Derivative Worksheet With Answers Page 13
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14S E C T I O N 4.1
Linear Approximation and Applications
367
n
1
2
Finally, from the following table, we see that for h D 10
, 1 Ä n Ä 4, E Ä
Kh
.
2
1
2
h
E D j tan.
C h/
1
2hj
Kh
4
2
1
2
2
10
2:305
10
3:10
10
2
4
4
10
2:027
10
3:10
10
3
6
6
10
2:003
10
3:10
10
4
8
8
10
2:000
10
3:10
10
Further Insights and Challenges
1
5
3
71. Compute dy=dx at the point P D .2; 1/ on the curve y
C 3xy D 7 and show that the linearization at P is L.x/ D
x C
.
3
3
Use L.x/ to estimate the y-coordinate of the point on the curve where x D 2:1.
3
Differentiating both sides of the equation y
C 3xy D 7 with respect to x yields
SOLUTION
dy
dy
2
3y
C 3x
C 3y D 0;
dx
dx
so
dy
y
D
:
2
dx
y
C x
Thus,
ˇ
ˇ
dy
1
1
ˇ
D
D
;
ˇ
2
dx
1
C 2
3
.2;1/
and the linearization at P D .2; 1/ is
1
1
5
L.x/ D 1
.x
2/ D
x C
:
3
3
3
Finally, when x D 2:1, we estimate that the y-coordinate of the point on the curve is
1
5
y
L.2:1/ D
.2:1/ C
D 0:967:
3
3
5
72. Apply the method of Exercise 71 to P D .0:5; 1/ on y
2x D 1 to estimate the y-coordinate of the point on the curve
C y
where x D 0:55.
5
Differentiating both sides of the equation y
C y
2x D 1 with respect to x yields
SOLUTION
dy
dy
4
5y
C
2 D 0;
dx
dx
so
dy
2
D
:
4
dx
5y
C 1
Thus,
ˇ
ˇ
dy
2
1
ˇ
D
D
;
ˇ
2
dx
5.1/
C 1
3
.0:5;1/
and the linearization at P D .0:5; 1/ is
Â
Ã
1
1
1
5
L.x/ D 1 C
x
D
x C
:
3
2
3
6
Finally, when x D 0:55, we estimate that the y-coordinate of the point on the curve is
1
5
y
L.0:55/ D
.0:55/ C
D 1:017:
3
6
4
4
73. Apply the method of Exercise 71 to P D . 1; 2/ on y
C 7xy D 2 to estimate the solution of y
7:7y D 2 near y D 2.
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