Calculus

Applications of the Derivative

Applications of Derivative Logo

Linear Approximation

Solved Problems

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Example 11

Derive the approximate formula \[\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}},\] where a > 0. Using this identity calculate approximately the value of \(\sqrt {150} .\)

Example 12

Derive the approximate formula \[\sqrt[n]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n - 1}}}},\] where a > 0. Using this formula, calculate \(\sqrt[8]{{250}}.\)

Example 13

Find an approximate value for \(\cos 46^\circ.\)

Example 14

Find the linearization of the function \[f\left( x \right) = {x^2} + 2\cos x\] at \(a = 0.\)

Example 15

Find an approximate value for \(\sin 179^\circ.\)

Example 16

Find the linearization of the natural logarithm \[f\left( x \right) = \ln x\] at \(x = 1.\)

Example 17

Find an approximate value of \(\ln 20.\)

Example 18

Calculate \({e^{0,1}}.\)

Example 19

Find an approximate value for \(\arccos 0,51.\)

Example 20

Find an approximate value for \(\arctan 0,95.\)

Example 21

Given the function \[f\left( x \right) = \frac{3}{{{x^2}}}.\] Use linear approximation at \(a = 2\) to estimate the value of \(f\left( {1.99} \right).\)

Example 22

Find the approximate value of the function \[f\left( x \right) = \sqrt {{x^2} + 3x} \] at \(x = 1,02.\)

Example 23

Find the approximate value of the function \[f\left( x \right) = \sqrt {5x - 1} \] at \(x = 1,99.\)

Example 24

Find the linear approximation to \[f\left( x \right) = \frac{{{x^2} + 1}}{{x - 2}}\] near \(a = 3.\)

Example 25

The function \[f\left( x \right) = - {x^2} + 100x\] is linearized at a point \(P\left( {a,f\left( a \right)} \right).\) The linear approximation line \(L\left( x \right)\) passes through \(P\) and intersects the \(x-\)axis at \(x = 180.\)

  1. Find the coordinates \(\left( {a,f\left( a \right)} \right)\) of the point \(P.\)
  2. Write the linearization equation \(L\left( x \right).\)

Example 11.

Derive the approximate formula \[\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}},\] where \({a \gt 0}.\) Using this identity calculate approximately the value of \(\sqrt {150} .\)

Solution.

Consider the function \(y = \sqrt x .\) If the independent variables changes by \(\Delta x\), the increment of the function is written as

\[\Delta y = \sqrt {x + \Delta x} - \sqrt x .\]

This increment for small \(\Delta x\) can be approximated by the differential so that

\[\Delta y = \sqrt {x + \Delta x} - \sqrt x \approx dy = f'\left( x \right)\Delta x = {\left( {\sqrt x } \right)^\prime }\Delta x = \frac{1}{{2\sqrt x }}\Delta x.\]

Thus,

\[\sqrt {x + \Delta x} \approx \sqrt x + \frac{1}{{2\sqrt x }}\Delta x.\]

We denote \(x = {a^2}\), \(\Delta x = h.\) Then we obtain the following approximate equation:

\[\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}}.\]

Using this formula we can estimate the value of \(\sqrt {150}:\)

\[\sqrt {150} = \sqrt {144 + 6} = \sqrt {{{12}^2} + 6} \approx 12 + \frac{1}{4} = 12,25.\]

The exact value (up to \(3\) digits after the decimal point) is equal to \(12,247\). As it can be seen, the relative error when using the approximate formula is

\[\frac{{12,250 - 12,247}}{{12,247}} = \frac{{0,003}}{{12,247}} = 2,4 \cdot {10^{ - 4}} \lt 0,03\% .\]

Example 12.

Derive the approximate formula \[\sqrt[n]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n - 1}}}},\] where \({a \gt 0}.\) Using this formula, calculate \(\sqrt[8]{{250}}.\)

Solution.

