(d) Draw the secant line of s(t) from t = 1 to t = 1.01. Without doing any additional work, what is the slope of that line?

Question

 model rocket with position function s(t) = −16t^2 + 200t

Answer 1

f(t) = -16t^2 +200t from t= 1 to 1.01

virtually the same as slope of the tangent line = the derivative of f(t) at t=1 equals f'(t) = -32t+200 =-32+200=168 or -32(1.01)+200=-32.32+200=167.68

try the average = (168 + 167.68)/2 = 167.84, as the secant line approaches the tangent line as change in t approaches zero.

or rewrite in vertex form -16(t^2 -200t/16 + 625/16) +625/16

-16(t +25/4)^2 +25/4 with vertex at (-25/4, 25/4) = maximum point = (-6.25, 6.25), then as t increases form -6.25 to t = 1 and 1.01, the slope increases very rapidly It's a parabola, downward opening. a very large positive slope, likely in the hundreds

or

slope of secant line = change in f(t) over change in t = (f(1.01) - f(1))/(1.01 -1) = (-16(1.01)^2 +200(1.01) - (-16 +200))/.01

= (-16.3216 +202 + 16 -200)(100) = (--0,3216 +2)(100)= 100(1.6784) = 167.84 which is the same identical answer, calculated in a far more tedious manner

s' = velocity of the rocket at time t, acceleration = s" = -32 feet per second per second due to gravity