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.

Answer 1

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 $T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}$