he daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function T(x)=94−10cos[π12(x−2)], where x is hours

Question

he daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function T(x)=94−10cos[π12(x−2)], where x is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

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Skylar 2 weeks 2021-11-20T11:08:57+00:00 1 Answer 0 views 0

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    2021-11-20T11:10:27+00:00

    Answer:

    The rate at which the temperature is changing at 4 p.m. is T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}

    Step-by-step explanation:

    The derivative,  f'(a) is the instantaneous rate of change of y = f(x) with respect to x when x = a.

    We know that the daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function

    T(x)=94-10\cos[\frac{\pi }{12}(x-2 )],

    where x is hours after midnight.

    To find the rate at which the temperature is changing at 4 p.m. First, we must find the derivative of this function.

    T'(x)=\frac{d}{dx}\left(94-10\cos \left[\frac{\pi \:\:}{12}\left(x-2\:\right)\right]\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\T'(x)=\frac{d}{dx}\left(94\right)-\frac{d}{dx}\left(10\cos \left(\frac{\pi }{12}\left(x-2\right)\right)\right)

    \mathrm{Derivative\:of\:a\:constant}:\quad \frac{d}{dx}\left(a\right)=0\\\\\frac{d}{dx}\left(94\right)=0

    \mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=\cos \left(u\right),\:\:u=\frac{\pi }{12}\left(x-2\right)\\\\

    10\frac{d}{du}\left(\cos \left(u\right)\right)\frac{d}{dx}\left(\frac{\pi }{12}\left(x-2\right)\right)\\\\10\left(-\sin \left(u\right)\right)\frac{\pi }{12}\\\\\mathrm{Substitute\:back}\:u=\frac{\pi }{12}\left(x-2\right)\\\\10\left(-\sin \left(\frac{\pi }{12}\left(x-2\right)\right)\right)\frac{\pi }{12}\\\\\frac{d}{dx}\left(10\cos \left(\frac{\pi }{12}\left(x-2\right)\right)\right)=-\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(x-2\right)\right)

    So,

    T'(x)=\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(x-2\right)\right)

    Then, at 4 p.m means x=16 because x is hours after midnight.

    Thus,

    T'(16)=\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(16-2\right)\right)\\\\T'(16)=-\frac{5\pi }{12}\approx-1.309

    The rate at which the temperature is changing at 4 p.m. is T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}

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