Page 292 - Calculus Demystified

Chapter 2
position of the body at time t is given by 279
2
p(t) =−16t − 10 · t + 100.
The body hits the ground when
2
0 = p(t) =−16t − 10t + 100
or t ≈ 2.207 seconds.
(c) Since the ball has initial velocity 10 feet/second straight up, we
know that v 0 = 10. The initial height is h 0 = 100. Therefore the
position of the body at time t is given by
2
p(t) =−16t + 10 · t + 100.
The body hits the ground when
2
0 = p(t) =−16t + 10t + 100
or t ≈ 2.832 seconds.
d 1
6. (a) sin(ln(cos x)) = cos(ln(cos x)) · · (− sin x)
dx cos x
− sin x
= cos(ln(cos x)) · .
cos x
d sin(cos x) sin(cos x)
(b) e = e · cos(cos x) · (− sin x).
dx
d 1
(c) ln(e sin x + x) = · (cos x + 1).
dx e sin x + x
d 1
2
2
(d) arcsin(x + tan x) = ·[2x + sec x].
dx 1 −[x + tan x] 2
2
x
d −1 1 e
x
(e) arccos(ln x − e /5) = · − .
x
dx 1 −[ln x − e /5] 2 x 5
d 2 x 1 x
(f) arctan(x + e ) = ·[2x + e ].
2
x 2
dx 1 + (x + e )

7. Of course v(t) = p (t) = 12t − 5so v(4) = 43 feet/second. The average
velocity from t = 2to t = 8is
p(8) − p(2) 364 − 34
v av = = = 55.
6 6
Thederivativeofthevelocityfunctionis(v ) (t) = 12.Thisderivativenever

vanishes, so the extrema of the velocity function on the interval [5, 10] occur
at t = 5 and t = 10. Since v(5) = 55 and v(10) = 115, we see that the
maximum velocity on this time interval is 115 feet per second at t = 10.