Math 1300 - 401 Worksheet: Section 2.1 Name: Solutions Patrick ...

Math 1300 - 401 Worksheet: Section 2.1 Name: SolutionsPatrick Marleau scores a hat trick and Erica is so happy that she throws her hat onto the ice. Theheight of the hat above the ice is given by the functions(t) = −4t 2 + 8t + 12where height is in meters and time is in seconds after the hat was thrown.(a) At what time does the hat hit the ice?We set the height equal to zero and solve for t:0 = −4t 2 + 8t + 120 = −4(t 2 − 2t − 3)0 = −4(t − 3)(t + 1)t = 3 or − 1. . . but -1 doesn’t make sense as an answer since the hat was thrown at time t = 0. So the hatmust hit the ice at time t = 3 seconds.(b) Find the average velocity of the hat. . .(i) . . . from t = 2 to t = 3. (Why is this number negative?)s(3) − s(2)= 0 − 12 = −12 m/s3 − 2 1This number is negative since the hat is falling downward.(ii) . . . from t = 2.9 to t = 3.s(3) − s(2.9)= 0 − 1.56 = −15.6 m/s3 − 2.9 0.1(iii) . . . from t = 2.99 to t = 3.s(3) − s(2.99)= 0 − 0.1596 = −15.96 m/s3 − 2.99 0.01(iv) . . . from t = 2.999 to t = 3.s(3) − s(2.999)= 0 − 0.015996 = −15.996 m/s3 − 2.999 0.001(c) Guess the instantaneous velocity of the hat at t = 3 based on your computations.-16 m/s, perhaps?

Math 1300 - 401 Worksheet: Section 2.1 Name: Solutions(d) Use a limit to calculate the instantaneous velocity of the hat at t = 3. Were you right?s ′ s(3 + h) − s(3)(3) = limh→0 h−4(3 + h) 2 + 8(3 + h) + 12 − ( −4(3) 2 + 8(3) + 12 )= limh→0 h−24h − 4h 2 + 8h= limh→0 h= lim(−16 − 4h)h→0= lim(−16) − 8 lim(h)h→0 h→0= −16 m/s(e) How fast was the hat going at the instant when it was first thrown into the air? Was it thrownupward or downward?It was thrown upward!s ′ s(0 + h) − s(0)(0) = limh→0 h−4h 2 + 8h + 12 − 12= limh→0 h= lim(−4h + 8)h→0= −4 lim(h) + limh→0= 8 m/sh→0(8)(f) What was the hat’s average velocity from the time it was thrown until the time it hit the ice?s(3) − s(0)3 − 0= 0 − 123 − 0= −4 m/s