Problem 24

Question

Suppose a gas station has an underground tank in the shape of a rectangular parallelepiped that measures 10 feet by 8 feet by 6 feet, with error of at most $0.005$ feet in each measurement. If the tank is filled with gasoline costing $$\$ 10$$ per cubic foot, estimate by how much the cost of the gasoline can vary from $$\$ 4800$$.

Step-by-Step Solution

Verified

Answer

The gasoline cost can vary by about $\$11.80$.

1Step 1: Calculate Volume of the Tank

First, calculate the volume of the tank without considering any errors. The volume $ V $ of a rectangular parallelepiped is given by $ V = l \cdot w \cdot h $, where $ l = 10 \text{ ft} $, $ w = 8 \text{ ft} $, and $ h = 6 \text{ ft} $. So, $ V = 10 \times 8 \times 6 = 480 \text{ cubic feet} $.

2Step 2: Determine Error in Volume

Next, evaluate how the measurement errors affect the volume. The possible errors in each dimension are $ \Delta l = \Delta w = \Delta h = 0.005 \text{ ft} $. Using linear approximation, the error in volume $ \Delta V $ can be estimated as $ \Delta V = \left| \frac{\partial V}{\partial l}\right| \Delta l + \left| \frac{\partial V}{\partial w}\right| \Delta w + \left| \frac{\partial V}{\partial h}\right| \Delta h = (wh) \Delta l + (lh) \Delta w + (lw) \Delta h $.

3Step 3: Calculate Partial Derivatives

Where the partial derivatives are calculated as: $ \frac{\partial V}{\partial l} = wh = 8 \times 6 = 48 $,$ \frac{\partial V}{\partial w} = lh = 10 \times 6 = 60 $,$ \frac{\partial V}{\partial h} = lw = 10 \times 8 = 80 $. So, $ \Delta V = 48 \times 0.005 + 60 \times 0.005 + 80 \times 0.005 = 0.48 + 0.3 + 0.4 = 1.18 \text{ cubic feet} $.

4Step 4: Calculate Limit on Cost Variability

Finally, calculate the associated variability in cost. Since the cost is $ \\(10 $ per cubic foot, the error in the cost is $ 10 \times \Delta V = 10 \times 1.18 = \$11.80 \). Thus, the cost of the gasoline can vary from $ \\(4800 - 11.80 $ to $ \$4800 + 11.80 \).

Key Concepts

Volume CalculationLinear ApproximationPartial Derivatives

Volume Calculation

Imagine you have a box-like underground tank at a gas station. To find out how much gas it can hold, you need to calculate the tank's volume. Volume is the space inside a 3D object like this tank. For a rectangular parallelepiped, volume is calculated by multiplying the length, width, and height together.
In this case:

Length ($ l $) is 10 feet
Width ($ w $) is 8 feet
Height ($ h $) is 6 feet

By multiplying these dimensions, the volume ($ V $) is found: \[ V = l \cdot w \cdot h = 10 \times 8 \times 6 = 480 \] cubic feet.
This means the tank can theoretically hold 480 cubic feet of gas, assuming all dimensions were measured perfectly.
But in real life, measurements always have a little error, which could affect how much gas the tank can truly hold.

Linear Approximation

When measurements aren't perfect, errors can add up. That's why we use something called linear approximation in calculus. This is like a quick math estimate to check how small changes affect the outcomes, in this case, the tank's volume.
With small measurement errors in each dimension, the actual volume of the tank can change slightly. Linear approximation helps estimate this change without having to re-measure everything.
Let's say the tank's length, width, and height each have an error of at most $0.005$ feet. We use the calculated partial derivatives and sum them up with these small errors to estimate how the volume might change:\[ \Delta V = (wh) \Delta l + (lh) \Delta w + (lw) \Delta h \]This tells us that the total change in volume ($ \Delta V $) is a combination of small changes in each dimension. So, linear approximation saves time and effort by quickly predicting errors' impact on volume.

Partial Derivatives

Partial derivatives are a tool in calculus used to measure change along one dimension while keeping the others constant. When calculating the error in our tank's volume, partial derivatives help us see how the volume changes if just one dimension grows slightly.
Each dimension (length, width, and height) has its own partial derivative:

For length: $ \frac{\partial V}{\partial l} = wh = 48 $
For width: $ \frac{\partial V}{\partial w} = lh = 60 $
For height: $ \frac{\partial V}{\partial h} = lw = 80 $

These values tell us how much the volume changes if there are slight increases in any one dimension while keeping the others fixed. Multiplying these derivatives by the errors shows the impact of small measurement errors on the total volume.
This is crucial for understanding overall error in scenarios where precision is important, like estimating gas tank capacity and subsequent costs.

Problem 24

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