Superprof

Exercise 1

f(x) = \frac{x + 2} {x ^2}

 

Exercise 2

f(x) = 7 ^ {3x ^2 - 1}

 

Exercise 3

f(x) = ln sin \sqrt{x}

 

Exercise 4

f(x) = log_x \sqrt{x}

 

Exercise 5

A square has a side of 2 m. Determine the increase in area of the square when its sides increase by 1 milimeter. Calculate the error made when using differentials rather than increases.

Exercise 6

Find the change in the volume of a cube of with edges of 20 centimeters, when each edge increases by 0.2 cm in length..

Exercise 7

Calculate the absolute and relative error made when calculating the volume of a sphere of 12.51 milimeters in diameter, measured with an instrument that shows thousandths of a centimeter.

Exercise 8

If  \sqrt{0.80} is replaced by \sqrt{0.81} = 0.9 , what are the approximations of absolute and relative error?

 

Solution of exercise 1

Calculate the differential:

f(x) = \frac {x + 2} {x^ 2}

df(x) = \frac{x ^2 - (x + 2) \cdot 2x} { x ^4} dx = \frac{x ^2 - 2x ^2 - 4x} {x ^4} dx

= -\frac{x + 4} {x ^3} dx

 

Solution of exercise 2

Calculate the differential:

f(x) = 7 ^ {3x ^2 - 1}

df(x) = 6x \cdot 7 ^ {3x ^2 - 1} \cdot ln 7dx

 

Solution of exercise 3

Calculate the differential:

f(x) = ln sin \sqrt {x}

df (x) = \frac{1} {sin\sqrt{x}} \cdot cos \sqrt{x} \frac {1} {2 \sqrt{x}} dx

= \frac {cot \sqrt{x}} { 2 \sqrt{x}} dx

 

Solution of exercise 4

Calculate the differential:

f(x) = log_x \sqrt{x}

y = log_x \sqrt{x}

x ^ y = \sqrt {x}

ln x ^y = ln \sqrt{x}

y lnx = \frac{1}{2} lnx

f(x) = \frac{1}{2}

df(x) = 0

 

Solution of exercise 5

A square has a side of 2 m. Determine the increase in area of the square when its sides increase by 1 milimeter. Calculate the error made when using differentials rather than increases.

S = x^2

\triangle S = (x + h) ^2 - x ^2 = 2.001 ^2 - 4 = 0.004001 m ^2

dS = 2x \cdot dx = 4.0.001 - 0.004 m ^2

Error = \triangle S - dS = 10 ^ {-6} m ^2

 

Solution of exercise 6

Find the change in the volume of a cube of with edges of 20 centimeters, when each edge increases by 0.2 cm in length.

V = x ^3

dV = 3x ^2 dx

dV = 3 \cdot 20 ^2 \cdot  0.2 = 240 cm ^2

 

Solution of exercise 7

Calculate the absolute and relative error made when calculating the volume of a sphere of 12.51 milimeters in diameter, measured with an instrument that shows thousandths of centimeter.

V = \frac{4 } {3} \pi r ^3

dV = 4 \pi r ^2 dr

\frac{dV} {V} = \frac{4 \pi r ^2 dr} {\frac{4}{3} \pi r ^3} = \frac {3dr}{r}

dV = 4 \pi \cdot 6.255 ^2 \cdot 0.01 = 4.917 mm ^3

dr = 0.01 mm

\frac {dV} {V} = \frac { 3 \cdot 0.01} {6.255} = 0.0048

 

Solution of exercise 8

If  \sqrt{0.80} is replaced by \sqrt{0.81} = 0.9  , what are the approximations of absolute and relative error?

f(x) = \sqrt{x}

df (x) = \frac{1} {2 \sqrt{x}} dx

\frac {df(x)} {f(x)} = \frac {\frac {1} {2 \sqrt{x}} dx} { \sqrt{x}} = \frac {dx} {2x}

dx = 0.81 - 0.80 = 0.01

df(x) = \frac{1} {2 \sqrt {0.81}} \cdot 0.01 = \frac {1} {180}

\frac{df(x)}{f(x)} = \frac{0.01}{2 \cdot 0.81} = \frac{1}{162}

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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