $$

Now, integrate $ |\mathbf{v}(t)| $ with respect to $ t $ from $ 0 $ to $ 1 $:

$$ \int_{0}^{1} \sqrt{2e^{2t}} \, dt $$

This integral can be evaluated directly, yielding the distance traveled by the object from $ t = 0 $ to $ t = 1 $.

How do you find the distance travelled from #0&lt;=t&lt;=1# by an object whose motion is #x=e^tcost, y=e^tsint#?

How do you find the distance travelled from #0<=t<=1# by an object whose motion is #x=e^tcost, y=e^tsint#?