Question

$$ \text { If } V=D^{1 / 4} T^{-5 / 6} \text { find } \frac{\partial V}{\partial T} \text { and } \frac{\partial V}{\partial D} $$

Short answer

Question: Find the partial derivatives of the function, V = D^(1/4) * T^(-5/6), with respect to T and D. Answer: The partial derivative of V with respect to T is: ∂V/∂T = -5/6 D^(1/4) T^(-11/6). The partial derivative of V with respect to D is: ∂V/∂D = 1/4 D^(-3/4) T^(-5/6).

Step-by-step solution

Understand the function

We are given the function: $$ V = D^{\frac{1}{4}}T^{-\frac{5}{6}} $$

Find the partial derivative with respect to T

To find the partial derivative of V with respect to T, we will treat D as a constant. So, applying the power rule of differentiation, we get: $$ \frac{\partial V}{\partial T} = D^{\frac{1}{4}} \cdot (-\frac{5}{6}) T^{-\frac{5}{6} - 1} $$ which simplifies to: $$ \frac{\partial V}{\partial T} = -\frac{5}{6} D^{\frac{1}{4}} T^{-\frac{11}{6}} $$

Find the partial derivative with respect to D

To find the partial derivative of V with respect to D, we will treat T as a constant. So, applying the power rule of differentiation, we get: $$ \frac{\partial V}{\partial D} = \frac{1}{4} D^{\frac{1}{4} - 1} T^{-\frac{5}{6}} $$ which simplifies to: $$ \frac{\partial V}{\partial D} = \frac{1}{4} D^{-\frac{3}{4}} T^{-\frac{5}{6}} $$ As a result, we have found the required partial derivatives: $$ \frac{\partial V}{\partial T} = -\frac{5}{6} D^{\frac{1}{4}} T^{-\frac{11}{6}} $$ and $$ \frac{\partial V}{\partial D} = \frac{1}{4} D^{-\frac{3}{4}} T^{-\frac{5}{6}} $$

Key concepts

Power Rule

The power rule in calculus is essential for differentiating functions of the form \(x^n\), where \(n\) is any real number. It states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). This rule is invaluable as it allows us to find how a function changes as its input changes.

In the context of partial differentiation, the power rule assists in finding the rate of change of a multivariable function with respect to one variable, keeping the others constant. For example, when differentiating \(V = D^{1/4}T^{-5/6}\) with respect to \(T\), we treat \(D\) as a constant. The power rule tells us that the derivative of \( T^{-5/6}\) with respect to \( T\) is \(-5/6 \cdot T^{-11/6}\).

This simplification is key to understanding complex functions in multivariable calculus and is foundational for tackling more intricate problems in engineering and scientific mathematics.

Multivariable Calculus

Multivariable calculus extends single-variable calculus to functions of several variables. This branch of mathematics is crucial for solving real-world problems that involve multiple factors changing simultaneously. It introduces key concepts like partial derivatives, gradients, and Lagrange multipliers.

Partial differentiation, a central concept in multivariable calculus, focuses on finding the derivative of a function with respect to one variable, treating all other variables as constant. In our problem, when computing \(\frac{\partial V}{\partial T}\), we treat \(D\) as a constant and use the power rule for \(T^{n}\). Simultaneously, when computing \(\frac{\partial V}{\partial D}\), we treat \(T\) as constant and apply the power rule similarly for \(D\).

This concept not only helps describe how systems behave in various disciplines, including physics and economics, but also forms the backbone of many analytical techniques in engineering. Understanding these concepts deeply allows practitioners to model and predict system behaviors effectively.

Engineering Mathematics

Engineering mathematics harnesses mathematical tools and techniques essential for solving practical engineering problems. It encompasses a variety of mathematical areas including calculus, linear algebra, and differential equations.

Within this discipline, the application of multivariable calculus is critical. Engineers often deal with systems where multiple factors, such as temperature, pressure, and volume, change simultaneously. Partial differentiation, using concepts like the power rule, helps engineers model and analyze these changes accurately. For instance, understanding how a particular component reacts to changes in temperature and pressure can be vital for designing safe and efficient systems.

By mastering these concepts, engineers are better equipped to develop innovative solutions and improvements in technology, structures, and materials. The ability to mathematically model and analyze real-world problems is a cornerstone of successful engineering practices.