Partial Derivatives Step-by-Step Solutions And Explanations
In the realm of multivariable calculus, understanding partial derivatives is crucial for analyzing functions that depend on multiple variables. These derivatives allow us to examine the rate of change of a function with respect to one variable while holding others constant. This article delves into several problems involving partial derivatives, providing step-by-step solutions and explanations to enhance your understanding. Let's embark on this mathematical journey, exploring how to compute and interpret partial derivatives in various contexts.
Problem 27: Finding Partial Derivatives of z = ln(x² + 1)
Understanding the Problem
In this problem, we are given a function z = ln(x² + 1), where x is further defined as x = r cosθ. Our task is to find the partial derivatives of z with respect to r ( ∂z/∂r ) and θ ( ∂z/∂θ ). This requires applying the chain rule, a fundamental concept in calculus that allows us to differentiate composite functions. The chain rule, in essence, helps us break down complex derivatives into simpler, manageable parts. We will first differentiate z with respect to x, and then differentiate x with respect to both r and θ. Finally, we will combine these results to obtain the desired partial derivatives.
Step-by-Step Solution
1. Differentiate z with respect to x:
We begin by finding ∂z/∂x. Given z = ln(x² + 1), we apply the chain rule. The derivative of the natural logarithm function, ln(u), with respect to u is 1/u. Here, u = x² + 1. Thus, we have:
∂z/∂x = d/dx [ln(x² + 1)] = 1/(x² + 1) * d/dx [x² + 1]
The derivative of x² + 1 with respect to x is 2x. Therefore,
∂z/∂x = (1/(x² + 1)) * (2x) = 2x / (x² + 1)
This result gives us the rate of change of z with respect to x, which is a crucial component in finding the partial derivatives with respect to r and θ.
2. Differentiate x with respect to r and θ:
Next, we need to find ∂x/∂r and ∂x/∂θ. Given x = r cosθ, we differentiate x with respect to r while treating θ as a constant:
∂x/∂r = d/dr [r cosθ] = cosθ
Similarly, we differentiate x with respect to θ while treating r as a constant:
∂x/∂θ = d/dθ [r cosθ] = -r sinθ
These derivatives show how x changes with respect to changes in r and θ, providing the necessary link to find how z changes with these variables.
3. Apply the Chain Rule to find ∂z/∂r and ∂z/∂θ:
Now, we use the chain rule to find ∂z/∂r and ∂z/∂θ. The chain rule states that:
∂z/∂r = (∂z/∂x) * (∂x/∂r)
Substituting the previously found derivatives, we get:
∂z/∂r = (2x / (x² + 1)) * (cosθ)
Since x = r cosθ, we can substitute x in the above equation:
∂z/∂r = (2(r cosθ) / ((r cosθ)² + 1)) * (cosθ) = (2r cos²θ) / (r² cos²θ + 1)
This gives us the partial derivative of z with respect to r.
Similarly, for ∂z/∂θ, the chain rule gives us:
∂z/∂θ = (∂z/∂x) * (∂x/∂θ)
Substituting the derivatives, we have:
∂z/∂θ = (2x / (x² + 1)) * (-r sinθ)
Again, substituting x = r cosθ, we get:
∂z/∂θ = (2(r cosθ) / ((r cosθ)² + 1)) * (-r sinθ) = (-2r² cosθ sinθ) / (r² cos²θ + 1)
Thus, we have found the partial derivative of z with respect to θ.
Conclusion for Problem 27
We have successfully found the partial derivatives ∂z/∂r and ∂z/∂θ using the chain rule. The key was to break down the problem into smaller steps: differentiating z with respect to x, differentiating x with respect to r and θ, and then combining these results. This methodical approach is essential for solving multivariable calculus problems effectively.
Problem 28: Calculating Partial Derivatives of u = rs² ln t
Problem Overview
In this problem, we are given a function u = rs² ln t, where r = x², s = 4y + 1, and t = xy³. The objective is to find the partial derivatives of u with respect to x ( ∂u/∂x ) and y ( ∂u/∂y ). This problem also requires a strategic application of the chain rule, but with an additional layer of complexity due to the multiple nested functions. We will proceed by differentiating u with respect to r, s, and t, and then differentiating r, s, and t with respect to x and y. Finally, we will combine these derivatives using the chain rule to find the desired partial derivatives.
Step-by-Step Solution
1. Differentiate u with respect to r, s, and t:
First, we find the partial derivatives of u with respect to r, s, and t. Given u = rs² ln t, we have:
∂u/∂r = d/dr [rs² ln t] = s² ln t
∂u/∂s = d/ds [rs² ln t] = 2rs ln t
∂u/∂t = d/dt [rs² ln t] = rs²/t
These derivatives show how u changes with respect to changes in r, s, and t, which are essential components for applying the chain rule.
