In the previous step we found the following just before using a direct substitution to find the velocity at a particular time. (16(a + h) — 1.86(a + h)²) – (16a – 1.86a²) h v(a) = lim h→0 To find the velocity at a general time t = a, we need to first simplify as much as possible. (16(a + h) — 1.86(a + h)²) – (16a 1.86a²) h 16a+ 16h 1.86 a² + v(a) = lim h→0 = lim h→0 = lim h→0 16a + 16h 1.86a² 16h- = lim h→0 = lim (16-( h h h ah - 1.86h2 a 1.86h ]× )ah + h²) - - 16a + 1.86a² ah- 1.86h2 16a + 1.86a²

The deck of a bridge is suspended 285 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by  y = 285 − 16t2. (a) Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time. (i) 0.1 seconds  (ii)0.05 seconds  (iii)0.01 seconds  (b) Estimate the instantaneous velocity (in ft/s) of the pebble after 2 seconds.

Please graph the function of D(t)= 28sin(360/365 t - 1.4)

Suppose that an oil spill in a lake covers a circular area and that the radius of the circle is increasing according to the formula r = f(t) = 12 + t1.65, where t represents the number of hours since the spill was first observed and the radius r is measured in meters. (Thus when the spill was first discovered, t = 0 hr, and the initial radius was r = f(0) = 12 + 01.65 = 12 m.) (a) Let A(r) = πr2, as in Example 5. Compute a table of values for the composite function A ∘ f with t running from 0 to 5 in increments of 0.5. (Round each output to the nearest integer.) Then use the table to answer the questions that follow in parts (b) through (d). t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5   A                         (b) After one hour, what is the area of the spill (rounded to the nearest 10 m2)?  m2(c) Initially, what was the area of the spill (when t = 0)?  m2Approximately how many hours does it take for this area to double?  hr(d) Compute the average…