Mathematics, 19.11.2020 17:40, rubyfun58

Water is poured into a conical paper cup at the rate of 3 cubic centimeters (cm) per second (s). The paper cup measures 8 cm tall and the top has a
radius of 4 cm.
a.) Use similar triangles to express the radius r in terms of the height h.
Be sure to solve your equation for r.
ANSWER: r =
<
b.) Determine how fast the water level is rising at the instant the water is 4
cm deep.(The volume of a cone is V = r2h.) Be sure to round your
answer to 3 Decimal Places.
ANSWER:
dh
dt
m/s

Answers: 1

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Answers

The radius in terms of height is $\mathbf{r =\frac 12h}$ , while the water level increases at 0.239 cm per second when the height is 4 cm

(a) Radius in terms of height

The given parameters are:

$\mathbf{h = 8cm}$ --- height

$\mathbf{r = 4cm}$ --- radius

The above parameters can be written as:

$\mathbf{r :h =4cm : 8cm}$

$\mathbf{r :h =4 : 8}$

Express as fractions

$\mathbf{\frac rh =\frac 48}$

$\mathbf{\frac rh =\frac 12}$

Multiply both sides by h

$\mathbf{r =\frac 12h}$

(b) How fast the water is rising

The volume of a cone is:

$\mathbf{V = \frac 13 \pi r^2h}$

Differentiate with respect to time

$\mathbf{V' = \frac 13 \pi(2rh\ \frac{dr}{dt} + r^2\frac{dh}{dt})}$

When the water is 4cm deep, then the radius is 2cm using $\mathbf{r =\frac 12h}$

So, we have:

$\mathbf{V' = \frac 13 \pi(2 \times 2 \times 4\times \frac{dr}{dt} + 2^2\frac{dh}{dt})}$

$\mathbf{V' = \frac 13 \pi(16 \frac{dr}{dt} + 4\frac{dh}{dt})}$

Also, we have:

$\mathbf{r =\frac 12h}$

Differentiate with respect to time

$\mathbf{\frac{dr}{dt} = \frac{1}{2}\frac{dh}{dt}}$

So, we have:

$\mathbf{V' = \frac 13 \pi(16 \times \frac{1}{2}\frac{dh}{dt} + 4\frac{dh}{dt})}$

$\mathbf{V' = \frac 13 \pi(8\frac{dh}{dt} + 4\frac{dh}{dt})}$

$\mathbf{V' = \frac 13 \pi(12\frac{dh}{dt})}$

Expand

$\mathbf{V' = 4\pi\frac{dh}{dt}}$

The volume increases at 3 cm^3 per second.

So, we have:

$\mathbf{3 = 4\pi\frac{dh}{dt}}$

Divide both sides by 4

$\mathbf{0.75 = \pi\frac{dh}{dt}}$

Divide both sides by pi

$\mathbf{0.239 = \frac{dh}{dt}}$

Rewrite as:

$\mathbf{\frac{dh}{dt} = 0.239}$

Hence, the water level increases at 0.239 cm per second when the height is 4 cm

Read more about volumes at:

link

answered: heatherlemaster

The water level is rising at the rate 0.238 cm per second

Step-by-step explanation:

Let the radius and height of water at time t be r and h respectively.

Height of cup =H= 8 cm

Radius of cup = R = 4 cm

By symmetry

$\frac{r}{h}=\frac{4}{8} \\r=\frac{1}{2}h ---1$

Volume of water at time t , $V= \frac{1}{3} \pi r^2 h$

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

With 1

$\frac{dr}{dt}=\frac{1}{2} \frac{dh}{dt}$

So,

$\frac{dV}{dt}=\frac{1}{3}\pi (\frac{1}{2}h) (2 (\frac{1}{2} \frac{dh}{dt}h)+ r \frac{dh}{dt})\\\frac{dV}{dt}=\frac{1}{3}\pi (\frac{1}{2}h) ( \frac{dh}{dt}h+ r \frac{dh}{dt})\\\frac{dV}{dt}=\frac{1}{3}\pi (\frac{1}{2}h) ( h+\frac{1}{2}h) \frac{dh}{dt}\\\frac{dV}{dt}=\frac{1}{6}\pi h(\frac{3h}{2}) \frac{dh}{dt}\\\frac{dV}{dt}=\frac{1}{6}\pi (\frac{3h^2}{2}) \frac{dh}{dt}\\\frac{dV}{dt}=\frac{1}{12}\pi 3h^2 \frac{dh}{dt}$

We are given that Water is poured into a conical paper cup at the rate of 3 cubic centimeters (cm) per second (s).

We are supposed to find how fast the water level is rising at the instant the water is 4 cm deep.

So, $\frac{1}{12}\pi 3(4)^2 \frac{dh}{dt}=3\\\frac{dh}{dt}=\frac{3 \times 12}{\pi 3(4)^2 }=0.238$

Hence the water level is rising at the rate 0.238 cm per second

answered: Guest

i need the table so i can see

step-by-step explanation:

answered: Guest

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

$\tan\left(x\right)=\frac{\left[altitude\right]}{\left[base\right]}=\frac{p}{r}$

$\tan\left(x\right)=\frac{p}{r}$

$\frac{p}{r}=tan(x)$

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).

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Water is poured into a conical paper cup at the rate of 3 cubic centimeters (cm) per second (s). Th...