Respuesta :
Answer:
The rate at which the temperature is changing at 4 p.m. is [tex]T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}[/tex]
Step-by-step explanation:
The derivative, [tex]f'(a)[/tex] is the instantaneous rate of change of [tex]y = f(x)[/tex] with respect to [tex]x[/tex] when [tex]x = a[/tex].
We know that the daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function
[tex]T(x)=94-10\cos[\frac{\pi }{12}(x-2 )][/tex],
where [tex]x[/tex] 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.
[tex]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)[/tex]
[tex]\mathrm{Derivative\:of\:a\:constant}:\quad \frac{d}{dx}\left(a\right)=0\\\\\frac{d}{dx}\left(94\right)=0[/tex]
[tex]\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)\\\\[/tex]
[tex]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)[/tex]
So,
[tex]T'(x)=\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(x-2\right)\right)[/tex]
Then, at 4 p.m means [tex]x=16[/tex] because [tex]x[/tex] is hours after midnight.
Thus,
[tex]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[/tex]
The rate at which the temperature is changing at 4 p.m. is [tex]T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}[/tex]
Using derivatives, it is found that the temperature is changing at a rate of -1.31 degrees Fahrenheit per hour at 4 p.m.
The temperature, in x hours after midnight, is modeled by the following function:
[tex]T(x) = 94 - 10\cos{\left(\frac{\pi}{12}(x - 2)\right)}[/tex]
4 p. m. is 16 hours after midnight, hence, the rate at which the temperature is changing at 4 p.m is [tex]T^{\prime}(16)[/tex].
The derivative of the function is:
[tex]T^{\prime}(x) = \frac{10\pi}{12}\sin{\left(\frac{\pi}{12}(x - 2)\right)}[/tex]
Hence, at 4 p.m., the rate is:
[tex]T^{\prime}(16) = \frac{10\pi}{12}\sin{\left(\frac{\pi}{12}(16 - 2)\right)}[/tex]
[tex]T^{\prime}(16) = \frac{5\pi}{6}\sin{\left(\frac{14\pi}{12}\right)}[/tex]
[tex]T^{\prime}(16) = \frac{5\pi}{6}\sin{\left(\frac{7\pi}{6}\right)}[/tex]
[tex]T^{\prime}(16) = -1.31[/tex]
The temperature is changing at a rate of -1.31 degrees Fahrenheit per hour at 4 p.m.
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