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MATH 148 Fall 2006 Lecture Notes, Friday, 13 October 2006Stewart, 2.6, Tangents, Velocities, and Other Rates of ChangeExercise # 18, page 120: If an arrow is shot upward on the moon with a velocity of 58 m/s, its height(in meters) after t seconds is given by H D 58t  0:83t2.(a) Find the velocity of the arrow after one second.The velocity at a is given byv.a/ D limh ! 0H .a C h/  H .a/h;so the velocity at one second isv.1/ D limh ! 0H .1 C h/  H.1/h:Evaluating the difference quotient we getH .1 C h/  H .1/hDŒ58.1 C h/  0:83.1 C h/2  Œ58.1/  0:83.1/2hDŒ58.1 C h/  58.1/ C Œ0:83.1 C h/2 .0:83.1/2/hD58Œ.1 C h/  .1/  0:83Œ.1 C h/2 .1/2hD58Œh  0:83Œ1 C 2h C h2 1hD58h  0:83.2h C h2/hDhŒ58  0:83.2 C h/hD 58  0:83.2 C h/:Since the original difference quotient and the polynomial 58  0:83.2 C h/2have the same value forh near 0, they have the same limit at 0. Because polynomials are continuous, we may evaluate thislimit by direct substitution. Therefore,v.1/ D limh ! 0H .1 C h/  H .1/hD limh ! 058  0:83.2 C h/ D 58  0:83.2 C 0/ D 58  0:83.2/ m/s;so v.1/ D 58  1:66 D 56:34 m/s.(b) Find the velocity of the arrow when t D a.Since the limitv.a/ D limh ! 0H .a C h/  H .a/h;1MATH 148 Fall 2006 Lecture Notes, Friday, 13 October 2006has the form00, we must rewrite the limit quotient in a fashion similar to what was done to find thevelocity at one second. ThusH .a C h/  H .a/hDŒ58.a C h/  0:83.a C h/2  Œ58a  0:83a2hDŒ58.a C h/  58a C Œ0:83.a C h/2 .0:83a2/hD58Œ.a C h/  a  0:83Œ.a C h/2 a2hD58h  0:83Œ.a C h/ C a  Œ.a C h/  ahD58h  0:83.2a C h/  .h/hDhŒ58  0:83.2a C h/hD 58  0:83.2a C h/:Therefore, since the difference quotient is equal to a polynomial in h, we may evaluate the limit bydirect substitution,v.a/ D limh ! 058  0:83.2a C h/ D 58  0:83.2a C 0/ D 58  1:66a m/s:(c) When will the arrow hit the moon?The short answer is, when the height above the surface is zero.The longer answer is, find the time when the height function is zero. Let x denote the time when theheight function is zero. Thus, H.x/ D 0, so we find the time of impact by solving the equationH .x/ D 58x  0:83x2D 0for x.Now58x  0:83x2D x.58  0:83x/ so either x D 0 or 58  0:83x D 0. The time x D 0corresponds to the time the arrow was shot upwards. Solving 58  0:83x D 0 yields the time ofimpact, namely x D 58=0:83 69:879518 ::: seconds.(d) With what velocity will the arrow hit the moon?Part (c) tells us what time the arrow hit the moon. Part (b) tells us how fast the arrow is goingat any point in time, theoretically at least. Remember, this is only an exercise, sort of a “thoughtexperiment”. Real astronauts have not, as far as we know, actually shot any arrows on the surface ofthe moon. So we just substitute the time of impact into our velocity formulav.x/ D v.58=0:83/ D 58  1:66.58=0:83/ D 58  2.58/ D58 m/s:In a way, however, this answer actually makes sense. If the arrow is shot upwards at 58 m/s, sincethere is absolutely no air to impede the arrows motion, it should return to its original starting positionwith the same speed, only in the opposite


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