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Acceleration Problems Key
If they provide a laundry list of uncommented Maple commands, dock ’em five points!
- (Not to be turned in) By the way, this function describes UNIFORM CIRCULAR
MOTION. You can easily compute (by hand, even; simplify along the way, especially if you use Maple)
v(t) = < −q sin(qt/R), q cos(qt/R) >
‖v(t)‖ = q
T(t) =
v(t)
‖v(t)‖
=< − sin(qt/R), cos(qt/R) >
a(t) = −
q 2
R
< cos(qt/R), sin(qt/R) >
N(t) =
dT/dt
‖dT/dt‖
=< − cos(qt/R), − sin(qt/R) >
κ =
R
Now using the vector projection formulae we have
aT =
a · v
v · v
aN = a − aT = a = −
q 2
R
< cos(qt/R), sin(qt/R) >.
On the other hand, using the decomposition formulae we have
aT =
d(‖v‖)
dt
T = 0T =< 0 , 0 >
aN = κ‖v‖ 2 N = −
q 2
R
< cos(qt/R), sin(qt/R) >.
(note ‖v‖ = q is constant). Of course, both approaches give the same result.
- (5 points)
(a) You did this.
(b) Find v =< et(cos(t) − sin(t)), et(cos(t) + sin(t)) > and ‖v‖ =
2 et.
(c) Find ds =
2 e t dt. Length is ∫ (^2)
− 5
2 et^ dt =
2(e^2 − t−^5 ) ≈ 10. 44.
(d) Find T =< (cos(t) − sin(t))/
2 , (cos(t) + sin(t))/
(e) It’s not hard (even by hand) to compute κ = 1 et
√ 2
. On the range given κ is max at t = −5. It makes sense.
(f) Find N =< −(cos(t) + sin(t))/
2 , (cos(t) − sin(t))/
(g) You find r(0) =< 1 , 0 >, v(0) =< 1 , 1 >, T(0) =< 1 /
2 >, κ(0) =
N = − < 1 /
(h) You find aT =
2 so aT =< 1 , 1 > at time t = 0. Also, aN =
2 so aN =< − 1 , 1 > at time t = 0.
- (5 points) Let me start by noting that you CANNOT do this problem by assuming that
the particle’s motion is r(t) =< t, t^3 >. That will give you some right answers—namely, T, N, and κ, which depend only on the geometry of the trajectory—but it won’t give the correct speed, velocity, or accelerations. In short, I never said the particle moved in time according to r(t) =< t, t^3 >, so you can’t assume this! All you really know is that r(t) =< x(t), x^3 (t) > for some function x(t)!
The first step is to work out everything you can from the geometry, that is, T, N, and κ. These quantities do NOT depend on any particle motion, only the curve itself.
We can easily work out the unit tangent vector T purely from geometry, for we know ‖T‖ = 1 and T must be tangent to the curve. The slope of y = x 3 at x = 1 is m = 3, so the vector < 1 , 3 > is tangent. To get T, just take
T =< 1 , 3 > /
We can also work out N from geometric considerations—rotate T counter-clockwise 90 degrees (it has to point INTO the curve) and find
N =< − 3 /
We can compute the curvature using κ = |f ′′(x)|/(1+(f ′(x))^2 )^3 /^2 to find κ = 3
Now plug everything we know (including a =< − 1 , 2 >) into a = aT T + aN N to find
d(‖v‖)
dt
T +
‖v‖ 2 N
= < q 1 /
10 − 9 q 2 / 50 , 3 q 1 /
10 + 3q 2 / 50 >.
where q 1 =
d(‖v‖) dt and^ q^2 =^ ‖v‖
2 (just for notational convenience). This leads to
equations q 1 /
10 − 9 q 2 /50 = − 1 , 3 q 1 /
10 + 3q 2 /50 = 2, which is just two equations in two unknowns, with solution q 1 =
10 /2 and q 2 = 25/3. This yields
d(‖v‖)
dt
, ‖v‖ =
- (Extra Credit) If the angle between a and v is acute then a · v > 0 (since a · v =
‖a‖‖v‖ cos(θ) and cos(θ) > 0 exactly when 0 ≤ θ < π/2 for θ in the range 0 to π
radians). But if a · v > 0 then
d(‖v‖) dt =^
a·v ‖v‖ >^ 0, that is, speed is increasing.^ The converse is also true: if speed is increasing then
a · v
‖v‖
d(‖v‖)
dt
so that a · v > 0 and the angle is acute.
Virtually identical reasoning works when speed is decreasing—this is equivalent to a · v < 0, i.e., an obtuse angle.
And finally, identical reasoning shows that if speed is constant (so d(‖v‖) dt = 0) then a · v = 0, and the vectors are orthogonal.
