011. `query 11
Question:
`qQuery
class notes #06 If x is the
height of a sandpile and y the volume, what proportionality governs
geometrically similar sandpiles?
Why should this be the proportionality?
Your solution:
Confidence Assessment:
Given Solution:
** the proportionality is y = k x^3. Any proportionality of volumes is a y = k
x^3 proportionality because volumes can be filled with tiny cubes; surface areas
are y = k x^2 because surfaces can be covered with tiny squares. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf x is the radius of a spherical balloon and y the surface area,
what proportionality governs the relationship between y and x? Why should this be the proportionality?
Your solution:
Confidence Assessment:
Given Solution:
** Just as little cubes can be thought of as filling the
volume to any desired level of accuracy, little squares can be thought of as
covering any smooth surface. Cubes
'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2.
Surfaces can be covered as nearly as we like with tiny
squares (the more closely we want to cover a sphere the tinier the squares would
have to be). The area of a square is
proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y =
k x^2.
By contrast, for volumes or things that depend on volume,
like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the
cube of linear dimensions. Thus the
proportionality for a volume would be y = k x^3. **
Self-critique (if necessary):
Self-critique Rating:
Question: `q
(NOTE: This question is not for Mth 163 students, who may safely ignore it)
Explain how you would use the concept of the differential to find
the volume of a sandpile of height 5.01 given the volume of a geometrically
similar sandpile of height 5, and given the value of k in the y = k x^3
proportionality between height and volume.
Your solution:
Confidence Assessment:
Given Solution:
** The class notes showed you that the slope of the y = k x^3
graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the
given information, you can evaluate y' at x = 5. That gives you the slope of the line
tangent to the curve, and also the rate at which y is changing with respect to
x. When you multiply this rate by
the change in x, you get the change in y.
The differential is
3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the
differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx,
though what's involved isn't really simple algebra. The differential expresses the fact that
near a point, provided the function has a continuous derivative, the approximate
change in y can be found by multiplying the change in x by the derivative).
That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx
(approx), or `dy = 3 k x^2 `dx (approx).
The idea is that the derivative is the rate of change of the
function. We can use the rate of
change and the change in x to find the change in y.
The differential uses the fact that near x = 5 the change in
y can be approximated using the rate of change at x = 5.
Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k,
whatever k is. To estimate the
change in y corresponding to the change .01 in x, we will multiply y ' by .01,
getting a change of y ' `dx = 75 k * .01.
}
SPECIFIC EXAMPLE:
We don't know what k is for this specific question. As a specific example suppose our
information let us to the value k = .002, so that our proportionality is y =
.002 x^3. Then the rate of change
when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y
would be y = f(5) = .002 * 5^3 = .25.
This tells us that at x = 5 the function is changing at a rate of .15
units of y for each unit of x.
Thus if x changes from 5 to 5.01 we expect that the change
will be
change in y = (dy/dx) * `dx =
rate of change * change in x (approx) =
.15 * .01 = .0015,
so that when x = 5.01, y should be .0015 greater than it was
when x was 5. Thus y = .25 + .0015 =
.2515. This is the differential
approximation. It doesn't take
account of the fact that the rate changes slightly between x=5 and x = 5.01.
But we don't expect it to change much over that short increment, so we expect
that the approximation is pretty good.
Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515
approximation we got from the differential--the differential is off by .000003.
That's not much, and we expected it wouldn't be much because the derivative
doesn't change much over that short interval.
But it does change a little, and that's the reason for the discrepancy.
The differential works very well for decently behaved
functions (ones with smooth curves for graphs) over sufficiently short
intervals.**
Self-critique (if necessary):
Self-critique Rating:
Question: `q
(NOTE: This question is not for Mth 163 students, who may safely ignore it)
What would be the rate of depth change for the depth function y =
.02 t^2 - 3 t + 6 at t = 30? (instant response not required)
Your solution:
Confidence Assessment:
Given Solution:
** You saw in the class notes and in the q_a_ that the rate
of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be
evaluated to give you the rate.
