qa15

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course Mth 271

11/6 9p

015. The differential and the tangent line

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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.

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Your solution:

derivative is 5x^4 differential is 5x^4 dx

at x = 3 5(3)^4 d(x) = 405 dx

at x=3.1 5(3.1)^4 d(x) = 461.76 dx

461.76-405 = 56.7605

f(3) = 3^5 = 243

dx = .1 405 *.1 = 40.5 243+40.5 = 283.5

f(3.1) = 3.1^5 = 286.291

286.291 - 283.5 = 2.791

confidence rating #$&*:2

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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.

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Self-critique (if necessary):It seems like this method of estimation does not get very close

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Self-critique Rating:2

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The smaller the dx, the better the approximation.

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Estimating a change of about 40 within 1 unit constitutes an error of 1/40 = 2.5%.

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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.

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Your solution:

y’ = 1/e differential = 1/e* dx

e = 2.718

ln(e) = 1

change in x = 2.8-2.718 = .082

1/e(.082)=.030

confidence rating #$&*:2

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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.

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Self-critique (if necessary):

n/a

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Self-critique Rating:3

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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.

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Your solution:

differential 1/2x ^-.5 dx x= 1.3 sq. rt. of 1.3 = 1.14

½(1.14) ^-.5

corrected 1/(2* sq. rt. of(x)) * dx

1/(2*1.14) *(1.3-1)=.13

confidence rating #$&*:

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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.

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Self-critique (if necessary):I don’t really understand this other than to assume that the ½

in front of the differential means that it is half the distance.

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Self-critique Rating:1

@&

if y is the square root of the number x, then dy is the change in the y value, dx the change in the x value. The equation says that y changes by half as much as x.

If x = 1, then y = 1. If x isn't 1, then there is some difference dx between 1 and x, and there is a difference dy between 1 and y.

y being the square root of x, dy = 1/2 dx says that y is twice as close to 1 as x.

Thus the square root of a number close to 1 is very nearly twice as close to 1.

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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.

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Your solution:

y= x^2

dy = 2x dx

dy = 2(1.2) (1.2-1)= .48

????I thought I was supposed to plug in numbers to test. This is where I get confused:

At x=1, would it be dy = 2(1) (1.2-1) or dy = 2(1)?

There has to be a change in x for there to be a dx.????

confidence rating #$&*:2

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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.

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Self-critique (if necessary):

I will take your word for it that this is what is shows, but it seems like it would be better to show it with numbers. I still would like to know what you plug in for dx if x=1 and you are trying to show a change from 1.

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Self-critique Rating:1.5

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You can plug in number of your own, which will be better than me telling you what happens.

Our reference point is x = 1, y = 1. So for example if x = .99, then dx = -0.01 and dy should be 2 dx = -0.02.

dy = -0.02 implies that the value of y should be 1 - 0.02 = .98.

x is the number, and y is the square root of the number.

Check and see if y = .98 is close to the square of x = .99.

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If x = 1.03, what is dx, what then must be dy, and what would be the value of y?

You can follow this format to make up more numerical examples of your own.

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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?

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Your solution:

L’ = (-250)(-.02)e^(-.02t)

5e^(-.02t)

at t = 50, 5e^(-.02*50)= 1.83940

dt = 2

dL =L’*dt = 1.83940*2 = 3.6788

confidence rating #$&*:2

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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.

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Self-critique (if necessary):

I understand the calculation but what would be the unit of measure here?

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Self-critique Rating:2

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No units were specified.

However the model would be reasonable for a fairly strong 250-lb lifter. In that case `dL would be the change in pounds of lifting strength.

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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.

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Your solution:

l’(r) = -2r^-3 = -2/(r^3)

dl = -2/(r^3) dr dr = 10-10.3 dr = -.3

[ -2/(1000)]*-.3 = .0006

confidence rating #$&*:1

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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.

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Self-critique (if necessary):

I thought the constant dropped out in finding the derivative. Also, I made dr negative since it was moving away from the light. I thought making it negative would show a decrease in the light.

