If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
003. `query 3
Question:
`q Query class notes #04
explain how we can prove that the rate-of-depth-change function for depth
function y = a t^2 + b t + c is y' = 2 a t + b
Your solution:
Confidence Assessment:
Given Solution:
** You have to find the average rate of change between clock times t and t
+ `dt:
ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2
+ b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2
+ b t + c ) ] / `dt
= [ 2 a t `dt + a `dt^2 + b `dt ] / `dt
= 2 a t + b t + a `dt.
Now if `dt shrinks to a very small value the ave rate of
change approaches y ' = 2 a t + b. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q explain how we know that the depth function for
rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for
some constant c, and explain the significance of the constant c.
Your solution:
Confidence Assessment:
Given Solution:
** Student Solution:
If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent
to the given function y ' (t)=mt+b .
Since 2at+b=mt+b for all possible values of t the parameter b
is the same in both equations, which means that the coefficients 2a and m must
be equal also.
So if 2a=m then a=m/2.
The depth function must therefore be y(t)=(1/2)mt^2+bt+c.
c is not specified by this analysis, so at this point c is
regarded as an arbitrary constant. c
depends only on when we start our clock and the position from which the depth is
being measured. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q Explain why, given only the rate-of-depth-change function y' and a
time interval, we can determine only the change in depth and not the actual
depth at any time, whereas if we know the depth function y we can determine the
rate-of-depth-change function y' for any time.
Your solution:
Confidence Assessment:
Given Solution:
** Given the rate function y' we
can find an approximate average rate over a given time interval by averaging
initial and final rates. Unless the rate function is linear this estimate will
not be equal to the average rate (there are rare exceptions for some functions
over specific intervals, but even for these functions the statement holds over
almost all intervals).
Multiplying the average rate by the time interval we obtain
the change in depth, but unless we know the original depth we have nothing to
which to add the change in depth. So if
all we know is the rate function, have no way to find the actual depth at any
clock time.
ANOTHER EXPLANATION:
The average rate of change over a time interval is rAve = `dy
/ `dt. If we know rAve and `dt, then, we
can easily find `dy, which is the change in depth. None of this tells us anything about the
actual depth, only about the change in depth.
If we don't know rAve but know the function r(t) we can't use
the process above to get the exact change in depth over a given interval, though
we can often make a pretty good guess at what the average rate is (for a
quadratic depth function, as the quiz showed, you can actually be exact the
average rate is just the rate at the midpoint of the interval; it's also the
average of the initial and final rates; and all this is because for a quadratic
the rate function is linear--if you think about those statements you see that
they characterize a linear function, whose average on an interval occurs at a
midpoint etc.). For anything but a linear
rate function we can't so easily tell what the average is.
However we do know that the rate function is the derivative
of the depth function. So if we can find
an antiderivative of the rate function, all we have to do to find the change in
depth is find the difference in its values from the beginning to the end of the
interval. This difference will be the
same whichever antiderivative we find, because the only difference that can
exist between two antiderivatives of a given rate function is a constant (whose
derivative is zero).
We have to develop some machinery to prove this rigorously
but this is the essence of the Fundamental Theorem of Calculus. You might not understand it completely at this
point, but keep coming back to this explanation every week or so and you will
soon enough.**
Self-critique (if necessary):
Self-critique Rating:
Question:
`q In terms of the depth model explain the processes of
differentiation and integration.
Your solution:
Confidence Assessment:
Given Solution:
** Rate of depth change can be found from depth data.
This is equivalent to differentiation.
Given rate-of-change information it is possible to find depth
changes but not actual depth. This is
equivalent to integration.
To find actual depths from rate of depth change would require
knowledge of at least one actual depth at a known clock time. **
Self-critique (if necessary):
Self-critique Rating: