course Mth 173 lֶPĭxgֱassignment #013
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22:45:40 `qNote 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?
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RESPONSE --> T ' (t) = 120 * ( -.1 e^(-.1 t) ) = -12 e^(-.1 t) t = 5 T' (5) = -12 e^(-.1 * 5 ) = -12 e^-.5 = -7.3 degrees per minute confidence assessment: 3
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22:45:46 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 --> ok self critique assessment: 3
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22:47:09 `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 * ( .3 e^(.3 t) ) = .003 e^(.3 t) t = 10 W ' (10) = .003 e^(.3 * 10) = .03 e^(3) = .06 grams per day confidence assessment: 3
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22:47:17 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 --> ok self critique assessment: 3
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22:48:31 `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) = 40 * ( .2 cos(.2 t - 1.6) ) = 8 cos(.2 t - 1.6) t = 10 seconds y ' (10) = 8 cos( .2 * 10 - 1.6) = 8 cos( .4) = 7.4 feet per second confidence assessment: 3
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22:48:39 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 --> ok self critique assessment: 3
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22:51:03 `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) = 50 / ( 20 + t ) ^ 2 + 1 / (120 `sqrt( t / 60) ) t = 40 G ' (40) = 50 / ( 20 + 40 ) ^ 2 + 1 / (120 `sqrt( 40 / 60) ) = .024 .5 hours * .024 points / hour = .012 points G(40) = ( 10 + 3 * 40) / (20 + 40) + `sqrt( 40 / 60) = 2.97 If the student is aiming for a 3-point, a couple more hours would do but the .5 hours won't. confidence assessment: 3
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22:51:10 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 --> ok self critique assessment: 3
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݁ѮXyv| assignment #013 [v͔ddh Calculus I 10-12-2008
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22:52:54 query problem 2.4.37 . Which graph matches the graph of the bus and why?
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RESPONSE --> The bus makes periodic stops, the graph for bus III only comes to a stop once. The graph matches bus II.
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22:52:59 ** The bus only makes periodic stops, whereas the graph for III only comes to a stop once. I would matche the bus with II. **
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RESPONSE --> ok
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22:54:13 describe the graph of the car with no traffic and no lights
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RESPONSE --> The car matches up with bus I. Bus I is a continuous, straight, horizontal line that has constant velocity, and has no traffic and no lights.
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22:54:16 ** The car matches up with (I), which is a continuous, straight horizontal line representing the constant velocity of a car with no traffic and no lights. *&*&
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RESPONSE --> ok
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22:55:17 describe the graph of the car with heavy traffic
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RESPONSE --> The car in heavy traffic will sometimes speed and sometimes be slow.It matches the graph in III with its frequent increases and decreases.
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22:55:40 ** The car in heavy traffic would do a lot of speeding up and slowing down at irregular intervals, which would match the graph in III with its frequent increases and decreases. **
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RESPONSE --> ok
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22:56:13 query 2.5.10 (was 2.4.8) q = f(p) (price and quantity sold)what is the meaning of f(150) = 2000?
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RESPONSE --> q = 2000 when p = 150 When price is 150 I will sell 2000 units.
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22:56:18 *&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*&
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RESPONSE --> ok
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22:57:53 what is the meaning of f'(150) = -25?
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RESPONSE --> f' is the derivative f'(150) = -25 when price is$150 tit will change at a rate of -25 units per dollar increase.
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22:57:57 ** f' is the derivative, the limiting value of `df / `dp, giving the rate at which the quantity q changes with respect to price p. If f'(150) = -25, this means that when the price is $150 the price will be changing at a rate of -25 units per dollar of price increase. Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. **
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RESPONSE --> ok
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22:57:57 ** f' is the derivative, the limiting value of `df / `dp, giving the rate at which the quantity q changes with respect to price p. If f'(150) = -25, this means that when the price is $150 the price will be changing at a rate of -25 units per dollar of price increase. Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. **
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RESPONSE -->
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22:59:21 ** When you fall without a parachute v will inrease, most rapidly at first, then less and less rapidly as air resistance increases. When t = 0 we presume that v = 0. The graph of v vs. t is therefore characterized as an increasing graph beginning out at the origin, starting out nearly linear (the initial slope is equal to the acceleration of gravity) but with a decreasing slope. The graph is therefore concave downward. At a certain velocity the force of air resistance is equal and opposite to that of gravity and you stop accelerating; velocity will approach that 'terminal velocity' as a horizontal asymptote. The reason for the concavity is that velocity increases less and less quickly as air resistance increases; the approach of the velocity to terminal velocity is more and more gradual **
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RESPONSE --> This is an increasing graph beginning out at the origin, starting out nearly linear, and with a decreasing slope. The graph is concave downward
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23:00:07 What does the t = 0 acceleration indicate?
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RESPONSE --> t = 0 this is the acceleration under the force of gravity
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23:00:10 ** t = 0 acceleration is acceleration under the force of gravity, before you build velocity and start encountering significant air resistance. Acceleration is rate of velocity change, indicated by the slope of the v vs. t graph. **
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RESPONSE --> ok
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23:00:29 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> this assignment was very helpful.
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