Handout 6, Summer 2014
Math 1823-171
02 June 2014
1. Use linear approximation to estimate sin(31◦ ).
Solution. Let f (x) = sin(x). Note that 30◦ is near 31◦ , and that
f (30◦ ) = sin(30◦ ) = 21 is easy to calculate. Therefore, we estimate
sin(31◦ ) with the line tangent to f (x) at x = 30◦ . We have:
L(x) = f 0 (30)(x − 30) + f (30)
Note that f√(30) = 12 as above. Also, f 0 (x) = cos(x), so that f 0 (30◦ ) =
cos(30◦ ) = 23 . Therefore,
√
1
3
(x − 30) +
L(x) =
2
2
Therefore, it should be the case that L(31◦ ) is close to f (31◦ ) =
sin(31◦ ), so that:
√
√
1
3
3−1
sin(31◦ ) = f (31◦ ) ≈ L(31) =
(31 − 30) + =
2
2
2
We have the result.
2. Find the linearization of f (x) = x1/3 at a = 8.
Solution. Recall that the linearization of f (x) at a = 8 is simply the
equation of the line tangent to f (x) at x = 8. We came up with a nice
formula for this equation in class:
L(x) = f 0 (8)(x − 8) + f (8)
We have that f (8) = 2, and that f 0 (x) = 13 x−2/3 . Therefore, we have:
1
1
1
f 0 (8) = 8−2/3 =
√
2 =
3
3
12
3 8
Therefore, we have:
L(x) =
1
(x − 8) + 2
12
This is a sufficient result.
3. Use linear approximation to estimate
1
1
.
4.002
2
Solution. Let f (x) = x1 . Then f (4) = 14 is easy to calculate, and 4 is
quite near 4.002. Therefore, as in 1 above, we use the linearization of
f at 4 to estimate f (4.002). We have:
L(x) = f 0 (4)(x − 4) + f (4)
We have f (4) = 14 easily. Since f (x) = x−1 , we have that f 0 (x) =
−x−2 = −1
. Therefore, f 0 (4) = −1
. This gives:
x2
16
L(x) =
−1
1
(x − 4) +
16
4
Note that
1
−1
(4.002 − 4) +
16
4
−1
2
1
=
·
+
16 1000 4
−1
2000
=
+
8000 8000
1999
=
8000
L(4.002) =
Therefore, we have:
1999
1
= f (4.002) ≈ L(4.002) =
4.002
8000
√
4. Use linear approximation to estimate 3 1001.
√
Solution. Let f (x) = 3 x. Again, 1001 is near 1000, and f (1000) = 10
is easily calculated. Therefore, we proceed as above. We have:
L(x) = f 0 (1000)(x − 1000) + f (1000)
Again, f (1000) = 10. Further f 0 (x) = 31 x−2/3 =
2. Therefore, we have that f 0 (1000) =
L(x) =
1
√
2
3( 3 1000)
1
√ 2 , as in problem
3( 3 x)
1
= 300
. This gives:
1
(x − 1000) + 10
300
so that
L(1001) =
1
3001
+ 10 =
300
300
Therefore, we have:
√
3001
3
1001 = f (1001) ≈ L(1001) =
300
3
√
5. Use linear approximation to estimate 99.8.
√
Solution. Let
√ f (x) = x. Note that 100 is close to 99.8, and that
f (100) = 100 = 10 is easy to calculate. Therefore, we proceed as
above. We have:
L(x) = f 0 (100)(x − 100) + f (100)
We have that f (100) = 10. Also, f 0 (x) = 2√1 x , so that f 0 (100) =
1
. Therefore, we have:
20
1
L(x) = (x − 100) + 10
20
so that
1
L(99.8) = (−0.2) + 10
20
1 −1
=
·
+ 10
20 10
−1
2000
=
+
200
200
1999
=
200
Therefore, we have:
√
1999
99.8 = f (99.8) ≈ L(99.8) =
200
1
2·10
=