Question

Math

Posted 2 months ago

```
Vlad is playing on a swing set.
His horizontal distance $D(t)$ (in $\mathrm{m}$ ) from the center (where being behind the center means a negative distance) as a function of time $t$ (in seconds) can be modeled by a sinusoidal expression of the form $a \cdot \cos (b \cdot t)+d$.
At $t=0$, when he pushes off, he is $1 \mathrm{~m}$ behind the center, which is as far back as he goes. The swing reaches the center $\frac{\pi}{6}$ seconds later.
Find $D(t)$.
$t$ should be in radians.
$D(t)=$
$\square$
```

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Answer from Sia

Posted 2 months ago

Solution by Steps

step 1

Given the initial condition at $t=0$, Vlad is 1 meter behind the center, we can determine the value of $d$ in the equation $D(t) = a \cdot \cos(b \cdot t) + d$. Since this is the farthest point back he goes, $d = -1$

step 2

The swing reaches the center $\frac{\pi}{6}$ seconds later, which means $D\left(\frac{\pi}{6}\right) = 0$. We can use this to find the value of $b$. Since $d = -1$, we have $0 = a \cdot \cos\left(b \cdot \frac{\pi}{6}\right) - 1$

step 3

Solving for $a$ using the fact that the cosine of an angle cannot be greater than 1, we find that $a = 1$ because at $t = 0$, $D(0) = a \cdot \cos(0) - 1 = 1 - 1 = 0$

step 4

Now we solve for $b$ using the equation $0 = 1 \cdot \cos\left(b \cdot \frac{\pi}{6}\right) - 1$. This simplifies to $\cos\left(b \cdot \frac{\pi}{6}\right) = 1$. The smallest positive value for $b$ that satisfies this equation is $b = 6$, because $\cos(0) = 1$ and $b \cdot \frac{\pi}{6} = \pi$ when $b = 6$

step 5

The specific function that models the horizontal distance as a function of time is $D(t) = \cos(6t) - 1$

Answer

$D(t) = \cos(6t) - 1$

Key Concept

Modeling with sinusoidal functions

Explanation

The problem involves determining the parameters of a sinusoidal function given initial conditions. By understanding the behavior of cosine functions and applying the given conditions, we can find the specific function that models the scenario.

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