open qa 29

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course Mth 271

029. `query 29

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Question: `qQuery 1b 7th edition 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.

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Your solution:

Dy = dy/dx * dx

Dy = (6x^2)^ 1/3 * dx = 1/3 (6x^2) ^ -2/3 * 12x * dx = 4x * (6x^2)^ -2/3 * dx = 4x/(6x^2)^2/3 * dx

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Given Solution:

`a dy is the differential; `dy means 'delta-y' and is the exact change.

y = (6x^2)^(1/3)

y' = dy/dx = 1/3(6x^2)^(-2/3)(12x)

y' = dy/dx = 4x(6x^2)^(-2/3)

y' = dy/dx = 4x / (6x^2)^(2/3).

So

dy = (4x / (6x^2)^(2/3)) dx **

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Given Solution:

`a** Query 3b 7th edition 3.8.12 compare dy and `dy for y = 1 - 2x^2, x=0, `dx = -.1 **** What is the differential estimate dy for the given function and interval, and what is the actual change `dy?

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Your solution:

Dy/dx = -4x

Dy = -4x * dx = -4 * 0 * -0.1 = 0

D’y = (1 - 2 * -0.1^2) - 1 = 0.98 -1 = -0.02

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Given Solution:

`a y ' = dy /dx = - 4 x so

dy = -4x dx.

The differential estimate is dy = -4 * 0 * (-.1) = 0.

Actual change is y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02. **

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Self-critique (if necessary):

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Question: `qQuery 4 a 7th edition 3.8.22 equation of the tangent line to y=sqrt(52-x^2) at (3, 4); tan line prediction and actual fn value for `dx = -.01 and .01.

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What is the equation of the tangent line and how did you obtain it?

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What are your tangent-line approximations for `dx = -.01 and `dx = +.01?

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What are the corresponding values of the actual function?

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Your solution:

f’(x) = (25 - x^2) ^ ½ = ½(25 - x^2) ^ -1/2 (- 2x) = -x/sqrt(25 - x^2) = -3/4

(y - 4) = -3/4 ( x - 3)

y - 4 = -3/4 x + 9/4

4y = 3x + 25

Y = -3/4x + 25/4

X + dx = 3 + 0.01 = 3.01

Y =- ¾ * 3.01 + 25/4 = 3.9925 approx.

F(x) = sqrt (25 - 9.0601) = sqrt(15.9399) = 3.99

confidence rating #$&*:

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Given Solution:

`a f(x) = sqrt(25-x^2)

f' (x) = -x / sqrt(25-x^2) so

f ' (3) = -3 / sqrt(25 - 3^2) = -3/sqrt(16)

f ' (3) = -3/4 so that the tangent line has equation

y - 4 = -3/4(x - 3)

y - 4 = -3/4 x + 9/4

y = -3/4 x + 25/4.

Using `dx = .01 we get x + `dx = 3.01. The tangent-line approximation is thus

y = -3/4 * 3.01 + 6.25 = 3.9925.

The actual function value is sqrt(25-3.01^2) = 3.992480432. The difference between the actual and approximate values is .00002, approx.

A similar difference is found approximating the function for `dx = -.01, i.e., at 2.999.

We see that for this short interval `dx the tangent-line approximation is very good, predicting the change in the y value (approximately .00750) accurate to 5 significant figures.

COMMON ERROR: Students often round off to 3.99, or even 4.0, which doesn't show any discrepancy between the tangent-line approximation and the accurate value.

Taken to enough decimal places the values of the function and the tangent-line approximation do not coincide; this must be the case because the function isn't linear, whereas the tangent line approximation is. You should in general use enough significant figures to show the difference between two approximations. “

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Self-critique (if necessary):

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Question: `q **** Query 5a 7th edition 3.8.38 drug concentration C = 3t / (27+t^3) **** When t changes from t = 1 to t = 1.5 what is the approximate change in C, as calculated by a differential approximation? Explain your work.

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Your solution:

dC/dt = [3(27+ t^3) - 3t (3t^2)]/(27 + t^3) ^2 =( 81 + 3t^3 - 9t^3)/(27 + t^3)^2 = 81 - 6t^3/(27 + t^3) ^ 2

dC = (81 - 6)/28^2 * 0.5 = 0.0478

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Given Solution:

`a By the Quotient Rule we have C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2, or

C' = (81 - 6t^3) / (27 + t^3)^2.

The differential is therefore

dC =( (81 - 6t^3) / (27 + t^3)^2) dx.

Evaluating for t = 1 and `dt = .5 we get

dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5)

dC = (75 / 784) (.5)

dC = .0478 mg/ml **

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