Linear Approximation
Solved Problems
Click or tap a problem to see the solution.
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 \(x = 180.\)
- Find the coordinates \(\left( {a,f\left( a \right)} \right)\) of the point \(P.\)
- 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
This increment for small \(\Delta x\) can be approximated by the differential so that
Thus,
We denote \(x = {a^2}\), \(\Delta x = h.\) Then we obtain the following approximate equation:
Using this formula we can estimate the value of \(\sqrt {150}:\)
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
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:
Considering \(\Delta x\) as a small quantity, we can replace the increment of the function \(\Delta y\) by its differential:
Then
Denoting \(x = {a^n}\), \(\Delta x = h,\) we obtain the following relationship:
Using this formula, we find:
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
Express the increment of the independent variable \(\Delta x\) in radians:
Using the approximate formula for small \(\Delta x\)
we get:
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
Take the derivative:
Then
Substitute this into \(L\left( x \right):\)
Thus, the linear approximation of the given function is the horizontal line
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 \(a:\)
Replacing the increment of the function by its differential, we obtain:
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 \(a = 1.\)
Write the linear approximation function \(L\left( x \right):\)
So the linearization of the natural logarithm at \(x = 1\) is
Example 17.
Find an approximate value of \(\ln 20.\)
Solution.
Consider the natural logarithmic function \(y = \ln x.\) Given that
it is convenient to take the point \({a}\) such that
Calculate the derivative and its value at the point \({a}:\)
Hence, the approximate value of \(\ln 20\) is equal
Example 18.
Calculate \({e^{0,1}}.\)
Solution.
Let \(f\left( x \right) = {e^x}.\) By setting \(a = 0,\) we have:
To calculate \({e^{0,1}}\) we use the approximate formula
Then
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 \(\arccos 0,51:\)
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 \(a:\)
For the approximate calculation we use the formula
Consequently,
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
where \(a= 2.\)
Take the derivative:
The function and the derivative have the following values at \(a = 2:\)
Plugging \(a,\) \(f\left( a \right),\) and \(f^\prime\left( a \right),\) we obtain:
Then the approximate value of \(f\left( {1.99} \right)\) is equal to
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
Find the value of the derivative of the given function at \(a:\)
Hence, the approximate value of the function at \(x = 1,02\) is equal to
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,
The derivative at \(a = 2\) has the following value:
Estimating the approximate value of the function at \(x = 1,99\), we have
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.
Calculate \(f\left( a \right)\) and \(f^\prime\left( a \right):\)
Now we can write the linear approximation equation:
Thus, the linearization of the given function at \(a = 3\) is
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 \(x = 180.\)
- Find the coordinates \(\left( {a,f\left( a \right)} \right)\) of the point \(P.\)
- Write the linearization equation \(L\left( x \right).\)
Solution.
\(1.\) The linear approximation \(L\left( x \right)\) is given by the equation
Take the derivative:
At the point \(P,\) the equation for \(L\left( x \right)\) becomes
We find the value of \(a\) from the condition \(L\left( x \right) = 0\) at \(x = 180.\) This yields:
Solve the quadratic equation:
We see that only one root \(a = 60\) belongs to the interval \(\left( {0,100} \right),\) so the point \(P\) has the coordinates:
\(2.\) The equation of the linear approximation \(L\left( x \right)\) derived in Section \(A\) has the form
Substituting \(a = 60,\) we get