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.
015. The differential and the tangent line
Question: `q001. Using the differential and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx.
At x = 3 the differential is `dy = 5 * 3^4 * `dx, or
`dy = 405 `dx.
Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5.
Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx..
Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3.
Self-critique (if necessary):
Self-critique Rating:
Question: `q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential of the function y = f(x) = ln(x) is `dy = f ' (x) `dx or
`dy = 1/x `dx.
If x = e we have
`dy = 1/e * `dx.
Between e and 2.8, `dx = 2.8 - e = 2.8 - 2.718 = .082, approx.. Thus `dy = 1/e * .082 = .030, approx..
Since ln(e) = 1, we see that ln(2.8) = 1 + .030 = 1.030, approx..
STUDENT QUESTION
I see were the rate is coming from
but why this value of one is used??????
INSTRUCTOR RESPONSE
The function is very easy to evaluate
with pleanty of accurately if x = e. All you need to know is that, to four
significant figures, e = 2.718. So we very easily see that ln(e) = 2.718.
You can't accurately evaluate ln(2.8). However 2.8 is close to e, so if you know
how quickly the function y = ln(x) is changing when x = e, you can easily
extrapolate to x = 2.8.
Of course you can just plug 2.8 into your calculator and get an accurate answer,
but that provides no insight into the behavior of the function, or into the
nature of this approximation, and gives you nothing on which to build later
understanding.
Self-critique (if necessary):
Self-critique Rating:
Question: `q003. Using the differential verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the differential approximation for the function f(x) = `sqrt(x) at an appropriate point.
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore
`dy = 1 / 2 * `dx.
This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much.
Self-critique (if necessary):
Self-critique Rating:
Question: `q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number.
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential for this function is easily seen to be `dy = 2 x * `dx. At x = 1 the differential is therefore
`dy = 2 * `dx.
This shows that as we move away from x = 1, the change in y is double the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but by twice as much.
Self-critique (if necessary):
Self-critique Rating:
Question: `q005. The lifting strength of an athlete in training changes according to the function L(t) = 400 - 250 e^(-.02 t), where t is the time in weeks since training began. What is the differential of this function? At t = 50, what approximate change in strength would be predicted by the differential for the next two weeks?
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential is
`dL = L ' (t) * `dt =
-.02 ( -250 e^(-.02 t) ) `dt, so
`dL = 5 e^(-.02 t) `dt.
At t = 50 we thus have
`dL = 5 e^(-.02 * 50) `dt, or
`dL = 1.84 `dt.
The change over the next `dt = 2 weeks would therefore be approximately
`dL = 1.84 * 2 = 3.68.
Self-critique (if necessary):
Self-critique Rating:
Question: `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters.
Your solution:
Confidence Assessment:
Given Solution:
`aThe differential is
`dI = I ' (r) * `dr,
where I ' is the derivative of I with respect to r.
Since I ' (r) = - 2 k / r^3, we therefore have
`dI = -2 k / r^3 * `dr.
For the present example we have r = 10 m and `dr = .3 m, so
`dI = -2 k / (10^3) * .3 = -.0006 k.
This is the approximate change in illumination.
Self-critique (if necessary):
Self-critique Rating:
Question: `q007. A certain crystal grows between two glass plates by adding layers at its edges. The crystal maintains a rectangular shape with its length double its width. Its width changes by .1 cm every hour. At a certain instant its width is 5 cm. Use a differential approximation to determine its approximate area 1 hour later.
Your solution:
Confidence Assessment:
Given Solution:
`aIf the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5.
f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20.
The rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width. A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area. Thus the approximate area should be 50 + 2 = 52. This can easily be compared with the accurate value of the area which is 52.02.
Self-critique (if necessary):
Self-critique Rating:
Question: `q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm.
Your solution:
Confidence Assessment:
Given Solution:
`aThe volume of a sphere is V(r) = 4/3 * `pi * r^3. The rate at which the volume changes with respect to the radius is dV / dr = 4 * `pi * r^2.
Thus when r = 20 the volume is changing at a rate of 4 * `pi * 20^2 = 1600 `pi cm^3 volume per cm of radius. It follows that if the radius is changing by .3 cm / day, the volume must be changing at 1600 `pi cm^3 / (cm of radius) * .3 cm of radius / day = 480 `pi cm^3 / day.
Note that this is the instantaneous rate at the instant r = 20. This rate will increase as r increases.
STUDENT QUESTION
Were did the initial formula come from?
I think I see how the problem was solved after that but were did V(r) = 4/3 *
`pi
* r^3 come from?
INSTRUCTOR RESPONSE That is the formula for the volume of a sphere. This should
be general knowledge.
This formula was also used in some of the Orientation exercises. You should know
everything that was covered in those exercises.
Self-critique (if necessary):
Self-critique Rating:
"