Evaluating the rate of depth change function y ' = .04 t - 3
for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8.
COMMON ERROR: y =
.02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change
INSTRUCTOR COMMENT:
This is the depth, not the rate of depth change. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qmodeling project 3
problem a single quarter-cup of sand makes a cube 1.5 inches on a side.
How many quarter-cups would be required to make a cube with twice the scale, 3
inches on a side? Explain how you know this.
Your solution:
Confidence Assessment:
Given Solution:
** You can think of stacking single cubes--to double the
dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer.
Thus it would take 8 cubes 1.5 inches on a side to make a
cube 3 inches on a side.
Since each 1.5 inch cube containts a quarter-cup, a 3 inch
cube would contain 8 quarter-cups.
COMMON ERROR:
It would take 2 quarter-cups.
INSTRUCTOR COMMENT: 2 quarter-cups would make two 1.5
inch cubes, which would not be a 3-inch cube but could make a rectangular solid
with a square base 1.5 inches on a side and 3 inches high. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat value of the parameter a would model this situation?
How many quarter-cups does this model predict for a cube three inches on a side?
How does this compare with your previous answer?
Your solution:
Confidence Assessment:
Given Solution:
** The proportionality would be
y = a x^3,
with y = 1 (representing one quarter-cup) when x = 1.5. So we have
1 = a * 1.5^3, so that
a = 1 / 1.5^3 = .296 approx.
So the model is y = .2963 x^3.
Therefore if x = 3 we have
y = .296 * 3^3 = 7.992, which is the same as 8 except for
roundoff error. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat would be the side measurement of a cube designed to hold 30
quarter-cups of sand? What equation
did you solve to get this?
Your solution:
Confidence Assessment:
Given Solution:
** You are given the number of quarter-cups, which
corresponds to y. Thus we have
30 = .296 x^3 so that
x^3 = 30 / .296 = 101, approx, and
x = 101^(1/3) = 4.7, approx..**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function
was y = .002 x^3. What function would have been obtained using 1/4 cup?
Your solution:
Confidence Assessment:
Given Solution:
** In this case, since it takes two quarter-cups to make a
half-cup, the person would need twice as many quarter-cups to get the same
volume y.
He would have obtained half as many half-cups as the actual
number of quarter-cups.
To get the function for the number of quarter-cups he would
therefore have to double the value of y, so the function would be y = .004 x^3.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 4. number of swings vs. length data. Which function fits best?
Your solution:
Confidence Assessment:
Given Solution:
** If you try the different functions, then for each one you
can find a value of a corresponding to every data point. For example if you use y = a x^-2
you can plug in every (x, y) pair and solve to see if your values of a are
reasonably consistent. Try this for
the data and you will find that y = a x^-2 does not give you consistent a
values—every (x, y) pair you plug in will give you a very different value of a.
The shape of the graph gives you a pretty good indication of
which one to try, provided you know the shapes of the basic graphs.
For this specific situation the graph of the # of swings vs.
length decreases at a decreasing rate.
The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7
all decrease at a decreasing rate.
In this case you would find that the a x^-.5 function works nicely,
giving a nearly constant value of a.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qproblem 7. time per swing model. For your data what expression represents
the number of swings per minute?
Your solution:
Confidence Assessment:
Given Solution:
** The model that best fits the data is a x^-.5, and with accurate data we find that for the case where length is in feet, the values of a is close to 55.
If you used alternative instructions you would have measured the pendulum length in cm or inches rather than feet. However the process will be identical, and the model y = a x^-.5 will fit the data just as well. You will just get a different value of a:
If your pendulum lengths were in centimeters, then your value of a would be close to 300.
If pendulum lengths are in inches then your value of a would be close to 190.
# per minute frequency = 55 x^-.5.