at 7 p.m. the temperature is 72 degrees F at 10 p.m. it is 52 degrees find the average rate of change in temperature

1 answer:

4 0

We will assume that we have a function relating between the time and the temperature. Any point in this function can be expressed as (x,y) where x is the time and y is the temperature (any point is expressed as (time, temperature)).

We have two points given:
<span>At 7 p.m. the temperature is 72 degrees F: First point is (7,72)
</span><span>At 10 p.m. the temperature is 52 degrees F: Second point is (10,52)

The rate of change is simply the slope of this function, this slope can be calculated as:
rate of change = (y2-y1) / (x2-x2)
we have y2= 52, y1= 72, x2= 10 and x1= 7

Substituting in the equation, we can calculate the rate of change as follows:
rate of change = (52-72) / (10-7) = -6.667 degrees/hours
Note: The negative sign indicated a DECREASE in the temperature.</span>

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$