The daily low temperature in guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric functio
1 answer:
Answer:
Step-by-step explanation:
The temperature peaks around july26 at 78°F, and has its minimum half year later at 49°F. Assuming a year is exactly 365 days, July 26 is 206/365 of a year after January 1.
Find the formula of the trigonometric function that models the daily low temperature T in Guangzhou t years after January 1, 2015. Define the function using radians
The function can be written as ...
where A is half the difference of the high and low temperature values, and B is the average of the high and low temperature values.
The value 206/365 is the horizontal right shift of the peak of the function. That makes the peak occur on July 26, as required.
Filling in these values gives us the function ...
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Check the picture below.
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,
![\bf \begin{cases} h=2\\ k=-3\\ a=6\\ c=2\sqrt{10} \end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2 \\\\\\ 4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2 \\\\\\ \sqrt{4}=b\implies 2=b\\\\ -------------------------------\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ah%3D2%5C%5C%0Ak%3D-3%5C%5C%0Aa%3D6%5C%5C%0Ac%3D2%5Csqrt%7B10%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5By-%20%28-3%29%5D%5E2%7D%7B%206%5E2%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%282%5Csqrt%7B10%7D%29%5E2%3D6%5E2%2Bb%5E2%5Cimplies%202%5E2%28%5Csqrt%7B10%7D%29%5E2%3D36%2Bb%5E2%0A%5C%5C%5C%5C%5C%5C%0A4%2810%29%3D36%2Bb%5E2%5Cimplies%2040%3D36%2Bb%5E2%5Cimplies%204%3Db%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%7B4%7D%3Db%5Cimplies%202%3Db%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%202%5E2%7D%3D1%5Cimplies%20%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%204%7D%3D1)
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,