queries 28 29

course mth 271

}KعYpKð掜_assignment #028

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028. `query 28

Applied Calculus I

12-10-2007

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20:24:16

Query 3.7.12 sketch y = -x^3+3x^2+9x-2 **** Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.

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RESPONSE -->

this doesnt factor... trying numbers you get x= -2

so we have (x+2)... but then what else? i cant remember how to long divide a polynomial

confidence assessment: 0

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20:26:41

First we find the zeros:

You can find the zero at x = -2 by inspection (i.e., try a few simple values of x and see if you 'hit' one).

Knowing that x = -2 is a zero, you know that (x - (-2) ) = x + 2 is a factor of the expression. You can find the other factor by long division of x + 2 into -x^3 + 3 x^2 + 9x - 2. You get quotient x^2 - 5 x + 1.

Thus -x^3 + 2 x^2 + 9 x - 2 = - (x + 2)(x^2 - 5x + 1).

This expression is zero when x+2 = 0 or when x^2 - 5 x + 1 = 0.

The first equation we already know gives us x = -2.

The second is solved by the quadratic formula. We get

x = [ -(-5) +- `sqrt( (-5)^2 - 4 * 1 * 1 ) ] / (2 * 1) = [ 5 +- `sqrt(21) ] / 2.

Simplifying we get approximate x values .21 and 4.7.

Then we find maxima and minima using 1st and 2d derivative:

The derivative is -3x^2+6x + 9, which gives the equation -3x^2+6x + 9 = 0 for critical points. Dividing thru by -3 we get

x^2 - 2x - 3 = 0 or(x-3)(x+1) = 0 so

x = 3 or x = -1.

Second derivative is -6x + 6, which is negative when x = 3 and positive when x = -1.

At x = 3 we have a maximum. Evaluating y = -x^3+3x^2+9x-2 at x = 3 we get y = 25. The negative second derivative indicates that (3,25) is a maximum.

At x = -1 we have a minimum. Evaluating y = -x^3+3x^2+9x-2 at x = -1 we get y = -7. The positive second derivative indicates that this (-1,-7) is a minimum.

Finally we analyze 2d derivative for concavity and pts of inflection:

The second derivative -6x + 6 is zero when x = 1; at this point the derivative, which is linear in x, changes from positive to negative. Thus the x = 1 point (1, 9) is a point of inflection.

The derivative is positive and the function therefore concave up on (-infinity, 1).

The derivative is negative and the function therefore concave down on (1, infinity).

The function is defined for all x so there are no vertical asymptotes.

As | x | -> infinity the magnitude of the function -> infinity so there are no horizontal asymptotes. **

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RESPONSE -->

okay so you use 1st deriv to find max and min and the 2nd deriv to find concavity and pts of inflection?

and you only test concavity on the intervals using the points of inflection?

self critique assessment: 1

Critical points are points where the derivative is zero. Any local or relative maximum or minimum will occur at a critical point (the words 'local' and 'relative' in this context mean the same thing).

Just because a point is a critical point doesn't mean that it's a local max or min. For example the graph of y = x^3 has a critical point at x = 0, but it doesn't have a max or a min at this point. The graph is increasing to the left of x = 0, and increasing to the right of x = 0.

To test whether a critical point gives a local or relative max or min you have to see whether the function is increasing on one side and decreasing on the other side of the critical point. If it's increasing as you approach from the left, and decreasing as you move to the right of the point, then you have a maximum; if it's decreasing before the point then increasing after the point you have a minimum.

This is equivalent to saying that if the derivative goes from positive to negative the point is a max, and if from negative to positive the point is a min. So this is the first-derivative test.

The second derivative tells you how the derivative is changing. If the derivative is increasing then the second derivative is positive (and the graph of the function is concave up), and if the derivative is decreasing then the second derivative is negative (and the graph of the function is concave down). If the derivative goes from positive to negative then the derivative is decreasing (and the graph is concave down), so the second derivative is negative; this is the situation when the critical point is a relative maximum. If the derivative goes from negative to positive then the derivative is increasing (and the graph is concave up), so the second derivative is positive; this is the situation when the critical point is a relative minimum.

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20:30:51

Query 3.7.34 sketch (x^2+1)/(x^2-1)

Note: The problem in the text might be (x^2+1)/(x^2-2). If so the solution given below can be easily adapted to that function.

Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.

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RESPONSE -->

looking at the numerator, there are no zeros

f'(x)=-4x/(x^2-1)^2)

-4x=0, x=0 is crit pt

f''(x)=(4(3x^2+1))/(x^2-1)^3)

vert asymp at x= 1, -1

2nd deriv test:

(-i,-1) (-1,1) (1,i)

+ - +

up down up

confidence assessment: 2

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20:32:17

First we look for zeros and intercepts:

The numerator is never zero, being the sum of the positive number 1 and the nonnegative quantity x^2. So the function has no zeros, i.e., never crosses the x axis.

When x=0 we have y = (0^2+1)/(0^2-1) = -1 so the y intercept is (0,-1).

