Water is leaking out of an inverted conical tank at a rate...

Water is leaking out of an inverted conical tank at a rate of $10.000 \mathrm{cm}^{3} / \mathrm{min}$ at the same time that water is being pumped into the tank at a constant rate. The tank has height $6 \mathrm{m}$ and the diameter at the top is 4 m. If the water level is rising at a rate of $20 \mathrm{cm} / \mathrm{min}$ when the height of the water is $2 \mathrm{m}$, find the rate at which water is being pumped into the tank.

Key Concepts

Chain Rule in Differentiation

The chain rule is a method for differentiating composite functions and is crucial in related rates problems. It allows one to differentiate an equation that relates volume to height, and then relate the derivative of the volume with respect to time to the derivative of the height with respect to time.

Unit Conversion

Ensuring consistency in units across all components of a problem is vital. This includes converting measurements, such as from meters to centimeters or vice versa, so that all parts of the calculation are compatible, thereby avoiding errors in the final result.

Similar Triangles and Geometric Relationships in Cones

When dealing with a conical tank, the geometry of the cone enables the establishment of a relationship between the water’s height and the radius at that height using similar triangles. This proportionality allows the simplification of the volume formula by expressing the radius in terms of the height.

Related Rates

Related rates problems involve determining the rate at which one quantity changes by relating it to other quantities whose rates of change are known. This typically requires establishing a relationship among variables and then using differentiation with respect to time to connect their rates.

Volume of a Cone

The volume of a cone is given by the formula V = (1/3)?r²h. In problems involving changing water levels in a conical tank, this formula is essential. By substituting the expression for the radius derived from similar triangles, the volume can be expressed solely in terms of the height, facilitating the differentiation process.

Transcript

00:01 So here is a diagram of the situation right here.

00:05 So we have an inverted cone with a radius at the top of 4 meters.

00:10 That's a maximum radius when the height is 6 meters.

00:13 And we have a variable h, which we can use to represent the height of the water at any given time.

00:23 So the volume of a cone can be represented using the following equation.

00:29 V equals pi r squared h over three however we can represent r in terms of h so in order to do that let's look at sort of a slice of the cone sort of a half cross section half of the cross section of the cone which looks like this so um at its fullest height the height of the triangle or of the triangle and therefore the cone is six and the other base has a, or the base has a length of four.

01:05 Now, at any height, we can draw a line representing the radius of the cone at that point.

01:17 And we know that since these two triangles are similar, because the three angles are all equal, we know that the ratio between the height and the radius is always six to four, six to five.

01:32 6 to 4 right here.

01:35 So we can solve for r.

01:36 R is equal to 4h over 6, which is equal to 2h over 3.

01:43 So let's rewrite this equation for volume, replacing r with 2h over 3.

01:52 So i skipped a few steps to make it a bit easier, but essentially i squared this term, and that became 4h squared over 9.

02:02 The 1 over 9, in the denominator moved over to the denominator of the entire fraction.

02:12 So multiplied by 3, that's going to be 27.

02:16 And i multiplied 4h squared by h, and that became 4h cubed.

02:21 So this is the equation for volume in relation to height, any given height.

02:27 However, we want an equation that defines the relationships between current volume, current height, and the rates of change of volume and height at a specific time.

02:39 So in order to get that, we differentiate the entire equation by t, where t is time in seconds, in this case, i believe.

02:48 Or no, minutes.

02:50 Or it can be any unit of time.

02:52 It just has to say constant.

02:54 So differentiating the left side yields simply dv over dt because v differentiates to 1, and we multiply 1 by dv over dt.

03:07 Dv over dt is the variable that represents the rate of change of the volume...

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