course Mth 173 10-22-2007......!!!!!!!!...................................
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RESPONSE --> We can find the derivative of the function By breaking it down into it's seperate parts. 70 , 120e^(z) , -.1t so we can ""chain"" together the different dirivatives. T'(t)= 120 (-.1e^(-.1t)) T'(t)= -12e^(-.1t) T'(5)= -12e^(-.1(5)) T'(5)= -7.28* F / m confidence assessment: 2
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21:26:08 The rate of temperature change is given by the derivative function ( T ( t ) ) ', also written T ' (t). Since T(t) is the sum of the constant function 70, whose derivative is zero, and 120 times the composite function e^(-.1 t), whose derivative is -.1 e^(-.1 t), we see that T ' (t) = 120 * ( -.1 e^(-.1 t) ) = -12 e^(-.1 t). Note that e^(-.1 t) is the composite of f(z) = e^z and g(t) = -.1 t, and that its derivative is therefore found using the chain rule. When t = 5, we have T ' (5) = -12 e^(-.1 * 5 ) = -12 e^-.5 = -7.3, approx.. This represents rate = change in T / change in t in units of degrees / minute, so at t = 5 minutes the temperature is changing by -7.3 degrees/minute.
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RESPONSE --> okay self critique assessment: 3
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21:31:04 `q002. The weight in grams of a growing plant is closely modeled by the function W(t) = .01 e^(.3 t ), where t is the number of days since the seed germinated. At what rate is the weight of the plant changing when t = 10?
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RESPONSE --> By breaking the function W(t)= .01e^(.3t) into it's parts we can find the derivative of the individual parts and chain them together. w(t)= .01, e^z, (.3t) W'(t)= .01(.3e^(.3t) W'(t)= .003e^(.3t) W'(t)= .003e^(.3(10)) W'(t)= .003e^(3) W'(t)= .06g / d confidence assessment: 2
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21:32:55 The rate of change of weight is given by the derivative function ( W ( t ) ) ', also written W ' (t). Since W(t) is .01 times the composite function e^(.3 t), whose derivative is .3 e^(.3 t), we see that W ' (t) = .01 * ( .3 e^(.3 t) ) = .003 e^(.3 t). Note that e^(.3 t) is the composite of f(z) = e^z and g(t) = .3 t, and that its derivative is therefore found using the chain rule. When t = 10 we have W ' (10) = .003 e^(.3 * 10) = .03 e^(3) = .06. Since W is given in grams and t in days, W ' will represent change in weight / change in clock time, measured in grams / day. Thus at t = 10 days the weight is changing by .06 grams / day.
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RESPONSE --> okay self critique assessment: 3
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21:42:10 `q003. The height above the ground, in feet, of a child in a Ferris wheel is given by y(t) = 6 + 40 sin ( .2 t - 1.6 ), where t is clock time in seconds. At what rate is the child's height changing at the instant t = 10?
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RESPONSE --> The function y(t)= 6+40 sin(.2t - 1.6) With individual functions of 6, 40(n), sin(z), .2t -1.6 within. y'(t)= 40(.2 cos (.2t-1.6)) y'(t)= 8 cos (.2t - 1.6) y'(t)= 8 cos (.2(10) -1.6) y'(t)= 8 cos (.4) y'(t)= 7.37 fps confidence assessment: 2
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21:43:29 The rate of change of altitude is given by the derivative function ( y ( t ) ) ', also written y ' (t). Since y(t) is the sum of the constant term 6, with derivative zero, and 40 times the composite function sin (.2 t - 1.6), whose derivative is .2 cos(.2 t - 1.6), we see that y ' (t) = 40 * ( .2 cos(.2 t - 1.6) ) = 8 cos(.2 t - 1.6). Note that sin(.2t - 1.6) is the composite of f(z) = sin(z) and g(t) = .2 t - 1.6, and that its derivative is therefore found using the chain rule. Thus at t = 10 seconds we have rate y ' (10) = 8 cos( .2 * 10 - 1.6) = 8 cos( .4) = 7.4, approx.. Since y represents altitude in feet and t represents clock time in seconds, this represents 7.4 feet per second. The child is rising at 7.4 feet per second when t = 10 sec.
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RESPONSE --> okay self critique assessment: 3
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22:13:25 `q004. The grade point average of a certain group of students seems to be modeled as a function of weekly study time by G(t) = ( 10 + 3t ) / (20 + t ) + `sqrt( t / 60 ). At what rate does the grade point average go up as study time is added for a typical student who spends 40 hours per week studying? Without calculating G(40.5), estimate how much the grade point average for this student would go up if she spend another 1/2 hour per week studying.
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RESPONSE --> G(t) = (10 +3t) / (20 +t) +'sqrt( t/60) I'm having trouble finding the derivative of this function so I'm going to have to do it the slope way. G(39) = (10 +3(39)) / (20 +(39)) + 'sqrt( (39) / 60) G(39) = 2.96 G(41) =(10 +3(41)) / (20 +(41)) + 'sqrt((41) / 60) G(41)= (2.99 (2.99 - 2.96) / (41- 39) = .015 points / hour of studying / week Had I been able to figure out the derivative I would have been able to plug 40.5 into the rate function and find out the slope at that point and then use differentiation to find out the point. By averaging the rate at 40 hrs. / week and at 40.5 hrs/week Then plugging 'dt and to the average rate to find the ammount change and add that to the G(t)=40. confidence assessment: 1
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22:18:31 The rate of change of grade point average is given by the derivative function ( G ( t ) ) ', also written G ' (t). Since G(t) is the sum of the quotient function (10 + 3 t ) / ( 20 + t), with derivative 50 / ( 20 + t ) ^ 2, and the composite function `sqrt( t / 60) , whose derivative is 1 / (120 `sqrt( t / 60) ), we see that G ' (t) = 50 / ( 20 + t ) ^ 2 + 1 / (120 `sqrt( t / 60) ). Note that `sqrt(t / 60) is the composite of f(z) = `sqrt(z) and g(t) = t / 60, and that its derivative is therefore found using the chain rule. Thus if t = 40 we have rate G ' (40) = 50 / ( 20 + 40 ) ^ 2 + 1 / (120 `sqrt( 40 / 60) ) = .024, approx.. Since G represents grade point and t represents weekly study time in hours, this represents .024 grade points per hour of weekly study time. The grade point is rising by .024 per additional hour of study. To estimate G(40.5) we assume that the .024 grade point rise per additional hour of study time remains valid as we increase study time from 40 to 40.5 hours. This is in increase of .5 hours in weekly study time so we would expect the grade point to go up by grade point change = .5 hours * .024 points / hour = .012 points. Since G(40) = ( 10 + 3 * 40) / (20 + 40) + `sqrt( 40 / 60) = 2.97 approx, the additional half-hour per week will tend to raise this by .012 to around 2.97. If the student is aiming for a 3-point, a couple more hours would do but the .5 hours won't.
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RESPONSE --> How did you find the derivative of the quotient function? Could the derivative of the 'sqrt function work like so: 'sqrt(t/60) = t/60 ^ (.5) y=x^n y'= nx^(n-1) so .5 (t/60) ^ (-.5) would this be okay for finding the function for the 'sqrt function and if no, then why not? self critique assessment: 2
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