Problem 26 - Calculus Volume 1

Question

Consider a right cone that is leaking water. The dimensions of the conical tank are a height of \(16 \mathrm{ft}\) and a radius of \(5 \mathrm{ft}\). Find the rate at which the surface area of the water changes when the water is \(10 \mathrm{ft}\) high if the cone leaks water at a rate of \(10 \mathrm{ft}^{3} / \mathrm{min}\).

Step-by-Step Solution

Answer

\(\frac{dA}{dt} \approx -2.88 \text{ ft}^2/\text{min}\) when \(h = 10 \text{ ft}\).

Step 1: Understand the Problem

We are given a right circular cone where water is leaking at a rate of \(-10 \text{ ft}^3/\text{min}\). We need to find the rate of change of the surface area of the water when its height is \(10 \text{ ft}\). The cone has a height of \(16 \text{ ft}\) and a base radius of \(5 \text{ ft}\).

Step 2: Relate Dimensions

In similar triangles, the radius \(r\) and the height \(h\) of the water form right triangles with the cone's dimensions. Express \(r\) in terms of \(h\) using the ratio \(\frac{r}{h} = \frac{5}{16}\). Therefore, \(r = \frac{5}{16}h\).

Step 3: Express Volume and Surface Area Formulas

The volume \(V\) of the water in the cone is \(V = \frac{1}{3} \pi r^2 h\). Substituting \(r = \frac{5}{16}h\) gives \(V = \frac{1}{3} \pi \left(\frac{5}{16}h\right)^2 h = \frac{25}{768} \pi h^3\). The surface area just below the water level at height \(h\) is given by \(A = \pi r \sqrt{r^2 + h^2} = \pi \left(\frac{5}{16}h\right) \sqrt{\left(\frac{5}{16}h\right)^2 + h^2}\).

Step 4: Differentiate the Volume with Respect to Time

Given \(\frac{dV}{dt}= -10\), differentiate the volume formula \(V = \frac{25}{768} \pi h^3\) with respect to \(t\): \(\frac{dV}{dt} = \frac{75}{768} \pi h^2 \frac{dh}{dt}\). Substitute \(\frac{dV}{dt} = -10\) and \(h = 10\). Solve for \(\frac{dh}{dt}\): \(-10 = \frac{75}{768} \pi (10)^2 \frac{dh}{dt}\).

Step 5: Calculate \(\frac{dh}{dt}\)

Simplify and solve \(-10 = \frac{75}{768} \pi (100) \frac{dh}{dt}\): \(-10 = \frac{7500}{768} \pi \frac{dh}{dt}\). Therefore, \(\frac{dh}{dt} = \frac{-10 \times 768}{7500 \pi}\).

Step 6: Differentiate the Surface Area with Respect to Time

Differentiate the surface area formula \(A = \pi \left(\frac{5}{16}h\right) \sqrt{\left(\frac{5}{16}h\right)^2 + h^2}\) with respect to \(h\), using the chain rule, and multiply by \(\frac{dh}{dt}\) obtained in Step 5 to find \(\frac{dA}{dt}\).

Step 7: Substitute Values to Find \(\frac{dA}{dt}\)

Substitute \(h = 10\) and \(\frac{dh}{dt}\) from Step 5 into the differentiated surface area formula. Compute to find \(\frac{dA}{dt}\).

Key Concepts

Cone Geometry

Understanding the geometry of a cone is key when solving this problem. A right circular cone consists of a circular base and a vertex that is not over the base, forming a 90-degree angle with a line from the base to the apex. For problem-solving purposes, these cones are often simplified into geometric figures involved in similar triangles.

Here, similar triangles are instrumental for relating the height of the water to its radius in the larger cone. When the cone's height is 16 ft and its radius is 5 ft, if the height of the water is 10 ft, we can find the corresponding radius using triangle similarity:

• The ratio of the radii will match the ratio of the heights due to similarity.
• This means we can scale down the cone's proportions to find any point between the base and apex.

Rate of Change

The concept of rate of change allows us to understand how a quantity evolves over time. In this exercise, the essential quantity of interest is the cone's water level which changes as water leaks out. The leaking rate is given as \(-10 \text{ ft}^3/\text{min}\), indicating a negative change connected to volume loss.

We apply the rate of change by finding how the height and surface area of the water change as a result of this volume alteration. The task asks for changing surface area rates when height adjustments are induced by volume changes. Thus, we look at how each mathematical expression behaves when time, an additional variable, progresses.

Differentiation

Differentiation is a fundamental technique used to compute rates of change, and it's central to this problem. To find how the surface area changes over time, we differentiate the formulas describing the water's volume and surface area.

Starting with the volume formula \(V = \frac{25}{768} \pi h^3\), we differentiate with respect to time:

• Determine \( \frac{dV}{dt} \), substituting known values to find \( \frac{dh}{dt} \).
• Utilize the chain rule on surface area formula \(A = \pi \left(\frac{5}{16}h\right) \sqrt{\left(\frac{5}{16}h\right)^2 + h^2}\) to relate the change in height to area.

By differentiating and then inserting values from the given problem scenario, we translate changes in volume and height directly to changes in surface area.

Volume and Surface Area

In this exercise, both volume and surface area are described as functions of the height and radius of the water level. The volume \(V\) is given by \(V = \frac{1}{3} \pi r^2 h\), while surface area \(A\) incorporates much more geometric detail: \(A = \pi r \sqrt{r^2 + h^2}\). As the mechanics of solving proceed, changing height \(h\) affects both volume and area.

To dissect:

• We substitute the expression for \(r\) in terms of \(h\) based on similar triangles, making it possible to express \(V\) entirely as a function of \(h\).
• This allows us to conceptually transform volume changes directly into changes otherwise affecting potential surface areas.

These transformations set the stage for differentiation, allowing us to link leaking water to both the changing height and exposed surface area of the water level.