course Mth271 xPcakʆhassignment #025
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13:15:47
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RESPONSE --> I accidently hit enter and saw the solution to this problem. The only thing I messed up was my primary equation was S= y + 3x, so my answers were x=8 and y = 24 Is this okay? confidence assessment: 3
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13:19:20 If we let x stand for the number of trees added to the 80 then the yield per tree is 400 - 32 x, and there would be 80 + x trees. The total yield would therefore be total yield = number of trees * yield per tree = (80 + x) * ( 400 - 32 x) = -32 x^2 + -2160 x + 32000, which is maximized when x = -34 approx.; this indicates -34 trees in addition to the 80, or 46 trees total. Another approach is to assume that only the additional trees experience the decrease. However it doesn't make sense for the yield decrease to apply only to the added trees and not to the original 400. If you're gonna crowd the orchard every tree should suffer. In any case, if we make the unrealistic assumption that the original 80 trees maintain their 400-apple-per-tree yield, and that the x additional trees each have a yield of 32 x below the 400, we have x added trees each producing 400 - 32 x apples, so we produce x (400 - 32 x) = 400 x - 32 x^2 additional apples. We therefore maximize the expression y = 400 x - 32 x^2. We obtain y ' = 400 - 64 x, which is 0 when -64 x + 400 = 0 or x = 6.25. Since y ' is positive for x < 6.25 and negative for x > 6.25 we see that 6.25 will be our maximizing value. We can't plant 6.25 trees, so the actual maximum must occur for either 6 or 7 trees. We easily see that the max occurs for 6 additional trees. So according to this interpretation we plant 86 trees. **
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RESPONSE --> I set my equation up completely wrong. This is something I struggle with, word problems are very difficult for me, and I can't figure out the primary equations. Now the equation is worked out for me, I understand the process. self critique assessment: 2]
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13:20:12 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The optimization problems give me problems. I found that if I have a primary equation to work off of, I can find the answer without much difficulty. However, on the problems that I have to set up myself, is where I run into major problems. self critique assessment: 1
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˩̴h⫩Oxģ assignment #026 026. `query 26 Applied Calculus I 05-13-2009
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13:22:57 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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RESPONSE --> P = R - C P = x(50-.1*sqrtx) - (35x + 500) P = 50x - .1x^2/3 - 35x + 500 P = 15x - .1x^2/3 + 500 f'(x) = 15 - .067x^-1/3 f'(x) = -.067x^1/3 = 15 x^1/3 = 223.89 s = 6.07 confidence assessment: 1
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13:24:32 Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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RESPONSE --> I was on the right track, but messed a couple signs up and didn't solve right, but I do see where I messed up and how to fix it. self critique assessment: 2
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13:24:41 ** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate?
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RESPONSE --> self critique assessment: 0
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13:26:30 According to my note here amount deposited A is proportional to the square of interest rate r so A = k r^2 for some proportionality constant k. The interest paid at rate r on amount A is A * r. The bank can reinvest at 12% so it gets return A * .12. The bank therefore nets .12 * A - r * A = (.12 - r) * A. Since A = k r^2 the bank nets profit P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3). We maximize this expression with respect to r: dP/dr = k * (.24 r - 3 r^2). dP/dr = 0 when .24 r - 3 r^2 = 0, when 3 r ( .08 - r) = 0, i.e., when r = 0 or r = .08. The second derivative is -6 r + .24, which is negative for r > .06. This shows that the critical point at r = .08 is a maximum. The max profit is thus P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k. In order to find the optimal interest rate it is not necessary to find the proportionality constant k. However if the proportionality constant was known we could find the max profit. **
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RESPONSE --> This is another problem I could find the equation to set up to solve the problem, but after seeing it, i can follow the process and come up with an answer. self critique assessment: 2
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ڢǯy{] assignment #027 027. `query 27 Applied Calculus I 05-13-2009
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17:41:54 Query 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
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RESPONSE --> lim = 0 confidence assessment: 2
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17:42:36 As you approach the vertical line x = -2 from the left (i.e., x -> -2-) y values drop asymptotically into unbounded negative values. So the limit is -infinity. **
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RESPONSE --> OK, I see that I looked to far into the question. I graphed the equation, and should have seen that it was - infinity. self critique assessment: 2
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17:43:04 Query 3.6.24 lim{x->infinity}( (5x^3+1) / (10x^3 - 3x^2 + 7) ) **** What is the desired limit and how did you obtain it?
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RESPONSE --> This equation stumped me. I had a hard time solving for x. confidence assessment: 0
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17:43:47 You can compare the leading terms of both numerator and denominator, which are both x^3 terms. The limit is therefore just the ratio of leading coefficiets: 5 / 10 = 1/2. A more rigorous algebraic treatment is sometimes called for if you are asked for a proof. See below: lim (x -> infinity) (5x^3 + 1)/(10x^3 - 3x^2 +7) = lim(x -> infinity) (5 x^3) / (10 x^3) = 5/10 or 1/2. You find this limit using the horizontal asymptote rules. Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is equal to a/b with a being the leading coefficient of the numerator and b being the leading coefficient of the denominator. You can also rearrange the expression by dividing numerator and denominator both by x^3 to get lim(x -> infinity) ( 5 + 1 / x^3) / ( 10 - 3 / x + 7 / x^3). Since the limits of 1/x^3, -3/x and 7 / x^3 are all zero you end up with just 5/10 = 1/2. **
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RESPONSE --> Again, I looked at the problem from the wrong angle. 5/10 = 1/2. self critique assessment: 2
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17:44:47 ** Query 3.6.50 sketch f(x) = (x-2) / (x^2-4x+3). **** Describe your graph, including a description of all intercepts, extrema, asymptotes and concavity.
