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d1i1m1o1n [39]
3 years ago
15

the graph of f(x)=2^x+1 is shown below explain how to find the average rate of change between x=1 and x=3

Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0
First find the values of f(x) for x = 1 and for x = 3.
Reading off the graph they are f(x) = 3 and f(x) = 9 respectively.
You could also have found thess values by substitution of x in f(x) - eg for x = 1 f(x) =  2^1 + 1 =  3.

Average rate of change  =   difference in f(x) / difference in x = (9 - 3) / (3 - 1) = 6/2 = 3 answer

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The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi
storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

V = \frac{1}{3}\pi r^{2}h

Differentiate the equation with respect to time t, such that

\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)

\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)

To differentiate the product,

Let r² = u, so that

\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)

Then, using product rule

\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]

Since u = r^{2}

Then, \frac{du}{dr} = 2r

Using the Chain's rule

\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}

∴ \frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]

Then,

\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]

Now,

From the question

\frac{dr}{dt} = 7 m/min

\frac{dV}{dt} = 236 m^{3}/min

At the instant when r = 99 m

and V = 180 m^{3}

We will determine the value of h, using

V = \frac{1}{3}\pi r^{2}h

180 = \frac{1}{3}\pi (99)^{2}h

180 \times 3 = 9801\pi h

h =\frac{540}{9801\pi }

h =\frac{20}{363\pi }

Now, Putting the parameters into the equation

\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]

236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]

236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]

708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}

708 = 30790.75 \frac{dh}{dt} + 76.36

708 - 76.36 = 30790.75\frac{dh}{dt}

631.64 = 30790.75\frac{dh}{dt}

\frac{dh}{dt}= \frac{631.64}{30790.75}

\frac{dh}{dt} = 0.021 m/min

Hence, the rate of change of the height is 0.021 meters per minute.

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