Student Questions 9-16-14
How would you solve for at and an given no t-value?
Lets look at the problem we did in class. Suppose I said ~r(t) = ht − 1, 2t, t2 i is a vector
valued function and asked you to find at and an . (Notice that I didn’t give you a specific
t-value so you will have to find them in general.) Well the formulas are the same:
~
~
||V~ × A||
V~ · A
~ = h0, 0, 2i. So
and an =
. We have V~ (t) = h1, 2, 2ti and A(t)
at =
~
~
||V ||
||V ||
√
4t
~ = h4, −2, 0i. So an = √ 20 . And were done!! If we want
at = √
. Now V~ × A
2
5 + 4t
5 + 4t2
√
20
~ = √ 4t
~ . The only difference between doing
we could write A
· T~ + √
·N
2
5 + 4t
5 + 4t2
things in general and doing things for a specific t-value is that in the first case we never
plug anything into t and in the second case we actually plug in something for t (whether
along the way or at the very end). So if you plug in t = 1 to what we have above you
will get the answer for what we did in recitation.
Can you work out a problem from 13.6?
Let me summarize the idea of 13.6. First if I give you a curve that is described by a
vector valued function ~r(t) = hx(t), y(t)i, you know how to compute velocity and
acceleration. It’s really easy, V~ (t) = hx0 (t), y 0 (t)i = x0 (t)~i + y 0 (t)~j and
~ = hx00 (t), y 00 (t)i = x00 (t)~i + y 00 (t)~j. Notice how we write V~ and A
~ in terms of the
A(t)
~
~
standard unit vectors i = h1, 0i and j = h0, 1i. We can do this because ~i and ~j are
perpindicular and both of length 1. But we don’t have to write a 2d vector in terms of ~i
and ~j, any two vectors that are length 1 and perpindicular will do. For instance, T~ and
~ are both length 1 and perpindicular and you have seen that we can write A
~ as
N
~ = at T~ + an N
~ . Ok so you know that if you are given a cuve described as ~r(t) you can
A
find velocity and acceleration.
However, there are many different ways to describe a curve. For instance I can describe
a curve in polar coordinates as r = r(t) and θ = θ(t). So as t changes, r and θ change
and they trace out a curve in polar coordinates. The goal of 13.6 is to figure out how to
~ if we have been given a curve desctribed in polar coordinates (instead of
find V~ and A
as a vector valued function).
Now it turns out that for every point (r(t), θ(t)) on the curve, the vectors
u~r = cos(θ)~i + sin(θ)~j and u~θ = −sin(θ)~i + cos(θ)~j are both length 1 and perpindicular.
~ in terms of ~i and ~j, we will
(See page 757 for a picture.) So instead of writing V~ and A
write them in terms of u~r and u~θ . The book does a derivation and tells us how to do
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that. It says
V~ = r0 (t)u~r + r(t)θ0 (t)~
rθ
and
~ = (r00 (t) − r(t)[θ0 (t)]2 )r~u + (r(t)θ00 (t) + 2r0 (t)θ0 (t))u~θ .
A
For problem 5 they give us r(t) = 2cos(4t) and θ(t) = 2t. I will find V~ .
r0 (t) = −8sin(4t) and θ0 (t) = 2. So V~ = (−8sin(4t))u~r + 4cos(4t)~
rθ . I hope this helped!
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