Problem 52 - Calculus Early Transcendentals

Question

A box with a square base of length \(x\) and height \(h\) has a volume \(V=x^{2} h\) a. Compute the partial derivatives \(V_{x}\) and \(V_{h}\) b. For a box with \(h=1.5 \mathrm{m},\) use linear approximation to estimate the change in volume if \(x\) increases from \(x=0.5 \mathrm{m}\) to \(x=0.51 \mathrm{m}\) c. For a box with \(x=0.5 \mathrm{m},\) use linear approximation to estimate the change in volume if \(h\) decreases from \(h=1.5 \mathrm{m}\) to \(h=1.49 \mathrm{m}\) d. For a fixed height, does a \(10 \%\) change in \(x\) always produce (approximately) a \(10 \%\) change in \(V\) ? Explain. e. For a fixed base length, does a \(10 \%\) change in \(h\) always produce (approximately) a \(10 \%\) change in \(V\) ? Explain.

Step-by-Step Solution

Answer

Answer: No, a 10% change in the base length (x) results in a 20% change in the volume (V) of the box.

Step 1: Compute the partial derivatives

To find the partial derivatives, differentiate the volume V with respect to x and h. \(V = x^2h\) \(V_x = \frac{dV}{dx} = 2xh\) \(V_h = \frac{dV}{dh} = x^2\)

Step 2: Linear approximation for V (x = 0.5m, x = 0.51m,

Now, we have to estimate the change in volume if x increases from 0.5m to 0.51m and h = 1.5m. Using the linear approximation formula, we have: \(\Delta V \approx V_x \Delta x\) Where \(\Delta x = 0.01\) and \(V_x = 2(0.5)(1.5)\). Let's calculate the change in volume: \(\Delta V \approx (2(0.5)(1.5))(0.01)\) \(\Delta V \approx 0.015\) m^3

Step 3: Linear approximation for V (h = 1.5m, h = 1.49m)

Now, we have to estimate the change in volume if h decreases from 1.5m to 1.49m and x = 0.5m. Using the linear approximation formula, we have: \(\Delta V \approx V_h \Delta h\) Where \(\Delta h = -0.01\) and \(V_h = (0.5)^2\). Let's calculate the change in volume: \(\Delta V \approx ((0.5)^2)(-0.01)\) \(\Delta V \approx -0.0025\) m^3

Step 4: Percentage change in V for a fixed height

To find out if a 10% change in x always produces approximately a 10% change in V, we can use the relationship between percentage change in V and percentage change in x. \(\frac{\Delta V}{V} = \frac{V_x\Delta x}{x^2h}\) \(\frac{\Delta V}{V} = \frac{2xh\Delta x}{x^2h}\) \(\frac{\Delta V}{V} = \frac{2\Delta x}{x} = 2\frac{\Delta x}{x}\) For a 10% change in x, \(\frac{\Delta x}{x} = 0.1\). Let's substitute this into the equation: \(\frac{\Delta V}{V} = 2(0.1)\) \(\frac{\Delta V}{V} = 0.2\) A 10% change in x results in a 20% change in V, so a 10% change in x does not produce approximately a 10% change in V.

Step 5: Percentage change in V for a fixed base length

To find out if a 10% change in h always produces approximately a 10% change in V, we can use the relationship between percentage change in V and percentage change in h. \(\frac{\Delta V}{V} = \frac{V_h\Delta h}{x^2h}\) \(\frac{\Delta V}{V} = \frac{x^2\Delta h}{x^2h}\) \(\frac{\Delta V}{V} = \frac{\Delta h}{h}\) For a 10% change in h, \(\frac{\Delta h}{h} = 0.1\). Let's substitute this into the equation: \(\frac{\Delta V}{V} = 0.1\) A 10% change in h results in a 10% change in V, so a 10% change in h does produce approximately a 10% change in V.

Key Concepts

Volume of a Box

When talking about the volume of a box, especially one with a square base, we are referring to the amount of space contained within that three-dimensional shape. The formula to determine this is rather straightforward. For a box with a base length of \(x\) and height \(h\), the volume \(V\) is given by the product of its area of base and its height: \[ V = x^2h \]This formula captures how both the base and the height contribute to the overall volume. The base length \(x\) is squared because the box has a square base. This means, if you know the dimensions of a box, plug them into this formula, and you can find the volume swiftly and accurately.

Linear Approximation

Linear approximation is a useful method when you want to estimate small changes in a function. It's particularly handy when calculating minor variations, so detailed computations aren't necessary.Consider you have a function, such as the volume mentioned earlier, and you want to know how a tiny change in one variable will affect the result. Linear approximation helps here by simplifying the calculations through derivatives.For example, if you know how the volume changes with respect to \(x\) and \(h\), you can estimate the change in volume using \[\Delta V \approx V_x \Delta x + V_h \Delta h\]where \(V_x\) and \(V_h\) are the partial derivatives of the volume with respect to \(x\) and \(h\), respectively. By plugging in small changes in \(x\) or \(h\), you quickly predict the effect on \(V\). This makes linear approximation a valuable tool in engineering and physics.

Percentage Change

Understanding percentage change helps to evaluate and compare the effects of alterations in variables, specifically how these changes impact the overall result.Let's see how it applies to the volume of a box. If you change the base length \(x\) by 10%, you observe how this affects the volume \(V\). The relationship derived shows that for a 10% increase in \(x\), the volume doesn't increase by just 10%; it doubles this effect to approximately a 20% rise in \(V\). \[\text{Percentage Change in } V = 2 \times \text{Percentage Change in } x\]In contrast, altering the height \(h\) by 10% directly results in a 10% change in the volume, as seen in the formula:\[\text{Percentage Change in } V = \text{Percentage Change in } h\]This insight is crucial when designing or altering box dimensions, allowing for strategic planning when the volume's sensitivity to dimension changes is a factor.

Mathematical Modeling

Mathematical modeling is all about creating a mathematical representation of a real-world situation. It involves using formulas, like those for volume, to predict and analyze outcomes based on different scenarios. In the context of the box with square base, modeling helps us understand how changes in its dimensions impact the volume. By applying partial derivatives and percentage change calculations, we can make informed predictions quickly and easily. These models are vital in fields such as architecture, engineering, and manufacturing, providing insights that save time and resources. Knowing how to model effectively ensures decisions are data-driven and based on solid predictions, improving the accuracy and efficiency of projects. Through mathematical modeling, complex real-world problems become manageable, allowing simulation and optimization to improve designs and processes.