294
Chapter 3 | Derivatives
Using the Chain Rule in a Velocity Problem
A particle moves along a coordinate axis. Its position at time t is given by s ( t ) = sin(2 t )+cos(3 t ). What is the velocity of the particle at time t = π 6 ?
Solution To find v ( t ), the velocity of the particle at time t , we must differentiate s ( t ). Thus, v ( t ) = s ′( t ) =2cos(2 t )−3sin(3 t ). Substituting t = π 6 into v ( t ), we obtain v ⎛ ⎝ π 6 ⎞ ⎠ =−2.
3.39 A particle moves along a coordinate axis. Its position at time t is given by s ( t ) = sin(4 t ). Find its acceleration at time t .
Proof At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example, we assume that g ( x ) ≠ g ( a ) for x ≠ a in some open interval containing a . We begin by applying the limit definition of the derivative to the function h ( x ) to obtain h ′( a ): h ′( a ) = lim x → a f ⎛ ⎝ g ( x ) ⎞ ⎠ − f ⎛ ⎝ g ( a ) ⎞ ⎠ x − a .
Rewriting, we obtain
f ⎛
⎝ g ( x ) ⎞ ⎠ g ( x )− g ( a ) · ⎞ ⎠ − f ⎛ ⎝ g ( a )
g ( x )− g ( a ) x − a .
h ′( a ) = lim x → a
Although it is clear that
g ( x )− g ( a ) x − a
lim x → a
= g ′( a ),
it is not obvious that
lim x → a ⎞ ⎠ . To see that this is true, first recall that since g is differentiable at a , g is also continuous at a . Thus, lim x → a g ( x ) = g ( a ). Next, make the substitution y = g ( x ) and b = g ( a ) and use change of variables in the limit to obtain lim x → a f ⎛ ⎝ g ( x ) ⎞ ⎠ − f ⎛ ⎝ g ( a ) ⎝ g ( x ) ⎞ ⎠ g ( x )− g ( a ) = ⎞ ⎠ − f ⎛ ⎝ g ( a ) f ′ ⎛ ⎝ g ( a ) ⎞ ⎠ g ( x )− g ( a ) = lim y → b f ( y )− f ( b ) y − b = f ′( b ) = f ′ ⎛ ⎝ g ( a ) ⎞ ⎠ . f ⎛
Finally,
f ⎛
⎝ g ( x ) ⎞ ⎠ g ( x )− g ( a ) · ⎞ ⎠ − f ⎛ ⎝ g ( a )
g ( x )− g ( a ) x − a
⎛ ⎝ g ( a )
⎞ ⎠ g ′( a ).
h ′( a ) = lim x → a
= f ′
□
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