Respuesta :
Answer:
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
[tex]\frac{r}{h}=\frac{4}{8} \\r=\frac{1}{2}h ---1[/tex]
Volume of water at time t , [tex]V= \frac{1}{3} \pi r^2 h[/tex]
[tex]\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})[/tex]
With 1
[tex]\frac{dr}{dt}=\frac{1}{2} \frac{dh}{dt}[/tex]
So,
[tex]\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}[/tex]
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,[tex]\frac{1}{12}\pi 3(4)^2 \frac{dh}{dt}=3\\\frac{dh}{dt}=\frac{3 \times 12}{\pi 3(4)^2 }=0.238[/tex]
Hence the water level is rising at the rate 0.238 cm per second
The radius in terms of height is [tex]\mathbf{r =\frac 12h}[/tex], 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:
[tex]\mathbf{h = 8cm}[/tex] --- height
[tex]\mathbf{r = 4cm}[/tex] --- radius
The above parameters can be written as:
[tex]\mathbf{r :h =4cm : 8cm}[/tex]
[tex]\mathbf{r :h =4 : 8}[/tex]
Express as fractions
[tex]\mathbf{\frac rh =\frac 48}[/tex]
[tex]\mathbf{\frac rh =\frac 12}[/tex]
Multiply both sides by h
[tex]\mathbf{r =\frac 12h}[/tex]
(b) How fast the water is rising
The volume of a cone is:
[tex]\mathbf{V = \frac 13 \pi r^2h}[/tex]
Differentiate with respect to time
[tex]\mathbf{V' = \frac 13 \pi(2rh\ \frac{dr}{dt} + r^2\frac{dh}{dt})}[/tex]
When the water is 4cm deep, then the radius is 2cm using [tex]\mathbf{r =\frac 12h}[/tex]
So, we have:
[tex]\mathbf{V' = \frac 13 \pi(2 \times 2 \times 4\times \frac{dr}{dt} + 2^2\frac{dh}{dt})}[/tex]
[tex]\mathbf{V' = \frac 13 \pi(16 \frac{dr}{dt} + 4\frac{dh}{dt})}[/tex]
Also, we have:
[tex]\mathbf{r =\frac 12h}[/tex]
Differentiate with respect to time
[tex]\mathbf{\frac{dr}{dt} = \frac{1}{2}\frac{dh}{dt}}[/tex]
So, we have:
[tex]\mathbf{V' = \frac 13 \pi(16 \times \frac{1}{2}\frac{dh}{dt} + 4\frac{dh}{dt})}[/tex]
[tex]\mathbf{V' = \frac 13 \pi(8\frac{dh}{dt} + 4\frac{dh}{dt})}[/tex]
[tex]\mathbf{V' = \frac 13 \pi(12\frac{dh}{dt})}[/tex]
Expand
[tex]\mathbf{V' = 4\pi\frac{dh}{dt}}[/tex]
The volume increases at 3 cm^3 per second.
So, we have:
[tex]\mathbf{3 = 4\pi\frac{dh}{dt}}[/tex]
Divide both sides by 4
[tex]\mathbf{0.75 = \pi\frac{dh}{dt}}[/tex]
Divide both sides by pi
[tex]\mathbf{0.239 = \frac{dh}{dt}}[/tex]
Rewrite as:
[tex]\mathbf{\frac{dh}{dt} = 0.239}[/tex]
Hence, the water level increases at 0.239 cm per second when the height is 4 cm
Read more about volumes at:
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