MHB Calculus: Related Rates - Inverted Pyramid Filling w/ Water | Yahoo Answers

AI Thread Summary
The problem involves calculating the rate at which the water level rises in an inverted square pyramid with a height of 10 m and a base side of 13 m, filled at a rate of 10 m³ per second. The volume of the pyramid is expressed as V = (1/3)s²h, where s is the side length of the water level at height h. By establishing the relationship between s and h, the volume can be rewritten in terms of h, leading to V = (169/300)h³. Differentiating this volume with respect to time and substituting the known rate of volume change allows for the calculation of the water level rise rate, yielding dh/dt = 10/169 m/s when the pyramid is full. The final result indicates that the water level rises at approximately 0.059 m/s when the pyramid is completely filled.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

How do I do this calculus related rates problem? Full solution please.?

The Louvre museum in Paris features an inverted square pyramid with a height of 10 m. The side of the square is 13 m. Supervillains decide to fill the pyramid with water, and do so at a rate of 10 m3 per second. How quickly is the water level rising when it reaches the top? *Please note that the pyramid is inverted* Thanks guys

I have posted a link there to this thread so the OP can view my work.
 
Mathematics news on Phys.org
Hello Temitope,

Let's begin with the formula for the volume of a pyramid:

$$V=\frac{1}{3}bh$$

Now, the base $b$ is a square, and so let's let $s$ be the measure of the sides of the square, and so our volume formula becomes:

$$V=\frac{1}{3}s^2h$$

At any point in time, the volume of water will be a square pyramid that is similar to the container, and so we know we will always have:

$$\frac{s}{h}=\frac{13}{10}\implies s=\frac{13}{10}h$$

Now we may express the volume in terms of one variable $h$:

$$V=\frac{1}{3}\left(\frac{13}{10}h\right)^2h=\frac{169}{300}h^3$$

Now, if we differentiate with respect to time $t$, we obtain:

$$\frac{dV}{dt}=\frac{169}{100}h^2\frac{dh}{dt}$$

Now, we are told:

$$\frac{dV}{dt}=10\frac{\text{m}^3}{\text{s}}$$

And so (given that our units of length are in meters and our units time in seconds) we may write:

$$10=\frac{169}{100}h^2\frac{dh}{dt}$$

Solving for $$\frac{dh}{dt}$$, we obtain:

$$\frac{dh}{dt}=\frac{1000}{169h^2}$$

And so we find that when the pyramid is full, or when $h=10$, we have:

$$\bbox[7px,border:2px solid #207498]{\left.\frac{dh}{dt}\right|_{h=10}=\frac{1000}{169(10)^2}\frac{\text{m}}{\text{s}}=\frac{10}{169}\frac{\text{m}}{\text{s}}}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top