Page 168 - Calculus Workbook For Dummies

152 Part III: Differentiation

Reject the negative solution because it’s outside the interval of interest: t = 0 to t = 5. So, the
only zero velocity occurs at t = 1.236 seconds.
Because v 0 = .25 meters/second and v 5 . - .037, the first derivative test tells you that
^ h
^ h
s ^ . 1 236h must be a local max and therefore a turnaround point.
The sloth thus goes right from t = 0 till t = 1.236 seconds, then turns around at s 1 ^ .236h, or
about .406 meters to the right of the trunk, and goes left till t = 5.
b. His total distance is the sum of the lengths of the two legs:
going right = s ^ . 1 236 - ^h s 0h
= .405 - .25
. .155
going left = s 5 - ^h s . 1 236h
^
. .198
Total distance is therefore .155 + .198 = .353 meters. His average speed is thus .353/5, or
.071 meters/second. That’s roughly a sixth of a mile/hour — much more like it for a sloth.
That’s pretty darn slow, but how quick do you think you’d be with only three toes?
c. Total displacement is defined as final position minus initial position, so that’s
6 1
s 5 - ^h s 0 = -
h
^
29 4
. - .043 meters
-
And thus his average velocity is .043- /5, or .0086 meters / second. You’re done.
2
o Two lines through the point (1, –3) are tangent to the parabola y = x . Determine the points of
tangency. The points of tangency are (–1, 1) and (3, 9).
1. Express a point on the parabola in terms of x.
2
,
The equation of the parabola is y = x , so you can take a general point on the parabola x yi
_
2
2
,
and substitute x for y. So your point is x x i.
_
2. Take the derivative of the parabola.
y = x 2
y = l 2 x
y 2 - y 1
3. Using the slope formula, m = x 2 - x 1 , set the slope of the tangent line from ,1 - 3i to
_
2
_ , x x i equal to the derivative. Then solve for x.
2
x - - 3h 2
^
x - 1 = x
2
3
2
x + = 2 x - 2 x
2
x - 2 x - = 0
3
3 =
^ x + 1 ^h x - h 0
x = - 1 or 3
2
4. Plug these x-coordinates into y = x to get the y-coordinates.
2
1 =
y = - h 1
^
2
y = 3 = 9
So there’s one line through (1, –3) that’s tangent to the parabola at (–1, 1) and another
through (1, –3) that’s tangent at (3, 9). You may want to confirm these answers by graphing
the parabola and your two tangent lines:
2
1 =
y = - ^ x + h 1
3 +
6
y = ^ x - h 9