### ¿Todavía tienes preguntas de matemáticas?

Pregunte a nuestros tutores expertos
Algebra
Pregunta

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.

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.

${{L\left( t \right) = \cos \left( 1 \right) - 2\sin \left( 1 \right)\left( {t - \frac{1}{2}} \right) = 0.5403 - 1.6829\left( {t - \frac{1}{2}} \right)}}$

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}{ % \quad Error \quad : \quad (@[email protected]) (O_O) } & 27.6292\\ L\left( 9 \right) & = - 13.7647 & \hspace{0.5in}f\left( 9 \right) & = 0.660317 & \hspace{0.5in}{ % \quad Error \quad : \quad (@[email protected]) (O_O) } & 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.

Solución
View full explanation on CameraMath App.