Let \(y = \sqrt[n]{x}.\) If the variable \(x\) changes by \(\Delta x,\) the increment of the function has the form:

\[\Delta y = \sqrt[n]{{x + \Delta x}} - \sqrt[n]{x}.\]

Considering \(\Delta x\) as a small quantity, we can replace the increment of the function \(\Delta y\) by its differential:

\[\Delta y = \sqrt[n]{{x + \Delta x}} - \sqrt[n]{x} \approx dy = f'\left( x \right)\Delta x = {\left( {\sqrt[n]{x}} \right)^\prime }\Delta x = {\left( {{x^{\frac{1}{n}}}} \right)^\prime }\Delta x = \frac{1}{n}{x^{\frac{1}{n} - 1}}\Delta x = \frac{1}{n}{x^{\frac{{1 - n}}{n}}}\Delta x.\]

Then

\[\sqrt[n]{{x + \Delta x}} \approx \sqrt[n]{x} + \frac{1}{n}{x^{\frac{{1 - n}}{n}}}\Delta x.\]

Denoting \(x = {a^n}\), \(\Delta x = h,\) we obtain the following relationship:

\[\sqrt[n]{{{a^n} + h}} \approx a + \frac{1}{n}{\left( {{a^n}} \right)^{\frac{{1 - n}}{n}}}h = a + \frac{h}{{n{a^{n - 1}}}}.\]

Using this formula, we find:

\[\sqrt[8]{{250}} = \sqrt[8]{{256 - 6}} = \sqrt[8]{{{2^8} - 6}} \approx 2 + \frac{{\left( { - 6} \right)}}{{8 \cdot {2^7}}} = 2 - \frac{6}{{1024}} \approx 1,994.\]

Example 13.

Find an approximate value for \(\cos 46^\circ.\)

Solution.

We choose \(a = 45^\circ.\) The derivative of cosine at this point is equal to

\[f\left( x \right) = \cos x,\;\; \Rightarrow f'\left( x \right) = - \sin x,\;\; \Rightarrow f'\left( {a = 45^\circ} \right) = - \sin 45^\circ = - \frac{{\sqrt 2 }}{2}.\]

Express the increment of the independent variable \(\Delta x\) in radians:

\[\Delta x = 46^\circ - 45^\circ = 1^\circ = \frac{{2\pi }}{{360}} = \frac{\pi }{{180}}\;\text{radians}.\]

Using the approximate formula for small \(\Delta x\)

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\Delta x,\]

we get:

\[\cos 46^\circ \approx \cos 45^\circ + \left( { - \frac{{\sqrt 2 }}{2}} \right) \cdot \frac{\pi }{{180}} = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} \cdot \frac{\pi }{{180}} = \frac{{\sqrt 2 }}{2}\left( {1 - \frac{\pi }{{180}}} \right) \approx 0,7071 \cdot \left( {1 - 0,0175} \right) = 0,6948.\]

Example 14.

Find the linearization of the function \[f\left( x \right) = {x^2} + 2\cos x\] at \(a = 0.\)

Solution.

Use the linear approximation

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).\]

Take the derivative:

\[f^\prime\left( x \right) = \left( {{x^2} + \cos x} \right)^\prime = 2x - 2\sin x.\]

Then

\[f\left( a \right) = f\left( 0 \right) = {0^2} + 2\cos 0 = 2,\]
\[f^\prime\left( a \right) = f^\prime\left( 0 \right) = 2 \cdot 0 - 2\sin 0 = 0.\]

Substitute this into \(L\left( x \right):\)

\[L\left( x \right) = 2 + 0 \cdot \left( {x - 0} \right) = 2.\]

Thus, the linear approximation of the given function is the horizontal line

\[L\left( x \right) = 2.\]

Example 15.

Find an approximate value for \(\sin 179^\circ.\)

Solution.

