Calculus I - Linear Approximations

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Section 4.11 : Linear Approximations

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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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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.