Is it enough as an answer to the tangant equation to put the velocity in t=2?
ie: r'(t) = [1i + 2tj + (3t^2)k] at t=2. T = 1i + 4j + 12k or should I leave it in the form of dexter's?
Let C be the curve: r(t) = (t, t^2, t^3) in R3.
Find the equation of the tangent line at P = (2,4,8).
In this case should I calculate the unite tanget T= r'(t)/||r'(t)||?
Or just the velocity r'(t) = 1i + 2tj + (3t^2)k? How to calculate the tangent equation at the point P?
Do I need to go through this?
Can't I just use \Delta \omega = \int_{0}^{2} \alpha(t) dt
But do I need to subtract the initial omega that is given from the final omega?
One more question please: when I did multiply the force equation by the radius, should I do the dot product or just...
Since it started from rest V initial = 0m/s
So first Velocity V= 0m/s + (0.5m/s^2)(t1)
Second final velocity since it stoped
0m/s = V(initial) + (-0.75m/s^2)(t2 = 10s)
and then you plug first Velocity in the second equation.
0m/s = [0m/s + (0.5m/s^2)(t1)] + (-0.75m/s^2)(t2 = 10s)
A pulley of moment of inertia 0.021kg-m^2 and radius 0.12m is acted upon by a force which varies in time as F= 0.23t + 0.12t^2 where F is in Newton and t is in second. suppose that the pulley is initially rotating at 0.18 r/s and the force acts tangentially to the pulley. Find the magnitude of...
I did solve this differential equation (x^4)y'''' + 6x^3y''' + 9x^2 + 3xy' + y = 0 using Cauchy Euler Equation. I got X^m (m^2 + 1)^2 = 0
I'm not sure how to get the roots of (m^2 + 1)^2. In my calculation I got
m = -i, +i, -i, +i when I put m^2 = -1. In the book they have m = (+-)...