Let \(x = 179^\circ\), \(a = 180^\circ.\) Hence, \(\Delta x = x - a = 179^\circ - 180^\circ = - 1\) \(= - {\frac{\pi }{{180}}}\) radians. Calculate the value of the function and its derivative at the point \(a:\)

\[f\left( {a} \right) = \sin 180^\circ = 0,\;\;f'\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x,\;\;f'\left( {a} \right) = \cos 180^\circ = - 1.\]

Replacing the increment of the function by its differential, we obtain:

\[f\left( x \right) \approx f\left( {a} \right) + dy = f\left( {a} \right) + f'\left( {a} \right)\Delta x,\;\; \Rightarrow \sin 179^\circ = 0 - 1 \cdot \left( { - \frac{\pi }{{180}}} \right) = \frac{\pi }{{180}} \approx 0,0175.\]

Example 16.

Find the linearization of the natural logarithm \[f\left( x \right) = \ln x\] at \(x = 1.\)

Solution.

We need to calculate \(f\left( a \right)\) and \(f^\prime\left( a \right)\) where \(a = 1.\)

\[f\left( a \right) = f\left( 1 \right) = \ln 1 = {0},\]
\[f^\prime\left( x \right) = \left( {\ln x} \right)^\prime = \frac{1}{x},\]
\[f^\prime\left( a \right) = f^\prime\left( 1 \right) = \frac{1}{1} = {1}.\]

Write the linear approximation function \(L\left( x \right):\)

\[L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right) = 0 + 1\left( {x - 1} \right) = x - 1.\]

So the linearization of the natural logarithm at \(x = 1\) is

\[{\ln x \approx x - 1.}\]
Linear approximation of the natural logarithm
Figure 2.

Example 17.

Find an approximate value of \(\ln 20.\)

Solution.

Consider the natural logarithmic function \(y = \ln x.\) Given that

\[\ln {e^3} = 3\ln e = 3,\]

it is convenient to take the point \({a}\) such that

\[a = {e^3} \approx 20,086.\]

Calculate the derivative and its value at the point \({a}:\)

\[f'\left( x \right) = \left( {\ln x} \right)^\prime = \frac{1}{x},\;\; \Rightarrow f'\left( {a} \right) \approx \frac{1}{{20,086}} \approx 0,0498.\]

Hence, the approximate value of \(\ln 20\) is equal

\[\ln 20 \approx \ln {e^3} + 0,0498 \cdot \left( {20 - 20,086} \right) = 3 - 0,0043 = 2,9957.\]

Example 18.

Calculate \({e^{0,1}}.\)

Solution.

Let \(f\left( x \right) = {e^x}.\) By setting \(a = 0,\) we have:

\[f\left( {a} \right) = {e^0} = 1,\;\;f'\left( x \right) = \left( {{e^x}} \right)^\prime = {e^x},\;\;f'\left( {a = 0} \right) = {e^0} = 1.\]

To calculate \({e^{0,1}}\) we use the approximate formula

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right).\]

Then

\[{e^{0,1}} \approx 1 + 1 \cdot \left( {0,1 - 0} \right) = 1,1.\]

Example 19.

Find an approximate value for \(\arccos 0,51.\)

Solution.

Suppose that \(f\left( x \right) = \arccos x\) and \(a = 0,5.\) Replacing the increment of the function \(\Delta y\) by its differential, we can compute the approximate value of \(\arccos 0,51:\)

\[f\left( x \right) \approx f\left( {a} \right) + dy = f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\]
\[f'\left( x \right) = {\left( {\arccos x} \right)^\prime } = - \frac{1}{{\sqrt {1 - {x^2}} }},\;\; \Rightarrow f'\left( {a = 0,5} \right) = - \frac{1}{{\sqrt {1 - 0,{5^2}} }} = - \frac{1}{{\sqrt {0,75} }} \approx - 1,1547,\;\; \Rightarrow \arccos 0,51 \approx \arccos 0,5 + \left( { - 1,1547} \right) \cdot \left( {0,51 - 0,5} \right) = \frac{\pi }{3} - 0,0115 \approx 1,035\;\text{radians} \approx 59,34^\circ.\]

Example 20.