Next we analyze the derivative to see if we can find relative maxima and minima:

The derivative is - 4x/(x^2 - 1)^2, which has its only zero when x = 0. So (0, -1) is the only critical point.

The second derivative is 4(3x^2 + 1)/(x^2 - 1)^3, which is negative when x = 0. So the critical point gives us a maximum.

We analyze the second derivative to determine concavity:

The second derivative 4(3x^2 + 1)/(x^2 - 1)^3 has a numerator which is always positive, since x^2 is always positive. The denominator is negative where x^2 - 1 < 0, which occurs between x = -1 and x = 1.

So the 2d derivative is positive on (-infinity, -1) and on (1, infinity), where the graph will be concave up, and negative on (-1, 1), where the graph will be concave down.

Now we look for vertical and horizontal asymptotes:

The denominator of (x^2+1)/(x^2-1) is zero and the numerator isn't when x = +1 and also when x = -1. So we have vertical asymptotes at x = +1 and at x = -1.

}For very large positive x or for very large negative x the +1 in the numerator and the -1 in the denominator are both insignificant and the function value is very close to x^2 / x^2 = 1. The function approaches its horizontal asymptote y = 1 for both large positive x and large negative x.

We finally determine where the function is positive and where negative:

For x < -1 the function cannot change sign, since it is continuous and has no zeros. Testing any x < -1 gives us a positive value. On this interval the function is therefore positive.

The same is true for x > 1.

For -1 < x < 1 the same argument shows that the function is negative**

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RESPONSE -->

seems like excess verbiage... am i missing something in my answer?

self critique assessment: 2

You said most of this in abbreviated form. It's not necessary for you to explain things to the same extent I do; my goal is to give enough detail to cover whatever a student might not understand. Your goal is to summarize how you solved the problem and you can assume that if you communicate it, which you did here, then I'll understand it.

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assignment #029

029. `query 29

Applied Calculus I

12-10-2007

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20:42:35

Query 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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RESPONSE -->

chain rule

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

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

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

confidence assessment: 3

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20:42:48

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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RESPONSE -->

yup

self critique assessment: 3

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20:50:34

** Query 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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RESPONSE -->

y'=-4x dy=-4x(dx)

dy=-4(0)(-.1)

dy=0

actual change is f(x+dx)-f(x)

f(0-.1)-f(0)

f(-.1)-f(0)= (1-2(-.1)^2)-(1-0)= -.02

self critique assessment: 3

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20:50:43

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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RESPONSE -->

yup

self critique assessment: 3

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20:57:52

Query Extra Problem: Give the equation of the tangent line to y= 2 * x^(1/3) - 1 at (8,3); tan line prediction and actual fn value at `dx = -.01 and .01. **** What is the equation of the tangent line and how did you obtain it?

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RESPONSE -->

f'(x)=2/3x^(-2/3)

f'(8)=2/3(8)^(-2/3)

f'(8)=1/6

pt slope:

y-3=1/6(x-8)

y=1/6x+5/3

dx=.01, then x+dx=8.01

y=1/6(8.01)+5/3= 3.001666666 tan approx

actual 2(8.01)^(1/3)-1= 3.001665972

dx=-.01, then x+dx= 7.99

y=1/6(7.99)+5/3= 2.998333333 tan approx

actual 2(7.99)^(1/3)-1= 2.998332638

confidence assessment: 3

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20:58:09

f(x) = 2x^(1/3) - 1

f' (x) = 2/3x ^(-2/3)

f ' (8) = 2/3(8)^(-2/3)

f ' (8) = 1/6

y - 3 = 1/6(x - 8)

y - 3 = 1/6x - 8/6

y = 1/6x + 10/6

y = (1/6)x + (5/3) after simplification.

Using `dx = .01 we get x + `dx = 8.01. The tangent-line approximation is thus

y = 1/6 * 8.01 + 5/3 = 3.001666666.

The actual function value is 2 * 8.01^(1/3) - 1 = 3.001665972. The difference is .0000007, approx.

A similar difference is found approximating the function for `dx = -.01, i.e., at 7.99.

We see that for this short interval `dx the tangent-line approximation is very good, predicting the change in the y value (approximately .001666) accurate to 4 significant figures.

COMMON ERROR: Students often round off to 3.0017, or even 3.002, 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.

The difference should be no greater than -.02 * .01^2 = -.000002 (based on a Taylor's Theorem estimate I did in my head so don't hold me responsible for its accuracy, and you aren't responsible for Taylor's Theorem at this point of the course); the discrepancy might therefore appear in the 6th decimal place, almost certainly not later than the 7th. **

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RESPONSE -->

yup

self critique assessment: 3

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21:02:16

**** Query 3.8.42 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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RESPONSE -->

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

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

t=1 dt=.5

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

dc=.0478 mg/ml

confidence assessment: 3

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21:02:31

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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RESPONSE -->

yup

self critique assessment: 3

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This looks good. See my notes. Let me know if you have any questions. &#