On 8/7 I went thru the last 2 assignments and copied off your questions on the Query problems. I did this because I only have access at work. Copying the problems allowed me to work on them at home. I didn't want you to think I was trying to cheat. (You will know I didn't cheat after you review the attached assignments.) I had submitted a test taken form a couple of times (for test#2). Have you had a chance to grade the test yet? thank you
......!!!!!!!!...................................
12:52:26 Query 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
......!!!!!!!!...................................
RESPONSE --> as x approaches 2 from the negative side: from -inf to -2 the graph approaches -inf from 0 from -2 to 2 graph approaches 0.25 from - inf.
.................................................
......!!!!!!!!...................................
12:52:29 08:31:56
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
12:53:42 ** 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. **
......!!!!!!!!...................................
RESPONSE --> was the original question x-> pos 2?
.................................................
......!!!!!!!!...................................
17:23:12 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?
......!!!!!!!!...................................
RESPONSE --> (5x^3+1) / (10x^3 - 3x^2 + 7) try and factor numerator and denom. (5x^3 +1)/(5x^3+1)
.................................................
......!!!!!!!!...................................
17:23:16 08:48:03
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:24:23 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:19:13 **** 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.
......!!!!!!!!...................................
RESPONSE --> f(x) = (x-2) / (x^2-4x+3) x y -10000, -.00001 -1000 , -.0001 -100 , -.001 -10 ,-.01 -1 ,-.375 0 , -.666 1 , - inf 10 , .127 100 , .01 1000 , .001 10000 , .0001 graph never crosses the x axis, intercepts the y axis at x=2. graph is not continous and has no extrema. the asymptote is as x approaches 1 from right or left. the graph concaves down as it approaches 1 from the right and left sides.
.................................................
......!!!!!!!!...................................
18:20:43 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:20:47 SOLUTION TO PROBLEM #43:
......!!!!!!!!...................................
RESPONSE -->
.................................................
w{철Xz assignment #027 ݺӯD Applied Calculus I 08-07-2006
......!!!!!!!!...................................
18:22:51 Query 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:22:53 08:31:56
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:22:57 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:31 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:33 08:48:03
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:39 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:24:14 **** 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:24:16 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:01:37 SOLUTION TO PROBLEM #43:
......!!!!!!!!...................................
RESPONSE --> g(x)= x^2/(x^2-16) intercepts y= 0 when x= 0 asymptotes occur when x= -4 and 4
.................................................
......!!!!!!!!...................................
19:02:07 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
i\ٱxY assignment #028 ݺӯD Applied Calculus I 08-11-2006
......!!!!!!!!...................................
15:31:52 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.
......!!!!!!!!...................................
RESPONSE --> y = -x^3+3x^2+9x-2 factor using synthetic division (x+2) (-x^2 +5x -1) x intercepts are x=-2 f '(x) = (x+2)(-2x +5) + (-x^2 +5x -1) = -2x^2 +5x -4x +10 -x^2 +5x -1 = -3x^2 +6x +9 = -3(x^2 -2x -3) the critical # is 3 f ""(x) = -6x + 6 = 6(-x+1) -inf to -1 decreasing concave downward point of inflection = -1 -1 to 3 increasing concave upward relative max =25 3 to inf decreasing concave downward
.................................................
......!!!!!!!!...................................
15:55:30 ** 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
......!!!!!!!!...................................
RESPONSE --> i did not use the quadratic equation before taking the derivative. this skewed my data.
.................................................
......!!!!!!!!...................................
15:57:19 (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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
16:46:22 Query 3.7.32 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.
......!!!!!!!!...................................
RESPONSE --> (x^2+1)/(x^2-1) factor (x^2+1)/(x+1)(x-1) no y intercept asymptote -1 & 1 f '(x) = [(x^2 -1) 2x - (x^2+1) 2x]/ (x^2 -1)^2 = 2x^3-2x-2x^3-2x/(x^2 -1)^2 =0/(x^2 -1)^2 f "" (x)= 0
.................................................
......!!!!!!!!...................................
16:49:37 ** 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**
......!!!!!!!!...................................
RESPONSE --> i messed this up. i got the y intercept wrong, the derivative and the second derivative wrong. i think i understand the basic concept.
.................................................
wrݩd assignment #029 ݺӯD Applied Calculus I 08-10-2006
......!!!!!!!!...................................
12:51:24 Query 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.
......!!!!!!!!...................................
RESPONSE --> cube root of (6x^2) y=(6x^2)^1/3 dy/dx= 1/3(6x^2)^ -2/3 (12x) dy=1/3(6x^2)^ -2/3 (12x) (dx)
.................................................
......!!!!!!!!...................................
12:53:16 ** 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 **
......!!!!!!!!...................................
RESPONSE --> i think i got it. i did not carry my equation all the way out.
.................................................
......!!!!!!!!...................................
12:59:28 **** 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?
......!!!!!!!!...................................
RESPONSE --> f(x) = 1-2x^2 f ' (x)= -4x dy= f'(x)dx estimate dy= -4(0) -.1 =0 actual dy=f (x+dx) - f(x) = -.1- 0 = -.1 difference -.1-0 = -.1
.................................................
......!!!!!!!!...................................
13:01:25 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i did not plug in my origin equation when solving the actual formula.
.................................................
......!!!!!!!!...................................
16:40:59 Query 3.8.22 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?
......!!!!!!!!...................................
RESPONSE --> y=2 x^(1/3) - 1 dy/dy=2/3(x)^-2/3= 2/3(x)^2/3 = 2/3(8)^2/3 =2.67
.................................................
......!!!!!!!!...................................
16:44:50 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i messed up on my derivative in the beginning of the process.
.................................................
......!!!!!!!!...................................
17:02:58 **** Query 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.
......!!!!!!!!...................................
RESPONSE --> C = 3t / (27+t^3) dC/dt= {[(27+t^3) 3 - 3t(3t^2)]/ (27+t^3) ^2} dt = {81+ 3t^3 -3t^3 - 6t/(27+t^3) ^2}(0.5) ={ -6t + 81/ (27+t^3) ^2} (0.5) ={ -6(1) + 81/ (27+(1)^3) ^2} (0.5) =.048 f'(t) = 3(1.5) / (27+(1.5)^3) - 3(1) / (27+(1)^3) = .041 .048-.041= .007
.................................................
......!!!!!!!!...................................
17:03:47 ** 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 **
......!!!!!!!!...................................
RESPONSE --> i think i got it.
.................................................
"