course MTH 271 12/15, 1200 assignment #029029. `query 29
......!!!!!!!!...................................
11:24:40 Query 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.
......!!!!!!!!...................................
RESPONSE --> (6x^2)^(1/3) (1/3)(6x^2)^(-2/3)(12x) (4x)(6x^2)^(-2/3) dx confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:24:50 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:27:32 ** 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 --> y=1 - 2x^2 'dy=f(x + 'dx) - f(x) 'dy=f(0 + -.1) - f(0) 'dy=f(-.1) - f(0) 'dy=.98 - 1 'dy= -.02 dy=f'(x)dx dy=f'(0)dx dy=-4(0)(-.1) dy=0 ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:27:39 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:37:13 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?
......!!!!!!!!...................................
RESPONSE --> f(x)= 2x^(1/3) - 1 f'(x)=(1/3)(2x^(-2/3)) f'(x)=(2/3) x^(-2/3) At point (8,3): f'(8)=(2/3)(8^(-2/3)) f'(8)=(2/3)(1/4) f'(8)=(2/12)=1/6=m Using slope-intercept form: y - 3= (1/6) (x - 8) y-3=(1/6)x - (8/6) y=(1/6)x -(8/6) + (18/6) y=(1/6)x + (10/6) 'dx=-.01 x=8 - .01=7.99 y=(1/6)(7.99) + (10/6) y=2.998 'dx=.01 x=8 + .01=8.01 y=(1/6)(8.01) + (10/6) y=3.002 Both values are found using the 'dx are very close to the given value of y=3. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:38:37 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 --> Apparently, I committed the common error of rounding off my answer prematurely and not using enough significant figures to show the value difference. ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:44:05 **** 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.
......!!!!!!!!...................................
RESPONSE --> C=(3t)/ (27 + t^3) C'=((27 + t^3)(3) - (3t)(3t^2) )/ (27 + t^3)^2 C'=(81 + 3t^3 - 9t^3 )/ (27 + t^3)^2 C'=[(81 - 6t^3) / (27 + t^3)^2] dx [(81 - 6(1^3)) / (27 + (1^3))^2] (.5) [(81-6) / (28^2)] (.5) (75/784) (1/2) =.048 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:44:13 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
................................................."
course MTH 271 12/15, 1200 assignment #029029. `query 29
......!!!!!!!!...................................
11:24:40 Query 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.
......!!!!!!!!...................................
RESPONSE --> (6x^2)^(1/3) (1/3)(6x^2)^(-2/3)(12x) (4x)(6x^2)^(-2/3) dx confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:24:50 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:27:32 ** 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 --> y=1 - 2x^2 'dy=f(x + 'dx) - f(x) 'dy=f(0 + -.1) - f(0) 'dy=f(-.1) - f(0) 'dy=.98 - 1 'dy= -.02 dy=f'(x)dx dy=f'(0)dx dy=-4(0)(-.1) dy=0 ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:27:39 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:37:13 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?
......!!!!!!!!...................................
RESPONSE --> f(x)= 2x^(1/3) - 1 f'(x)=(1/3)(2x^(-2/3)) f'(x)=(2/3) x^(-2/3) At point (8,3): f'(8)=(2/3)(8^(-2/3)) f'(8)=(2/3)(1/4) f'(8)=(2/12)=1/6=m Using slope-intercept form: y - 3= (1/6) (x - 8) y-3=(1/6)x - (8/6) y=(1/6)x -(8/6) + (18/6) y=(1/6)x + (10/6) 'dx=-.01 x=8 - .01=7.99 y=(1/6)(7.99) + (10/6) y=2.998 'dx=.01 x=8 + .01=8.01 y=(1/6)(8.01) + (10/6) y=3.002 Both values are found using the 'dx are very close to the given value of y=3. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:38:37 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 --> Apparently, I committed the common error of rounding off my answer prematurely and not using enough significant figures to show the value difference. ------------------------------------------------ Self-critique Rating:ent: 3
.................................................
......!!!!!!!!...................................
11:44:05 **** 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.
......!!!!!!!!...................................
RESPONSE --> C=(3t)/ (27 + t^3) C'=((27 + t^3)(3) - (3t)(3t^2) )/ (27 + t^3)^2 C'=(81 + 3t^3 - 9t^3 )/ (27 + t^3)^2 C'=[(81 - 6t^3) / (27 + t^3)^2] dx [(81 - 6(1^3)) / (27 + (1^3))^2] (.5) [(81-6) / (28^2)] (.5) (75/784) (1/2) =.048 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
11:44:13 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
................................................."