Thomas Calculus 13th [Solutions]

214 Chapter 3 Derivatives
CHAPTER 3
ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2 2sin cos d (sin 2 )
d
2 2
cos 2 cos sin
2 2
(b) cos 2 cos sin d (cos 2 )
d
d
(2sin cos ) 2cos 2
d 2 2
(cos sin )
d
d
2sin 2 (2cos )( sin ) (2sin )(cos )
sin 2 cos sin sin cos sin 2 2sin cos
2[(sin )( sin ) (cos )(cos )]
2. The derivative of sin ( x a) sin x cos a cos xsin
a with respect to x is cos( x a) cos x cos a sin xsin a,
2
which is also an identity. This principle does not apply to the equation x 2x 8
2
0, since x 2x 8 0 is
not an identity: it holds for 2 values of x ( 2 and 4), but not for all x.
2
3. (a) f ( x) cos x f ( x ) sin x f ( x ) cos x, and g( x)
a bx cx g ( x ) b 2 cx g ( x ) 2 c;
also, f (0) g (0) cos(0) a a 1; f (0) g (0) sin(0) b b 0; f (0) g (0)
cos(0) 2c c 1
2 . Therefore, g( x) 1 1 x
2 .
2
(b) f ( x) sin( x a) f ( x ) cos( x a), and g( x ) bsin x c cos x g ( x) b cos x c sin x ; also
f (0) g(0) sin( a ) bsin(0) c cos(0) c sin a; f (0) g (0) cos( a) b cos(0) csin(0)
b cos a . Therefore, g( x) sin x cos a cos xsin a.
(4)
2
(c) When f ( x) cos x, f ( x)
sin x and f ( x ) cos x; when g( x)
1
(4)
1 x , g ( x)
0 and g ( x) 0.
2
Thus f (0) 0 g (0) so the third derivatives agree at x 0 . However, the fourth derivatives do not
(4) (4)
agree since f (0) 1 but g (0) 0. In case (b), when f ( x) sin( x a ) and
g( x) sin x cos a cos xsin
a , notice that f ( x) g( x ) for all x, not just x 0. Since this is an identity, we
( n) ( n)
have f ( x) g ( x ) for any x and any positive integer n.
4. (a) y sin x y cos x y sin x y y sin x sin x 0; y cos x y sin x
y cos x y y cos x cos x 0; y a cos x bsin
x y asin x b cos x
y a cos x bsin
x y y ( a cos x bsin x ) ( a cos x bsin x) 0
(b) y sin(2 x)
y 2cos(2 x)
y 4sin(2 x)
y 4y 4sin(2 x) 4sin(2 x ) 0. Similarly,
y cos(2 x) and y a cos(2 x) bsin(2 x ) satisfy the differential equation y 4y 0. In general,
2
y cos( mx),
y sin( mx) and y a cos ( mx) bsin ( mx ) satisfy the differential equation y m y
0.
2 2 2 2
5. If the circle ( x h) ( y k)
a and y x 1 are tangent at (1, 2), then the slope of this tangent is
m 2x 2 and the tangent line is y 2 x . The line containing (h, k) and (1, 2) is perpendicular to
(1, 2)
y 2 x k 2 1
2 2 2
h 5 2k the location of the center is (5 2 k, k). Also, ( x h) ( y k)
a
h 1 2
2
2
1 ( )
x h ( y k)
y 0 1 ( y ) ( y k)
y 0 y y
.
k y
At the point (1, 2) we know y 2 from the
tangent line and that y 2 from the parabola. Since the second derivatives are equal at (1, 2) we obtain
2
1 (2) 9
2
2
2
2 k . Then h 5 2k 4 the circle is ( x 4) y 9 a . Since (1, 2) lies on the circle
k 2 2
2
5 5
we have that a 2 .
2
6. The total revenue is the number of people times the price of the fare: r( x) xp x 3 x , where 0 x 60.
40
2
The marginal revenue is dr 3 x 2x 3 x 1 dr 3 x 3 x 2x
3 3 x 1 x .
dx 40 40 40 dx 40 40 40 40 40
Then dr 0 x 40 (since x 120 does not belong to the domain). When 40 people are on the bus the
dx
2
marginal revenue is zero and the fare is p (40) 3 x
40
$4.00.
( x 40)
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