womankrush538

womankrush538

23.08.2019 •

Mathematics

Parallelograms uvwx and hjik are congruent. write congruence statements comparing the corresponding parts. then determine which transformations map parallelogram uvwx onto parallelogram hjik

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Ответ:

mikayla62

02.05.2023

The rate of change of volume is 127.3cm^3/s .

Step-by-step explanation:

The surface area of the sphere of radius is given by

$(1).\: \: A = 4\pi R^2$ .

and the volume of the same is

$(2).\: \: V = \dfrac{4}{3}\pi R^3.$

Now we are told that there rate of change of surface area is 30cm^2/s or

$\dfrac{dA}{dt} = \dfrac{d}{dt} (4\pi R^2) = 30cm^2/s$

which means

$(3).\: \: \dfrac{dR^2}{dt} = \dfrac{30cm^2/s}{4\pi}$

From the product rule we see that

$\dfrac{dR^2}{dt} = R\dfrac{dR}{dt}+R\dfrac{dR}{dt} = 2R\dfrac{dR}{dt}$

which we substitute into equation (3) to get:

$(4). \: \; \dfrac{dR}{dt} = \dfrac{30}{8\pi R}$

Now, we find the rate of change of volume by taking the time derivative of equation (2) as follows:

$(5). \:\: \frac{dV}{dt} = \dfrac{4}{3}\pi\frac{dR^3}{dt}$

From the product rule

$\frac{dR^3}{dt} = R\frac{dR^2}{dt} + R^2\frac{dR}{dt}$

and since

$\dfrac{dR^2}{dt} =2R\dfrac{dR}{dt}$ ,

we have

$\frac{dR^3}{dt} = R(2R\frac{dR}{dt} )+ R^2\frac{dR}{dt}$

$\frac{dR^3}{dt} =3 R^2\frac{dR}{dt}$

putting this into equation (5) we get:

$(6). \:\: \frac{dV}{dt} =4\pi R^2\frac{dR}{dt}$

from equation (4) we substitute the value of $\dfrac{dR}{dt}$ to get:

$\frac{dV}{dt} =4\pi R^2 \dfrac{30}{8\pi R} =15R$

$(7). \:\: \frac{dV}{dt} =15R$

Now, when the area is $288\pi cm^2$ , the radius is

$4\pi R^2= 288\pi$

$R = 6\sqrt{2}$ ,

putting that into equation (7) we get

$(7). \:\: \frac{dV}{dt} =15(6\sqrt{2} )$

$\boxed{ \frac{dV}{dt} =127.3\: cm^3/s}$

Thus, the rate of change of volume is $127.3\: cm^3/s.$

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