Question
Find the minimum spced of a particle with parametric trajectory $c(t)=\left(t^{3}-4 t, t^{2}+1\right)$ for $t \geq 0 .$ Hint: It is easier to find the minimum of the square of the speed.
Step 1
The velocity vector is given by the derivative of the position vector. So, we differentiate $c(t)$ with respect to $t$ to get the velocity vector $v(t)$: \[v(t) = c'(t) = \left(3t^{2}-4, 2t\right).\] Show more…
Show all steps
Your feedback will help us improve your experience
Carson Merrill and 94 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the minimum speed of a particle with trajectory $c(t)=$ $\left(t^{3}-4 t, t^{2}+1\right)$ for $t \geq 0 .$ Hint: It is easier to find the minimum of the square of the speed.
PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR FUNCTIONS
Arc Length and Speed
Find the minimum speed of a particle with trajectory c(t) = (t^3 - 8t, t^2 + 1) for t ≥ 0. Hint: it is easier to find the minimum of the square of the speed.
Find the minimum speed of a particle with trajectory $c(t)=$ $\left(t^{3}, t^{-2}\right)$ for $t \geq 0.5$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD