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14:37:57
Note that there are 4 questions in this assignment. `q001. The Fahrenheit temperature T of a potato just taken from the oven is given by the function T(t) = 70 + 120 e^(-.1 t), where t is the time in minutes since the potato was removed from the oven. At what rate is the temperature changing at t = 5?......!!!!!!!!...................................
RESPONSE --> T(t) = 70 + 120 e^(-.1 t)the rate of change is the same as the actual function for e^t, so: T(t)' = 70 + 120 e^(-.5) T(t)' = 70 +72.78 T(t)' =142.78
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14:54:45 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.......!!!!!!!!...................................
RESPONSE --> oh. musta missed the part where the a*b=y function is applied to the der. rule. but that stilldoesn't tell me why the -.1 is put *e. i thought that applied to the x^ whatever deriv rule, but it doesn't fit that either-???
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15:00:14 `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 --> W(t) = .01 e^(.3 t ) W(t)' = .01 (.3)e^(.3 (10) ) W(t)' = .003e^3 W(t)' = .003(20.09) W(t)' = .060
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15:00:26 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.......!!!!!!!!...................................
RESPONSE --> ok.
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15:15:26 `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 --> y(t) = 6 + 40 sin ( .2 t - 1.6 )
y(t)' =40(.2)cos ( .2 t - 1.6 ) y(t)' =8cos ( .2 t - 1.6 ) ???.................................................
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15:25:40 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.......!!!!!!!!...................................
RESPONSE --> ok. so i got the der.formula right, kinda? i wasn't sure what the 1.6 had to do.
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15:41:58 `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 ) G(t)' = ( 3t ) / (t ) + `sqrt( t / 60 ) G(t)' = 3+ ( t / 60 )??????? ok.....im lost
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15:45:04 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.......!!!!!!!!...................................
RESPONSE --> ok. im still lost. i can'tfind the" ' of t/thatever " ????
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