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Self-critique Rating:2

@&

A constant term would have derivative zero.

However the derivative of a constant multiple of an expression is that multiple of the derivative of the expression.

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r is the distance from the light. If dr is negative it means your distance from the light is decreasing.

You basically overthought this (which is way better than underthinking). The light level is decreasing, but that's I and not r. I decreases when r increaeses.

*@

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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.

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Your solution:

area = w*2w = 2w^2

a’ = 2*2w = 4w

w= 5 4*5 = 20

change in t 5.1-5 = .1

confidence rating #$&*:1.5

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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.

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Self-critique (if necessary):

I understand the first part, but I don’t understand this part:

t he rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width.

@&

a is area, w is width.

w ' = da/dw is the rate of change of area a with respect to width w.

If w ' = 20, that means the rate of change of area with respect to width is 20.

Thus change in area / change in width = 20, and change in area = 20 * change in width.

*@

A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area.

@&

change in area = 20 * change in width

so if change in width is .1 the change in area is

20 * .1 = 2.

*@

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.

I know that the differential is related to the change in t which is .1, but I don’t see how using the differential is really useful here. It seems much simpler to just calculate the actual area.

@&

For this problem calculation of the area is easy enough.

In many situations the approximation is much easier than the calculation of the quantity, especially when multiple approximations are called for.

In economics, for example, the differential of the profit function is closely related to the marginal profit.

*@

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Self-critique Rating:2

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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.

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Your solution:

V(r) = (4/3) pi r^3

V’ = 3(4/3) pi r^2 = 4 pi r^2

r= 20 4 pi (20)^2 = 1600 pi cm^3

1600 pi * differential of .3 = 480 pi / cm^3 per day

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

I think I need prompting to remember to look up the formula and use it. I was trying to use r = .3^t.

It makes sense but I really wish I had access to the video notes. I never got a replacement set, and none of the video links in the notes work. The message I get is “The system cannot find the file specified.”

I am trying to use the book as a resource even if not directed to do so but the book doesn’t follow these qa’s and queries in the same order.

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self-critique Rating:2

@&

The links work, but not necessarily by clicking on them.

If you right-click that should work.

If you copy the links into the Address Box of your Internet browser that will work.

Be sure you are using the YouTube links, which are easily identified because they all include youtu in URL.

The differential is addressed in the Class Notes, easily identified on the Class Notes page in the lines:

#06: Project #3; Derivative of y = a x^3; The Differential

Linked Outline of Introductory Topics through 9/04/98: Study these for the upcoming 9/13/98 major quiz

Class Notes #7-#12: More Basic Concepts, Review of Selected Precalculus Topics

#07: The Differential; Tangent Line Approximation to Differential

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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.

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Your solution:

V(r) = (4/3) pi r^3

V’ = 3(4/3) pi r^2 = 4 pi r^2

r= 20 4 pi (20)^2 = 1600 pi cm^3

1600 pi * differential of .3 = 480 pi / cm^3 per day

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

I think I need prompting to remember to look up the formula and use it. I was trying to use r = .3^t.

It makes sense but I really wish I had access to the video notes. I never got a replacement set, and none of the video links in the notes work. The message I get is “The system cannot find the file specified.”

I am trying to use the book as a resource even if not directed to do so but the book doesn’t follow these qa’s and queries in the same order.

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self-critique Rating:2

@&

The links work, but not necessarily by clicking on them.

If you right-click that should work.

If you copy the links into the Address Box of your Internet browser that will work.

Be sure you are using the YouTube links, which are easily identified because they all include youtu in URL.

The differential is addressed in the Class Notes, easily identified on the Class Notes page in the lines:

#06: Project #3; Derivative of y = a x^3; The Differential

Linked Outline of Introductory Topics through 9/04/98: Study these for the upcoming 9/13/98 major quiz

Class Notes #7-#12: More Basic Concepts, Review of Selected Precalculus Topics

#07: The Differential; Tangent Line Approximation to Differential

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#