(e) Draw the tangent line to s(t) at t = 1. Without doing any additional work, estimate the slope of that line.

Question

rocket with position function s(t) = −16t^2 + 200t from Exercise 2.

Answer 1

tangent line's slope = derivative of f(1) = f'(1) = -32(1) + 200 = -32 +200 = 168

at the point (1, 168)

maximum point is when velocity = -32t +200 = 0, when t = 200/32 = 25/4 = 6 1/4 seconds

max = -16(25/4)^2 + 200(25/4) = -625 +1250= 625 = the point (6 1/4, 625)

it's a downward opening parabola in vertex form

-16(t^2 -200t/16 +25/4) +16(100^2)/16^2 = 625

= -16(t- 25/4)^2 + 100 with vertex = (6 1/4, 625) = maximum

tangent line's equation is y -184 = 168(x-6.25) or

y = 168x -168(25)/4 + 184

y = 168x +16 which has slope = 168 and y intercept =16 = the point (0, 17)

and has x intercept = -16/168 =-2/21 = the point (-2/21, 0)