Find an approximate value for \(\arctan 0,95.\)

Solution.

Let \(f\left( x \right) = \arctan x\), \(a = 1.\) Determine the value of the derivative at \(a:\)

\[f'\left( x \right) = \left( {\arctan x} \right)^\prime = \frac{1}{{1 + {x^2}}},\;\; \Rightarrow f'\left( {a = 1} \right) = \frac{1}{{1 + {1^2}}} = \frac{1}{2}.\]

For the approximate calculation we use the formula

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right).\]

Consequently,

\[\arctan 0,95 \approx \arctan 1 + \frac{1}{2} \cdot \left( {0,95 - 1} \right) = \frac{\pi }{4} + \frac{1}{2} \cdot \left( { - 0,05} \right) \approx 0,7604\;\text{radians} \approx 43,57^\circ.\]

Example 21.

Given the function \[f\left( x \right) = \frac{3}{{{x^2}}}.\] Use linear approximation at \(a = 2\) to estimate the value of \(f\left( {1.99} \right).\)

Solution.

We apply the linear approximation formula

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right),\]

where \(a= 2.\)

Take the derivative:

\[f^\prime\left( x \right) = \left( {\frac{3}{{{x^2}}}} \right)^\prime = \left( {3{x^{ - 2}}} \right)^\prime = - 6{x^{ - 3}} = - \frac{6}{{{x^3}}}.\]

The function and the derivative have the following values at \(a = 2:\)

\[f\left( a \right) = f\left( 2 \right) = \frac{3}{{{2^2}}} = \frac{3}{4},\]
\[f^\prime\left( a \right) = f^\prime\left( 2 \right) = - \frac{6}{{{2^3}}} = - \frac{3}{4}.\]

Plugging \(a,\) \(f\left( a \right),\) and \(f^\prime\left( a \right),\) we obtain:

\[L\left( x \right) = \frac{3}{4} + \left( { - \frac{3}{4}} \right)\left( {x - 2} \right) = - \frac{3}{4}x + \frac{9}{4} = \frac{3}{4}\left( {3 - x} \right).\]

Then the approximate value of \(f\left( {1.99} \right)\) is equal to

\[f\left( {1.99} \right) \approx L\left( {1.99} \right) = \frac{3}{4}\left( {3 - 1.99} \right) = 0.7575\]

Example 22.

Find the approximate value of the function \[f\left( x \right) = \sqrt {{x^2} + 3x} \] at \(x = 1,02.\)

Solution.

Choose the point \(a = 1.\) Then

\[f\left( {a} \right) = \sqrt {{1^2} + 3 \cdot 1} = 2.\]

Find the value of the derivative of the given function at \(a:\)

\[f'\left( x \right) = \left( {\sqrt {{x^2} + 3x} } \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 3x}} \cdot \left( {{x^2} + 3x} \right)^\prime } = \frac{{2x + 3}}{{2\sqrt {{x^2} + 3x} }},\;\; \Rightarrow f'\left( {a = 1} \right) = \frac{{2 \cdot 1 + 3}}{{2\sqrt {{1^2} + 3 \cdot 1} }} = \frac{5}{4} = 1,25.\]

Hence, the approximate value of the function at \(x = 1,02\) is equal to

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow \sqrt {1,02} \approx 2 + 1,25 \cdot \left( {1,02 - 1} \right) = 2,025.\]

Example 23.

Find the approximate value of the function \[f\left( x \right) = \sqrt {5x - 1} \] at \(x = 1,99.\)

Solution.

