QA 15

course MTH 271

10/26, 1800

assignment #015015. The differential and the tangent line

10-26-2009

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17:10:50

`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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RESPONSE -->

f(x)=x^5

f'(x)=5x^4

'dy/'dx=f'(x)

'dy=f'(x) * 'dx

'dy=(5x^4) * 'dx

'dy(3)=5(3^4)

'dy(3)=405

'dx=3.1-3

'dx=.1

'dy(3.1)=405*.1

'dy(3.1)=40.5

f(3)=3^5

f(3)=243

f(3.1)=243 + 40.5=283.5

confidence rating: 2

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17:10:57

The 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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RESPONSE -->

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Self-critique Rating:ent: 3

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17:11:21

`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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RESPONSE -->

f(x)=ln(x)

'dy= 1/x * 'dx

x=e

'dx=(2.8 - e)

'dx=.082

'dy=1/e * .082

'dy=.030

y=ln(e)

y=1

y=ln(2.8)

1 + .030=1.030

confidence rating: 2

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17:11:28

The 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..

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Self-critique Rating:ent: 3

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17:17:57

`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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RESPONSE -->

f(x)= x^(1/2)

=1/2 x^(-1/2)

f'(x)=1 / 2x^(1/2)

'dy=f'(x) * 'dx

'dy=1 / 2x^(1/2) * 'dx

That the square root of a number close to one is twice as close to 1 as the number is proven by the presence of 1/2 in the derivative.

confidence rating: 2

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17:18:04

The 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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RESPONSE -->

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Self-critique Rating:ent: 3

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17:20:09

`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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RESPONSE -->

f(x)=x^2

f'(x)=2x

'dy=f'(x) * 'dx

'dy= 2x * 'dx

The derivative of a square gives 2x causes the number close to one to be twice as far from one as the number.

confidence rating: 2

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17:20:20

The 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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RESPONSE -->

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Self-critique Rating:ent: 3

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17:21:49

`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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RESPONSE -->

'dL = L'(t)*'dt

-.02(-250 e^(-.02t)) * 'dt

5e^(-.02t) * 'dt

t=50

5e^(-.02*50)*'d

1.84 * 'dt='dL

'dt=2 weeks

1.84 * 2='dL

3.7='dL

confidence rating: 2

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17:21:56

The 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 Rating:ent: 3

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17:23:48

`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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RESPONSE -->

'dI= I'(r) * 'dr

'dr=10.3-10=.03

I'(r)=- (2r)(k) / (r^2)^2

-2k/r^3

'dI=-2 k /r^3 * .03

'dI=-2k / (10^3) * .03

'dI=-.0006k

confidence rating: 2

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17:23:54

The 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 Rating:ent: 3

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17:23:59

`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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RESPONSE -->

area= l * w

l=2w

area= 2w * w

area=2w^2

f(x)=2w^2

f'(x)=4w

f'(5)=4*5=20

y=2(5^2)=50

'dy=f'x * 'dx

'dy=20 * .1

'dy=2

y + 'y

50 + 2=52

confidence rating: 2

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17:24:04

If 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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RESPONSE -->

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Self-critique Rating:ent: 3

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17:24:11

`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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RESPONSE -->

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

V'=(4/3)(3) pi (r^2)

v'=4 pi r^2

V'=4 pi (20^2)

V'=1600 pi

'dr=.3

'dv=v' * 'dr

=1600 pi * .3

=480 pi

confidence rating: 2

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17:24:17

The 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.

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Self-critique Rating:ent: 3

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