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Posted 2 months ago

Vlad is playing on a swing set.
His horizontal distance D(t)D(t) (in m\mathrm{m} ) from the center (where being behind the center means a negative distance) as a function of time tt (in seconds) can be modeled by a sinusoidal expression of the form acos(bt)+da \cdot \cos (b \cdot t)+d.

At t=0t=0, when he pushes off, he is 1 m1 \mathrm{~m} behind the center, which is as far back as he goes. The swing reaches the center π6\frac{\pi}{6} seconds later.

Find D(t)D(t).
tt should be in radians.

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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
Given the initial condition at t=0t=0, Vlad is 1 meter behind the center, we can determine the value of dd in the equation D(t)=acos(bt)+dD(t) = a \cdot \cos(b \cdot t) + d. Since this is the farthest point back he goes, d=1d = -1
step 2
The swing reaches the center π6\frac{\pi}{6} seconds later, which means D(π6)=0D\left(\frac{\pi}{6}\right) = 0. We can use this to find the value of bb. Since d=1d = -1, we have 0=acos(bπ6)10 = a \cdot \cos\left(b \cdot \frac{\pi}{6}\right) - 1
step 3
Solving for aa using the fact that the cosine of an angle cannot be greater than 1, we find that a=1a = 1 because at t=0t = 0, D(0)=acos(0)1=11=0D(0) = a \cdot \cos(0) - 1 = 1 - 1 = 0
step 4
Now we solve for bb using the equation 0=1cos(bπ6)10 = 1 \cdot \cos\left(b \cdot \frac{\pi}{6}\right) - 1. This simplifies to cos(bπ6)=1\cos\left(b \cdot \frac{\pi}{6}\right) = 1. The smallest positive value for bb that satisfies this equation is b=6b = 6, because cos(0)=1\cos(0) = 1 and bπ6=πb \cdot \frac{\pi}{6} = \pi when b=6b = 6
step 5
The specific function that models the horizontal distance as a function of time is D(t)=cos(6t)1D(t) = \cos(6t) - 1
D(t)=cos(6t)1D(t) = \cos(6t) - 1
Key Concept
Modeling with sinusoidal functions
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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