Let \(a = 2\). Hence,

\[f\left( {a = 2} \right) = \sqrt {5 \cdot 2 - 1} = \sqrt 9 = 3.\]

The derivative at \(a = 2\) has the following value:

\[f'\left( x \right) = {\left( {\sqrt {5x - 1} } \right)^\prime } = \frac{1}{{2\sqrt {5x - 1} }} \cdot {\left( {5x - 1} \right)^\prime } = \frac{5}{{2\sqrt {5x - 1} }},\;\; \Rightarrow f'\left( {a = 2} \right) = \frac{5}{{2\sqrt {5 \cdot 2 - 1} }} = \frac{5}{6} \approx 0,833.\]

Estimating the approximate value of the function at \(x = 1,99\), we have

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\; \Rightarrow {\sqrt {1,99}} \approx {3 + 0,833 \cdot \left( {1,99 - 2} \right) } = 3 - 0,0083 \approx 2,992.\]

Example 24.

Find the linear approximation to \[f\left( x \right) = \frac{{{x^2} + 1}}{{x - 2}}\] near \(a = 3.\)

Solution.

Let's linearize this function at \(a = 3.\) First, take the derivative using the quotient rule and the chain rule.

\[f^\prime\left( x \right) = \left( {\frac{{{x^2} + 1}}{{x - 2}}} \right)^\prime = \frac{{2x\left( {x - 2} \right) - \left( {{x^2} + 1} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2{x^2} - 4x - {x^2} - 1}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x - 1}}{{{{\left( {x - 2} \right)}^2}}}.\]

Calculate \(f\left( a \right)\) and \(f^\prime\left( a \right):\)

\[f\left( 3 \right) = \frac{{{3^2} + 1}}{{3 - 2}} = 10,\]
\[f^\prime\left( 3 \right) = \frac{{{3^2} - 4 \cdot 3 - 1}}{{{{\left( {3 - 2} \right)}^2}}} = - 4.\]

Now we can write the linear approximation equation:

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right) = 10 - 4\left( {x - 3} \right) = - 4x + 22.\]

Thus, the linearization of the given function at \(a = 3\) is

\[y = - 4x + 22.\]

Example 25.

The function \[f\left( x \right) = - {x^2} + 100x\] is linearized at a point \(P\left( {a,f\left( a \right)} \right).\) The linear approximation line \(L\left( x \right)\) passes through \(P\) and intersects the \(x-\)axis at \(x = 180.\)

  1. Find the coordinates \(\left( {a,f\left( a \right)} \right)\) of the point \(P.\)
  2. Write the linearization equation \(L\left( x \right).\)

Solution.

Linear approximation of a quadratic function
Figure 3.

\(1.\) The linear approximation \(L\left( x \right)\) is given by the equation

\[L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).\]

Take the derivative:

\[f^\prime\left( x \right) = \left( { - {x^2} + 100x} \right)^\prime = - 2x + 100.\]

At the point \(P,\) the equation for \(L\left( x \right)\) becomes

\[L\left( x \right) = - {a^2} + 100a + \left( { - 2a + 100} \right)\left( {x - a} \right) = - {a^2} + \cancel{100a} - 2ax + 100x + 2{a^2} - \cancel{100a} = {a^2} + \left( {100 - 2a} \right)x.\]

We find the value of \(a\) from the condition \(L\left( x \right) = 0\) at \(x = 180.\) This yields:

\[{a^2} + \left( {100 - 2a} \right) \cdot 180 = 0,\;\; \Rightarrow {a^2} - 360a + 18000 = 0.\]

Solve the quadratic equation:

\[D = {360^2} - 4 \cdot 18000 = 57600.\]
\[a = \frac{{360 \pm \sqrt {57600} }}{2} = \frac{{360 \pm 240}}{2} = 60,300.\]

We see that only one root \(a = 60\) belongs to the interval \(\left( {0,100} \right),\) so the point \(P\) has the coordinates:

\[a = 60,\; f\left( a \right) = - {60^2} + 100 \cdot 60 = 2400.\]

\(2.\) The equation of the linear approximation \(L\left( x \right)\) derived in Section \(A\) has the form

\[L\left( x \right) = {a^2} + \left( {100 - 2a} \right)x.\]

Substituting \(a = 60,\) we get

\[L\left( x \right) = {60^2} + \left( {100 - 2 \cdot 60} \right)x = 3600 - 20x.\]
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