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Unformatted text preview: 25 m/sec for t ≥ 10 seconds is constant. Limit v(t) − v(10)
v(t) − v(10)
≠ right derivative t → 10+
left derivative Limit0−
right
t→1
t − 10
t − 10
we say : v(t) is not differentiable at t = 10 secs.
differentiable Even though v(t) is not differentiable at t = 10 secs, we can still compute the left
differentiable
left
derivative and right derivative and interpret them according to the situation.
deriv tiv
right deriv tiv Exer
ercise1:
Exercise1: Given f ’(x) = AVERAGE RATE OF CHANGE of f(x), what is f(x) ?
a) ex
b) x
c) constant
d) x
e) ax + b 90 Example 4: Is the function y(t), that describes the height of a bouncing ball,
bouncing
differentiable at instants t0, t1, t2, t3, . . . ?
y(t) SLOPE of tangent at t1 from the left is negative
left negative h
e
i
g
h
t
t0 SLOPE of tangent at t1 from the right is postive
right postive n
t1 t2 t3 t ertical
At t1 we have two instantaneous ver tical speeds : the speed on impact at t1 while
instantaneous ver
descending and the speed after impact at t1 while rising. We know the duration of
rising
flight T = 2u0. sinθ/ g (see page 75). To find the speed on impact at t1 while
descending we substitute T = 0 and t = T0 in y0’(t) = u0. sinθ0 − g(t − t0)
to get y0’(t− ) = −u0. sinθ0 .
1
After impact at t1 we know this speed on rising is y1’(t+ ) = +u1.sinθ1. Alternatively,
rising
1
we may substitute t = t1 in y1’(t) = u1.sinθ1 − g(t − t1) to get the same result. Geometrically
Geometrically, we can see that we have two tangents at t1.
left derivative y0’(t− ) = −u0.sinθ0 ≠
1 y1’(t+ ) = +u1.sinθ1 right derivative
right
1 we say : y(t) is not differentiable at t = t1 secs.
differentiable
Even though y(t) is not differentiable at t = t0, t1, t2, t3, . . . , we can still
differentiable
compute the left derivative and right derivative and interpret them according to
left deriv tiv
right deriv tiv
the situation.
91 Example 5 : is f(x) = + √x differentiable at x = 0 ?
differentiable
f(x)
f(x) = +√ x (0, 0) x We know that f(x) =...
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 Fall '09
 TAMERDOğAN

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