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RESPONSE --> My graph has an asymptote of y = 0. y' = -1/2x^-3/2 y' = -1/ x^3/2 self critique assessment: 0
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17:45:48 The function has zeros when the numerator is 0, at x = 2; only one x intercept which occurs at x = 2. The numerator is negative on (-inf,2) and positive on (2, inf). The y intercept is at x = 0; we get (0, -2/3). Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-inf,1), (1, 3) and (3, inf) the denominator is respectively positive, negative and positive. The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive. For large pos and neg x values f(x) -> 0 so + and - x axis is an asymptote. First derivative is - (x^2 - 4x + 5)/(x^2 - 4x + 3)^2 and 2d derivative is 2(x - 2)(x^2 - 4x + 7)/(x^2 - 4x + 3)^3. The function is critical when 1st derivative is 0; numerator of 1st derivative is never 0 (quad formula, discriminant is negative) so there are no critical points. 2d derivative: x^2 - 4 x + 7 is always positive (discriminant negative, expression positive for x=0). Numerator zero only when x = 2. The denominator is zero at x=3, x=1. The 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf). Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up. So the function starts with a horizontal asymptote below the negative x axis, remains negative and concave down to a vertical asymptote at x=1. It then becomes positive with upward concavity, descending from a vertical asymptote at x = 1 to a zero at x = 2, then becoming negative with downward concavity as it approaches a negative asymptote at x = 3. To the right of x = 3 the graph descends from a positive vertical asymptote to a horizontal asymptote at the positive x axis, remaining positive and concave up. **
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RESPONSE --> I thought the 1/x asymptote was zero. Maybe it was the wrong equation I solved for. self critique assessment: 1
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17:47:24 SOLUTION TO PROBLEM #43:
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RESPONSE --> The equation has to have x^3. So my equation ended up being -x^3-X^2 + x +3 self critique assessment: 1
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17:48:25 The graph of x y^2 = 4 is not defined for either x = 0 or y = 0. The function has horizontal and vertical asymptotes at the axes. The graph of x y^2 = 4 is not defined for negative x because y^2 cannot be a negative number. However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation. You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x). The graph of the first-quadrant function will be decreasing since its derivative is negative, and will as you say be asymptotic to the x axis. The graph of the fourth-quadrant function is increasing and also asymptotic to the x axis. **
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RESPONSE --> This was number 45 for me. So I didn't get the right answer. My number 43 was similar, but not as detailed. What I need to do is analyze from every angle to make it much easier to graph. self critique assessment: 2
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nPwɅ assignment #028 028. `query 28 Applied Calculus I 05-13-2009
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17:50:28 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 --> y' = -3x^2 + 6x + 9 y' = -3(x^2 - 2x - 3 (3, 25) (-1,11) y'' = -6x + 6 y'' = -6(x-1) x = 1 (1,9) confidence assessment: 2
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17:52:08 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 --> I got all right, except for the asymptotes which I forgot to do. The second derivative I messed up, and I don't really know why I did. I'll have to go back and rework to see if I can come up with the right answer. self critique assessment: 2
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17:54:31 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 --> I began by finding the derivative y' = [(x^2 - 2)(2x) - [(x^2+1)(2x)]] / (x^2 - 2)^2 = [2x^3 - 4x - 2x^3 - 2x]/ x^4 - 4x^2 + 4] = -2x / x^4 - 4x^2 + 4 From here I got stuck, I don't really know where to go from here. Asymptotes are: +-*sqrt2 confidence assessment: 2
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17:55:04 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 --> I still don't really understand where I messed up or how I messed up. self critique assessment: 1
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ɧَӺm] assignment #029 029. `query 29 Applied Calculus I 05-13-2009
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17:56:15 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 --> dy/dx = *cubedrootx^2 dy/dx = 2/3x^-1/3dx confidence assessment: 2
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17:57:02 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 --> I should have used the product rule and it would have given me the right answer. self critique assessment: 2
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17:58:22 ** 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 --> delta y = f(-.1) - f(0) = 1-2(-.1)^2 - 1 = -.02 dy = (1-4x)(-.1) -.1 + .4x +.1 + 4(-.1) = -.14 self critique assessment: 1
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17:58:46 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 --> I don't understand where I messed up on this problem. self critique assessment: 1
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18:01:43 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 --> To obtain the tangent line, we must take the derivative y' = 2/3 x^-2/3 y' = 2 / 3*sqrt(x^2) at x = 8, we obtain y = 1/2 so (y - 1/2) /(x-8) = 0 y - 1/2 = x-8 y = x -81/2 confidence assessment: 1
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18:02:31 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 --> I got 1/2 instead of 1/6 but other than that I was on the right track. And I forgot to do the dy and dx part. Sorry. self critique assessment: 2
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18:03:47 **** 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 --> I have to say that I don't know how to set this problem up, or where to go. confidence assessment: 0.
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18:04:34 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 --> I can see now that we start off with the derivative. Then take the dC. Then evaluate at the t values self critique assessment: 2
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