Checkpoint
ddt[r(t)·r′(t)]=8e4t
ddt[u(t)×r(t)]=−(e2t(cost+2sint)+cos2t)i+(e2t(2t+1)−sin2t)j+(tcost+sint−cos2t)k
s=5t, or t=s/5. Substituting this into r(t)=〈3cost,3sint,4t〉 gives
r(s)=〈3cos(s5),3sin(s5),4s5〉,s≥0.
κ=4[1+(4x−4)2]3/2
At the point x=1, the curvature is equal to 4. Therefore, the radius of the osculating circle is 14.
A graph of this function appears next:
The vertex of this parabola is located at the point (1,3). Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are (1,134). The equation of the osculating circle is
(x−1)2+(y−134)2=116.
v(t)=r′(t)=(2t−3)i+2j+ka(t)=v′(t)=2iv(t)=‖r′(t)‖=√(2t−3)2+22+12=√4t2−12t+14
The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.
Section 3.1 Exercises
One possibility is By increasing the coefficient of t in the third component, the number of turning points will increase.