project

www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/projects/ess_wp_0706a_stu.pdf


i have question on number 4 i dont know were to begin or dont know how to set it up i read the problem 5 times and still seem confused by it .i know you need to use separable equation.
Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectional area A(h) at height h. Then the volume of water up to height h is given by
\(\displaystyle V(h)\,\, =\,\, \int_0^h\,\, A(u)\,\, du\)
and so the Fundamental Theorem of Calculus gives dV/dh = A(h). It follows that
\(\displaystyle \frac{dV}{dt}\, =\, \frac{dV}{dh}\, \frac{dh}{dt}\, =\, A(h)\,\, \frac{dh}{dt}\)
and so Torricelli’s Law becomes
\(\displaystyle A(h)\,\, \frac{dh}{dt}\, =\, -a\, \sqrt{2\, g\, h} \)
(a) Suppose the tank has the shape of a sphere with radius 2 m and is initially half full of
water. If the radius of the circular hole is 1 cm and we take g=10 m/s2, show that h satisfies the differential equation
\(\displaystyle (4h\, -\, h^2)\, \frac{dh}{dt}\, =\, -0.0001\, \sqrt{20\,h}\)
(b) How long will it take for the water to drain completely?


I believe I've got the problem correct. In any case if the tank has the shape of a sphere, which is the cross sectional area? Please show your work and where you are stuck so we can help. You should also read
http://www.freemathhelp.com/forum/threads/77972-Read-Before-Posting!!
 
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would you start by find the separable equation? of(4h-h1/2)dh/dt=-0.0001*20h1/2
 
im stuck on "a" i dont know what to do
The first thing you need to do is to get the cross sectional area of a sphere. If we let x and y be the usual Cartesian coordinates and z the height, then the sphere's equation is
x2 + y2 + z2 = R2
if the center of the sphere is (x, y, z) = (0, 0,0). Now let a plane of constant height z = h intersect the sphere. The intersection of the plane and the surface of the sphere will be a circle. What is the radius of that circle? Given the radius, what is the area of that circle?
 
x2 + y2 + z2 = 22
since x,y,z is 0 that would mean the area of the circle would be 4
 
x2 + y2 + z2 = 22
since x,y,z is 0 that would mean the area of the circle would be 4
A circle in the x-y plane has the form of
x2 + y2 = r2
so using your equation [x2 + y2 + z2 = 22] and the equation for the circle what would r2 be if z were equal to h? Given r2 what would be the area of the circle?
 
A circle in the x-y plane has the form of
x2 + y2 = r2
so using your equation [x2 + y2 + z2 = 22] and the equation for the circle what would r2 be if z were equal to h? Given r2 what would be the area of the circle?

r=4
 
Suppose z=1, then the circle is
x2 + y2 + 12 = 4
or
x2 + y2 = 4 - 12 = 3 = \(\displaystyle (\sqrt{3})^2\)

Suppose z = 1.1, then the circle is
x2 + y2 + 1.12 = 4
or
x2 + y2 = 4 - 1.12 = 4 - 1.21 = 2.79 = \(\displaystyle (\sqrt{2.79})^2\)

Suppose z = 1.2, then the circle is
x2 + y2 + 1.22 = 4
or
x2 + y2 = 4 - 1.22 = 4 - 1.44 = 2.56 = \(\displaystyle (\sqrt{2.56})^2\)

Suppose z = h, then the circle is
x2 + y2 + h2 = 4
or ...

What is the aree of the circle in each of those cases?
 
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Suppose z=1, then the circle is
x2 + y2 + 12 = 4
or
x2 + y2 = 4 - 12 = 3 = \(\displaystyle (\sqrt{3})^2\)

Suppose z = 1.1, then the circle is
x2 + y2 + 1.12 = 4
or
x2 + y2 = 4 - 1.12 = 4 - 1.21 = 2.79 = \(\displaystyle (\sqrt{2.79})^2\)

Suppose z = 1.2, then the circle is
x2 + y2 + 1.22 = 4
or
x2 + y2 = 4 - 1.22 = 4 - 1.44 = 2.56 = \(\displaystyle (\sqrt{2.56})^2\)

Suppose z = h, then the circle is
x2 + y2 + h2 = 4
or ...

What is the aree of the circle in each of those cases?

are you talking about the one that x2+y2+h2=4???
 
r2=22=4
so the area of a circle would be A=r2*pi
matt757,
I believe you need much more help than is generally available here. You need to review your Basic/Beginning algebra and become very familiar with that before you can advance to Intermediate / Advanced algebra and then on to calculus.

Look at the equation
x2+y2=4-h2
Now look at the equation
x2 + y2 = r2
If those two equations are the same circle what is r2? That is, if
x2 + y2 = r2 = x2+y2 = 4-h2
what is r2? After that you will need the area of that cross sectional area. The a in the original equation is the area of the hole but you will need to change that to meters first since the rest of your physical measurements are also in m.
 
i get it now because since z=h right?? ok now you go back to x2+y2+h2=r2 now we would minus the h which would give us the x2+y2=r2-h2 since 2 is the radius then it would be x2+y2=4-h2 since the formula is x2+y2=r2 then it would be x2+y2=(4-h2)2
 
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i get it now because since z=h right?? ok now you go back to x2+y2+h2=r2 now we would minus the h which would give us the x2+y2=r2-h2 since 2 is the radius then it would be x2+y2=4-h2 since the formula is x2+y2=r2 then it would be x2+y2=(4-h2)2
x2 + y2 = 4-h2 = \(\displaystyle (\sqrt{4-h^2})^2\)
 
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