Problem 48 - Calculus

Question

When an ice dam for a glacial lake breaks, the maximal outflow rate, \(Q\) in meter \(^{3} /\) sec is a function of \(V,\) the volume of the lake (in millions of meter's). (a) What are the units of \(d Q / d V ?\) (b) Observation shows that \(d Q /\left.d V\right|_{V=12}=22 .\) About how much is the difference in maximal outflow when dams break in two lakes with volumes near \(12 \cdot 10^{6}\) meter \(^{3}\) if one of them has volume 500,000 meter \(^{3}\) greater than the other?

Step-by-Step Solution

Answer

\(dQ/dV\) has units of \(\text{m}^3/\text{sec per million m}^3\); \(\Delta Q = 11\, \text{m}^3/\text{sec}\).

Step 1: Understanding Units

First, we need to identify the units of \(dQ/dV\). The derivative \(dQ/dV\) represents the rate of change of the outflow rate \(Q\) with respect to the volume \(V\). Given \(Q\) is in \(\text{m}^3/\text{sec}\) and \(V\) is in millions of \(\text{m}^3\), the units of \(dQ/dV\) will be \(\frac{\text{m}^3/\text{sec}}{\text{millions of m}^3}\).

Step 2: Interpreting Given Derivative Value

We know \(dQ/dV = 22\) at \(V = 12\). This indicates that for a small change in volume, the outflow rate \(Q\) changes by 22 times the change in volume, provided \(V\) is near 12 million cubic meters.

Step 3: Calculating the Volume Difference in Millions

Given that one lake is 500,000 cubic meters larger than the other, we need to convert this difference into millions of cubic meters. Thus, \(\Delta V = \frac{500,000}{1,000,000} = 0.5\) million cubic meters.

Step 4: Calculating the Difference in Maximal Outflow

Using the linear approximation, the difference in maximal outflow \(\Delta Q\) is approximately \(dQ/dV \times \Delta V = 22 \times 0.5 = 11.0\) cubic meters per second.

Key Concepts

Derivative Units

Derivatives help us understand how a quantity changes with respect to another. In this exercise, we look at the derivative \(dQ/dV\) which signifies the rate at which the outflow rate \(Q\), measured in \(\text{m}^3/\text{sec}\), changes with respect to the lake volume \(V\) in millions of cubic meters.
Identifying units in derivatives can sometimes be tricky, yet it's an important step to understanding the underlying relationship between quantities. Here:

• The derivative \(dQ/dV\) has the units \(\frac{\text{m}^3/\text{sec}}{\text{millions of m}^3}\).
• This represents how much the outflow rate changes per million cubic meters of water volume.

By analyzing the units of the derivative, you can get insights into the scaling of the effect. In our case, we're understanding how the outflow rate of water would vary if the size of the lake changes slightly.

Rate of Change

The 'rate of change' is a core concept in calculus. For this specific exercise, it helps estimate how adjustments in our glacial lake's volume affect the water flow out of it. By observing that \(dQ/dV = 22\) at \(V = 12\) million cubic meters:

• We learn that each incremental million cubic meter boost in volume leads to a 22 \(\text{m}^3/\text{sec}\) alteration in outflow rate.
• This means the lake's increased water capacity intensifies its discharge capability proportionately.

The rate of change, thus, translates into real-world implications about water management in areas with fluctuating glacial lakes, assisting in decisions about infrastructure resilience against these water disasters.

Volume Difference

Changes in lake volumes, here described as a 'volume difference,' directly relate to outflow dynamics. Understanding this connection is pivotal for situations where minor volume modifications can lead to substantial flow changes.
Suppose one lake has a volume that's 500,000 cubic meters (or 0.5 million cubic meters) higher than another. Utilizing the derivative value, the difference in outflow rate, \(\Delta Q\), can be computed:

• \(\Delta Q = dQ/dV \times \Delta V = 22 \times 0.5 = 11.0 \ \text{m}^3/\text{sec}\).

This means that such a small increase in volume will lead to a 11 cubic meter per second rise in the outflow rate. It shows how small volume increments have significant effects, and it reflects why planners need these calculations to better predict flood conditions.