Q. 86 - Calculus

Question

For Exercises 85 and 86, suppose that Annie is planning a kayak trip around Orcas Island in August. The tides create strong currents in several places on the coast of that island.

After a bad experience on one trip, Annie models the tidal current velocity around Point Lawrence in a more accurate way as,

$C\left(t\right)=0.90\cos (0.51t+2.02)+0.49\cos \left(\frac{0.51t}{2}+2.13\right)$

1. Plot this function $C\left(t\right)$ together with the function $c\left(t\right)$ from the previous exercise on the same axes.
2. The maximum currents given by $c\left(t\right)$ occurred when $t=-4\pm 12.32n$ hours, for n an integer. Demonstrate that $C\left(t\right)$ does not have maxima at these points.

Step-by-Step Solution

Answer

Part(a) The graph is as follows,

Part(b) The function does not have maxima at $t=-4\pm 12.32n$ points.

Part(a) Step 1. Given Information

We are given a function, 

$C\left(t\right)=0.90\cos (0.51t+2.02)+0.49\cos \left(\frac{0.51t}{2}+2.13\right)$

Part(a) Step 2. Graphing the functions

Part(a) The graph is as follows,

Part(b) Step 1. Maxima at points

Assuming $t=-4\pm 12.32n$ be the point in which $C\left(t\right)$ is maximum.

The point is a critical point for given function,

$$\begin{gathered} C^{\text{'}}\left(t\right)=0 \\ \frac{d}{dt}\left[0.90\cos (0.51t+2.02)+0.49\cos \left(\frac{0.51t}{2}+2.13\right)\right]=0 \\ -0.90\sin (0.51t+2.02)\times 0.51-0.49\sin \left(\frac{0.51t}{2}+2.13\right)\times \frac{0.51}{2}=0 \\ 2\times 0.90\sin (0.51t+2.02)+0.49\sin \left(\frac{0.51t}{2}+2.13\right)=0 \\ 3.67\sin (0.51t+2.02)+\sin \left(\frac{0.51t}{2}+2.13\right)=0 \end{gathered}$$

From the equation, $t\ne -4\pm 12.32n$.

Thus, t is not the critical point for $C\left(t\right)$.

Hence, the function $C\left(t\right)$ does not have maxima at these points.