# Thread: How to calculate wetted surface area?

1. ## How to calculate wetted surface area?

A resevoir has the shape of an inverted cone with altitude 6m and diameter at the top 18m. Water runs in at $\displaystyle 3 m^3/s$. At the instant the maximum depth is 4m, determine the following.
a) the rate of increase of depth [completed Answer for $\displaystyle \frac{dh}{dt}$ is $\displaystyle \frac{1}{12 \pi}$ when h=4m
b) the rate of increase of wetted surface area of the reservoir [answer is
$\displaystyle \frac{\sqrt{13}}{2}$

b) since i am not given $\displaystyle \frac{ds}{dt}$ or $\displaystyle \frac{dr}{dt}$the only way i can solve for surface area is if i put r and s in terms of h then differentiate, plug in values and hopefully get the answer.

SA = $\displaystyle \pi r^2 + \pi rs$

SA = $\displaystyle \pi \left(\frac{3}{2}h\right)^2 + \pi \left(\frac{3}{2}h\right) \left(\frac{\sqrt{13}}{2}h\right)$

SA = $\displaystyle \frac{9}{4} \pi h^{2} + \frac{3\sqrt{13}}{4} \pi h^{2}$

$\displaystyle \frac{dSA}{dt} = \frac{9}{2} \pi h \frac{dh}{dt} + \frac{3 \sqrt{13}}{2} \pi h \frac{dh}{dt}$

$\displaystyle \frac{dSA}{dt} = \frac{9}{2} \pi (4) \left(\frac{1}{12 \pi} \right) + \frac{3 \sqrt{13}}{2} \pi (4) \left(\frac{1}{12 \pi} \right)$

$\displaystyle \frac{dSA}{dt} = \frac{36 \pi}{24 \pi} + \frac{12 \sqrt{13} \pi}{24 \pi}$

$\displaystyle \frac{dSA}{dt} = \frac{3}{2} + \frac{ \sqrt{13}}{2}$

$\displaystyle \frac{dSA}{dt} = \frac{3 + \sqrt{13}}{2}$

which is definenty not the answer. So what am i doing wrong???? my logic seems fine... and also I even tried direct subsitution for s in the SA formula and still didnt get the right answer. I am not given $\displaystyle \frac{dr}{dt}$ or $\displaystyle \frac{ds}{dt}$ the only thing i know is to write s and r in terms of h and even then as u see is still the incorrect answer. Please help. Im in dire need of assistance.

And if u are wondering how i got the slant height. Its from using phythagorean theorum and substituting $\displaystyle r = \frac{3}{2}h$ Then using phythagorean theorum:

$\displaystyle \left( \frac{3}{2}h \right)^2 + h^2 = s^2$

$\displaystyle \frac{9}{4} h^2 + h^2 = s^2$

$\displaystyle \sqrt{\frac{13}{4}h^2} = s$

$\displaystyle \frac{\sqrt{13}}{2}h = s$

And also the relation of r:h is from making a diagram. I do not know how to create pictures in LateX but from reading this problem and making a diagram this is what you should get:

$\displaystyle \frac{r}{h} = \frac{9}{6} = \frac{3}{2}$

$\displaystyle 2r = 3h$

$\displaystyle r = \frac{3}{2}h$

2. Originally Posted by ghostanime2001
A resevoir has the shape of an inverted cone with altitude 6m and diameter at the top 18m. Water runs in at $\displaystyle 3 m^3/s$. At the instant the maximum depth is 4m, determine the following.
a) the rate of increase of depth [completed Answer for $\displaystyle \frac{dh}{dt}$ is $\displaystyle \frac{1}{12 \pi}$ when h=4m
b) the rate of increase of wetted surface area of the reservoir [answer is
$\displaystyle \frac{\sqrt{13}}{2}$

b) since i am not given $\displaystyle \frac{ds}{dt}$ or $\displaystyle \frac{dr}{dt}$the only way i can solve for surface area is if i put r and s in terms of h then differentiate, plug in values and hopefully get the answer.

SA = $\displaystyle \pi r^2 + \pi rs$

SA = $\displaystyle \pi \left(\frac{3}{2}h\right)^2 + \pi \left(\frac{3}{2}h\right) \left(\frac{\sqrt{13}}{2}h\right)$

SA = $\displaystyle \frac{9}{4} \pi h^{2} + \frac{3\sqrt{13}}{4} \pi h^{2}$

$\displaystyle \frac{dSA}{dt} = \frac{9}{2} \pi h \frac{dh}{dt} + \frac{3 \sqrt{13}}{2} \pi h \frac{dh}{dt}$

$\displaystyle \frac{dSA}{dt} = \frac{9}{2} \pi (4) \left(\frac{1}{12 \pi} \right) + \frac{3 \sqrt{13}}{2} \pi (4) \left(\frac{1}{12 \pi} \right)$

$\displaystyle \frac{dSA}{dt} = \frac{36 \pi}{24 \pi} + \frac{12 \sqrt{13} \pi}{24 \pi}$

$\displaystyle \frac{dSA}{dt} = \frac{3}{2} + \frac{ \sqrt{13}}{2}$

$\displaystyle \frac{dSA}{dt} = \frac{3 + \sqrt{13}}{2}$

which is definenty not the answer. So what am i doing wrong???? my logic seems fine... and also I even tried direct subsitution for s in the SA formula and still didnt get the right answer. I am not given $\displaystyle \frac{dr}{dt}$ or $\displaystyle \frac{ds}{dt}$ the only thing i know is to write s and r in terms of h and even then as u see is still the incorrect answer. Please help. Im in dire need of assistance.

And if u are wondering how i got the slant height. Its from using phythagorean theorum and substituting $\displaystyle r = \frac{3}{2}h$ Then using phythagorean theorum:

$\displaystyle \left( \frac{3}{2}h \right)^2 + h^2 = s^2$

$\displaystyle \frac{9}{4} h^2 + h^2 = s^2$

$\displaystyle \sqrt{\frac{13}{4}h^2} = s$

$\displaystyle \frac{\sqrt{13}}{2}h = s$

And also the relation of r:h is from making a diagram. I do not know how to create pictures in LateX but from reading this problem and making a diagram this is what you should get:

$\displaystyle \frac{r}{h} = \frac{9}{6} = \frac{3}{2}$

$\displaystyle 2r = 3h$

$\displaystyle r = \frac{3}{2}h$
Kudos to you for showing all your work

Note that $\displaystyle \frac{dS}{dt} = \frac{dS}{dh} \cdot \frac{dh}{dt} = \frac{dS}{dh} \cdot \frac{1}{12 \pi}$.

Note that the wetted surface area of the reservoir does not include the flat circular top of the cone. It is only the sloped side. So $\displaystyle S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2$.

3. So now do i just differentiate with respect to h for this formula?

$\displaystyle S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2$

4. Originally Posted by ghostanime2001
So now do i just differentiate with respect to h for this formula?

$\displaystyle S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2$
Yes. Then substitute h = 4. Then substitute into the expression for dS/dt.

5. how do i eliminate the $\displaystyle \pi$

6. Originally Posted by ghostanime2001
how do i eliminate the $\displaystyle \pi$
Substitute everything into the expression for dS/dt. The pi's cancel. If you're having trouble doing this you will need to show your working.

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