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.

 

029. `query 29

 

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.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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 **

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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. “

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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 **

 

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: