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The movement of the crest of a wave is modelled with the equation $h(t)=0.3\cos(3t)+0.4\sin(3t)$. Find the maximum height of the wave and the time at which it occurs.

I tried finding the derivative of this function in order to find the maximum, but no matter how I tried I could not get the right answer.

I know that the maximum is at $(0.309,0.5)$ because I graphed the function and saw that was the highest it went.

Is there anything I am doing Wrong?

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    `I tried finding the derivative of this function in order to find the maximum` That's one way to do it. Someone would be better able to point out what went wrong if you posted your attempt. – 2017-02-05
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    The derivative I found was h'(t)=-0.9sin(3t)+1.2cos(3t), I then simplified that to tan(3t)=4/3, which gave me t=0.309, I subbed into the original function and got h(0.309)=0.407, which is not close to 0.5 – 2017-02-05
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    $h'(t)=0$ gives $\tan(3t) = 1.2/0.9 = 4/3\,$. Then with simple trig $\sin(3t)=\frac{4}{5}$ and $\cos(3t)=\frac{3}{5}\,$. – 2017-02-05
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    Try looking at my answer I explained what you do and why – 2017-02-05

3 Answers 3

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Hint:

First rewrite your function as $A\cos(3x+\varphi)$, where $A>0$, $A^2=0.3^2+0.4^2$, and $-\pi\le\varphi<\pi$.

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Perhaps you did not calculate the derivative properly.

We have: $$h(t) = 0.3\cos(3t) + 0.4\sin(3t)$$

We must apply chain rule to $\cos(3t)$ and $\sin(3t)$.

$\cos(3t)$ derivative is $-\sin(3t)(3)$

$\sin(3t)$ derivative is $\cos(3t)(3)$

So,

$$h'(t) = 0.3(-\sin(3t)(3)) + \cos(3t)(3)$$ $$h'(t) = -0.9\sin(3t) + 1.2\cos(3t)$$

Let's solve for when $h'(t) = 0$

$$0 = -0.9\sin(3t) + 1.2\cos(3t)$$ $$0.9\sin(3t) = 1.2\cos(3t)$$ $$\text {I will now divide both sides by 0.9}$$ $$\sin(3t) = \frac{4}{3}\cos(3t)$$ $$\text {I will now divide both sides by cos(3t)}$$ $$\tan(3t) = \frac{4}{3}$$ $$\text {We know sin divided by cos is tan. Now we can take tan inverse}$$ $$3t = \tan^{-1}(\frac{4}{3})$$ $$t = \frac{\tan^{-1}(\frac{4}{3})}{3} = 0.309098406001$$

(As the answer says).

We can also compute the second derivative, and make sure that this is indeed a maximum

$$f''(x) = -2.7\cos(3t) -3.6\sin(3t)$$ $$f''(0.309) < 0$$

Therefore, it is a maximum.

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This problem is trivialized by using vectors:

Let $x=3t$ and let $\vec{v}=(0.3, 0.4)$ and $\vec{w}=(\cos{x}, \sin{x})$. We get

$$0.3\cos{3t}+0.4\sin{3t} = 0.3\cos{x}+0.4\sin{x}=\vec{v} \cdot \vec{w}=||v||||w||\cos{\theta}$$ where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$. $\cos \theta$ is maximized when $\theta=0$ so we get $$||v||||w||\cos{0}=\sqrt{0.3^3+0.4^2}\sqrt{\sin^2{x}+\cos^2{x}}=\boxed{0.5}$$

Now to find $t$ we substitute to get $$0.5=0.3\cos 3t + 0.4\sin 3t \longrightarrow 5=3\cos3t + 4\sin 3t \longrightarrow 5=3\cos{3t}+4\sqrt{1-\cos 3t}$$

Letting $\cos{3t}=y$ we get $$5=3y+4\sqrt{1-y} \longrightarrow (5-3y)^2=(4\sqrt{1-y})^2$$

which we can easily solve as a quadratic in $y$, then solve for $t$