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Section 4.11 : Linear Approximations

4. Find the linear approximation to \(f\left( t \right) = \cos \left( {2t} \right)\) at \(t = \frac{1}{2}\). Use the linear approximation to approximate the value of \(\cos \left( 2 \right)\) and \(\cos \left( 18 \right)\). Compare the approximated values to the exact values.

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Start Solution

We’ll need the derivative first as well as a couple of function evaluations.

\[f'\left( t \right) = - 2\sin \left( {2t} \right)\hspace{0.5in}f\left( {\frac{1}{2}} \right) = \cos \left( 1 \right)\hspace{0.5in}f'\left( {\frac{1}{2}} \right) = - 2\sin \left( 1 \right)\] Show Step 2

Here is the linear approximation.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{L\left( t \right) = \cos \left( 1 \right) - 2\sin \left( 1 \right)\left( {t - \frac{1}{2}} \right) = 0.5403 - 1.6829\left( {t - \frac{1}{2}} \right)}}\]

Make sure your calculator is set in radians! Remember that we use radians by default in this class.

Show Step 3

Now, if we want to approximate \(\cos \left( 2 \right)\), that is equivalent to evaluating \(f\left( 1 \right) = \cos \left( 2 \right)\), we need to evaluate the linear approximation at \(t = 1\). Likewise, to approximate \(\cos \left( {18} \right)\) we need to evaluate the linear approximation at \(t = 9\).

So, here are the approximations of the values along with the exact values.

\[\begin{align*}L\left( 1 \right) & = - 0.301169 & \hspace{0.5in}f\left( 1 \right) & = - 0.416147 & \hspace{0.5in}{\mbox{% Error : }} & 27.6292\\ L\left( 9 \right) & = - 13.7647 & \hspace{0.5in}f\left( 9 \right) & = 0.660317 & \hspace{0.5in}{\mbox{% Error : }} & 2184.56\end{align*}\]

So, as we might have expected the farther from \(t = \frac{1}{2}\) we got the worse the approximation is. Recall that the approximation will generally be more accurate the closer to the point of the linear approximation.