Math  /  Calculus

QuestionHelp | Baran Uyan (6791123176) |Logout Gradebook External
12024 Remaining Time: 12:41:31
A water tank has the shape of a right circular cone, with the tip pointing downwards. The tank is 25 cm tall, and the radius (at the top) is 11 cm . If water is being drained from the tank at a rate of 6 cm3/s6 \mathrm{~cm}^{3} / \mathrm{s}, find the rate at which the water level is changing when the water is 6 cm deep.
Do not include units in your answer. Your answer can be exact or approximate. If it is approximate, round to three decimal places.

Studdy Solution

STEP 1

What is this asking? We have a cone-shaped tank leaking water, and we want to figure out how fast the water level is dropping at a specific moment. Watch out! Don't mix up the rate of change of the *volume* with the rate of change of the *height*!

STEP 2

1. Relate the radius and height
2. Express the volume in terms of height
3. Differentiate with respect to time
4. Plug in the known values

STEP 3

Alright, let's **visualize**!
We've got a cone, right?
The water inside forms a smaller cone *similar* to the tank itself.
This means the ratio of corresponding sides stays constant.

STEP 4

The tank's height is **25 cm**, and its radius is **11 cm**.
Let's call the water's height hh and its radius rr.
Because the cones are similar, we know that: rh=1125 \frac{r}{h} = \frac{11}{25}

STEP 5

We can express the radius rr in terms of the height hh: r=1125h r = \frac{11}{25}h This is super useful because it lets us relate the changing radius to the changing height!

STEP 6

The **volume** VV of a cone is given by: V=13πr2h V = \frac{1}{3} \pi r^2 h

STEP 7

Now, let's **substitute** the expression for rr we found earlier: V=13π(1125h)2h V = \frac{1}{3} \pi \left(\frac{11}{25}h\right)^2 h V=13π121625h3 V = \frac{1}{3} \pi \cdot \frac{121}{625} h^3 V=121π1875h3 V = \frac{121\pi}{1875} h^3 Boom! The volume is now entirely in terms of the height hh.

STEP 8

We're given that the volume is changing at a rate of 6 cm3/s-6 \ \text{cm}^3/\text{s}.
This is dVdt\frac{dV}{dt}.
The negative sign indicates the volume is *decreasing*.
We want to find dhdt\frac{dh}{dt}, the rate at which the height is changing.

STEP 9

Let's **differentiate** both sides of our volume equation with respect to time tt: dVdt=ddt(121π1875h3) \frac{dV}{dt} = \frac{d}{dt}\left(\frac{121\pi}{1875} h^3\right) dVdt=121π18753h2dhdt \frac{dV}{dt} = \frac{121\pi}{1875} \cdot 3h^2 \frac{dh}{dt} dVdt=363π1875h2dhdt \frac{dV}{dt} = \frac{363\pi}{1875} h^2 \frac{dh}{dt}

STEP 10

We know dVdt=6\frac{dV}{dt} = -6 and we're interested in the moment when h=6h = 6.
Let's **plug** these values into our equation: 6=363π1875(6)2dhdt -6 = \frac{363\pi}{1875} \cdot (6)^2 \frac{dh}{dt} 6=363π187536dhdt -6 = \frac{363\pi}{1875} \cdot 36 \frac{dh}{dt} 6=13188π1875dhdt -6 = \frac{13188\pi}{1875} \frac{dh}{dt}

STEP 11

Now, let's **solve** for dhdt\frac{dh}{dt}: dhdt=6187513188π \frac{dh}{dt} = \frac{-6 \cdot 1875}{13188\pi} dhdt=1125013188π \frac{dh}{dt} = \frac{-11250}{13188\pi} dhdt0.271 \frac{dh}{dt} \approx -0.271

STEP 12

The water level is changing at approximately 0.271-0.271 cm/s.

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