PRACTICE PROBLEMS WITH LINEAR APPROXIMATION

Problem 1 :

Let f (x)  =  x . Find the linear approximation at x = 27. Use the linear approximation to approximate ∛27.2

Solution

Problem 2 :

Use the linear approximation to find approximate values of

(i)  (123)2/3

(ii)  (15)1/4

(iii)  ∛26              Solution

Problem 3 :

Consider the implicit function defined by 3(x2 + y2)2 = 100xy . Use a tangent line approximation at the point (3,1) to estimate the value of y when x = 3.1.

Solution

Problem 4 :

Finding a local linear approximation at a given point is finding the equation of the tangent line at that point.

a) Find the local linear approximation of f(x) = x3 - 2x + 3 at the point where x = 2.

b) Use your approximation to estimate f(2.1), f(1.9) and f(1.99).

Solution

Answer Key

1)  L(27.2)  =  3.0074 (approximately)

2)  i)  the approximate value of (123)2/3 is 24.733.

ii)   the approximate value of (15)1/4 is  1.968.

iii)  the approximate value of ∛26 is 3.0370.

3)  1.14 to two decimals.

4)  L(2) = 10x - 13

i)  8     ii)  6    iii)  6.9

Problem 1 :

f(x)  =  x- 5x + 12 and x0  =  2

Solution

Problem 2 :

g(x)  =  √(x+ 9) and x0  =  -4

Solution

Problem 3 :

h(x)  =  x/(x + 1) and x0  =  1

Solution

Problem 4 :

Since there were no problems on linear approximation on the second practice prelim, we are including some separately.

Consider the function f(x) = e2x.

(a) Determine the linearization L(x) of f(x) at the point (0, 1).

(b) Use your result in (a) to approximate e0.2.

Solution

Problem 5 :

Find the linear approximation of f(x) = x sin (πx2) about x = 2. Use the approximation to estimate f(1.99)

Solution

Problem 6 :

Let f be a differentiable function such that f(3) = 2 and f'(3) = 5. If the tangent line at x = 3 is used to find an approximation to a zero of f , that approximation is

(A) 0.4   (B) 0.5    (C) 2.6    (D) 3.  4 (E) 5.5

Solution

Problem 7 :

The approximate value of y = √(4 + sin x) at x = 0.12 obtained from t he tangent to the graph at x  = 0

(A) 2   (B) 2.03    (C) 2.06    (D) 2.12  (E) 2.24

Solution

Answer Key

1)  L(2)  =  7x - 4

2)  L(-4)  =  (1/5) (9-4x)

3)  L(1)  =  (1/4)(1+x)

4)  a) 2x   b) 0.4

5)  L(2)  = 8 x - 16

approximate value of f(1.99) is  0.08

6) x = 2.6, option C

7)  2.03, option B

Problem 1 :

The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. find the following in calculating the area of the circular plate:

(i) Absolute error (ii) Relative error (iii) Percentage error

Solution

Problem 2 :

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9 8 . cm. Find approximations for the following :

(i) change in the volume (ii) change in the surface area

Solution

Problem 3 :

The time T , taken for a complete oscillation of a single pendulum with length l , is given by the equation

T  =  2π√(l/g)

where g is constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of 1.

Solution

Problem 4 :

Show that the percentage error in the nth root of a number is approximately 1/n times the percentage error in the number.

Solution

Problem 5 :

Sham was given three of the grades of her classmates. These grades were the three closest values to Sham’s grade. If Sham’s grade is considered the actual value, what is the mean absolute error of her classmates’ grades?

Sham’s Grade: 96.3

Classmate 1 : 95.6

Classmate 2 : 94.8

Classmate 3 : 95.2

Solution

Problem 6 :

Find the absolute and relative errors. The actual value is 125.68 mm, and the measured value is 119.66 mm.

Solution

Problem 7 :

The book’s length is 12.5cm, but Via measured only 12.4cm. Find the absolute and relative errors.

Solution

Problem 8 :

The thermometer measures up to the nearest 2 degrees. The temperature was measured at 38° C. Find the relative Error.

Solution

Answer Key

1)  i)  Absolute error = 0.0225π cm2

ii)  Relative error  =  0.0059 cm2

iii)  Percentage error =  0.6%

2)  i)  Volume decreases by 80π cm3.

ii)  Surface area decreases by 16π cm2.

3)  Percentage error  = 1%

4)  dy/y  =  (1/n) (1/x) dx

5)  the Mean Absolute Error is 1.1

6) the Absolute Value is 6.02mm with a Relative Error of 4.79%.

7)  the Absolute Value is 0.1cm with a Relative Error of 0.8%.

8)  Therefore, the Relative Error is 2.63%.

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