+79 Hello!~
I'm relatively new here on Web2.0calc, but I hope to become a member of this amazing community! My friend told me it was a great site for people who want to learn, genuinely.
Vlad is playing on a swing set.
His horizontal distance \(D(t)\) 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 \(a*cos(b*t)+d\)
At \(t=0\), when he pushes off, he is \(1\ m\), 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.
I am mostly stuck on the part, "At \(t=0\), when he pushes off, he is \(1\ m\), behind the center, which is as far back as he goes. The swing reaches the center \(\frac{pi}{6}\) seconds later."
Can somebody please clarify for me, if, that means, "(0,-1)" or "(-1,0)"?
And the second part of that sentence: "(\(\frac{pi}{6}\),0)" or "(0,\(\frac{pi}{6}\))"
I am so sorry if I'm being a nuisance...
Please don't be mean.
Thank you!~
+33614 Since you are only being given information about the horizontal distance, I suspect the 1m refers to the point (-1, 0).
So you would have \(D(0) = a*cos(b*0)+d \text{ or }D(0)=a+d\)
The pi/6 refers to a time, not a distance, so \(D(\frac{\pi}{6})=a*cos(b*\frac{\pi}{6})+d\) and because this is the centre where D = 0 this becomes \(0=a*cos(b*\frac{\pi}{6})+d\)
Does this answer your questions?
+79 So, if I'm correct, (-1,0) would be the minimum point, and (0,pi/6) would be the midline point?
+79 I solved it!~
:D
Here's the official solution!~
Thank you so much!~
First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph.
Then, we should use the given information to find the amplitude, midline, and period of the function's graph.
Finally, we should find aaa, b, and d in the expression \(a*cos(b*t)+d\) by considering the features we found.
At t=0, the swing is 1m, behind the center. This means the graph of the function passes through (0,-1).
We are given that this is the farthest point behind the center, which corresponds to a minimum point of the graph.
\(\frac{\pi}{6}\) seconds later the distance is 0m. This corresponds to the point \((\frac{\pi}{6},0)\).
We are given that this is the middle of the swing, which corresponds to the midline of the graph.
In conclusion, the graph has a minimum point at (0,-1) and then intersects its midline at \((\frac{\pi}{6},0)\).
The midline intersection is at y=0, so this is the midline.
The minimum point is 1 unit below the midline, so the amplitude is 1.
The minimum point is \((\frac{\pi}{6})\) units to the left of the midline intersection, so the period is \(4*\frac{\pi}{6} \\ \\=\frac{2\pi}{3}\).
The amplitude is -1.
The midline is y=0.
The period is 3.
The answer:
\(D(t)=-cos(3t)\)