2. Differentiate r, s, and t with respect to x and y:
Next, we need to find the partial derivatives of r, s, and t with respect to x and y. Given r = x², s = 4y + 1, and t = xy³, we have:
∂r/∂x = d/dx [x²] = 2x
∂r/∂y = d/dy [x²] = 0
∂s/∂x = d/dx [4y + 1] = 0
∂s/∂y = d/dy [4y + 1] = 4
∂t/∂x = d/dx [xy³] = y³
∂t/∂y = d/dy [xy³] = 3xy²
These derivatives illustrate how r, s, and t change with respect to changes in x and y, providing the necessary links for the chain rule.
3. Apply the Chain Rule to find ∂u/∂x and ∂u/∂y:
Now, we use the chain rule to find ∂u/∂x and ∂u/∂y. The chain rule for this problem states:
∂u/∂x = (∂u/∂r) * (∂r/∂x) + (∂u/∂s) * (∂s/∂x) + (∂u/∂t) * (∂t/∂x)
Substituting the previously found derivatives, we get:
∂u/∂x = (s² ln t) * (2x) + (2rs ln t) * (0) + (rs²/t) * (y³)
∂u/∂x = 2xs² ln t + (rs²y³) / t
Substituting r = x², s = 4y + 1, and t = xy³, we have:
∂u/∂x = 2x(4y + 1)² ln(xy³) + (x²(4y + 1)²y³) / (xy³)
∂u/∂x = 2x(4y + 1)² ln(xy³) + x(4y + 1)²
This gives us the partial derivative of u with respect to x.
Similarly, for ∂u/∂y, the chain rule gives us:
∂u/∂y = (∂u/∂r) * (∂r/∂y) + (∂u/∂s) * (∂s/∂y) + (∂u/∂t) * (∂t/∂y)
Substituting the derivatives, we have:
∂u/∂y = (s² ln t) * (0) + (2rs ln t) * (4) + (rs²/t) * (3xy²)
∂u/∂y = 8rs ln t + (3rs²xy²) / t
Substituting r = x², s = 4y + 1, and t = xy³, we get:
∂u/∂y = 8x²(4y + 1) ln(xy³) + (3x²(4y + 1)²xy²) / (xy³)
∂u/∂y = 8x²(4y + 1) ln(xy³) + 3x²(4y + 1)² / y
Thus, we have found the partial derivative of u with respect to y.
Conclusion for Problem 28
We have successfully found the partial derivatives ∂u/∂x and ∂u/∂y by carefully applying the chain rule. The process involved differentiating the function with respect to intermediate variables and then with respect to the final variables, followed by a combination of these derivatives. This problem highlights the importance of methodical application of the chain rule in complex multivariable functions.
Problem 29: Determining Partial Derivative of w = 4x² + 4y² + z²
Understanding the Problem Context
In this problem, we are given the function w = 4x² + 4y² + z², where x = ρ sinφ cosθ, y = ρ sinφ sinθ, and z = ρ cosφ. The goal is to find the partial derivative of w with respect to ρ ( ∂w/∂ρ ). This problem involves transforming from Cartesian coordinates to spherical coordinates and requires a thorough application of the chain rule. The key here is to first differentiate w with respect to x, y, and z, and then differentiate x, y, and z with respect to ρ. Finally, we combine these derivatives to obtain the partial derivative of w with respect to ρ.
Step-by-Step Solution
1. Differentiate w with respect to x, y, and z:
We begin by finding the partial derivatives of w with respect to x, y, and z. Given w = 4x² + 4y² + z², we have:
∂w/∂x = d/dx [4x² + 4y² + z²] = 8x
∂w/∂y = d/dy [4x² + 4y² + z²] = 8y
∂w/∂z = d/dz [4x² + 4y² + z²] = 2z
These derivatives show how w changes with respect to changes in x, y, and z. They are crucial components for applying the chain rule in the subsequent steps.
2. Differentiate x, y, and z with respect to ρ:
Next, we need to find the partial derivatives of x, y, and z with respect to ρ. Given x = ρ sinφ cosθ, y = ρ sinφ sinθ, and z = ρ cosφ, we differentiate each with respect to ρ:
∂x/∂ρ = d/dρ [ρ sinφ cosθ] = sinφ cosθ
∂y/∂ρ = d/dρ [ρ sinφ sinθ] = sinφ sinθ
∂z/∂ρ = d/dρ [ρ cosφ] = cosφ
These derivatives illustrate how x, y, and z change with respect to changes in ρ, providing the necessary links for the chain rule.
3. Apply the Chain Rule to find ∂w/∂ρ:
Now, we apply the chain rule to find ∂w/∂ρ. The chain rule for this problem is expressed as:
∂w/∂ρ = (∂w/∂x) * (∂x/∂ρ) + (∂w/∂y) * (∂y/∂ρ) + (∂w/∂z) * (∂z/∂ρ)
Substituting the previously found derivatives, we get:
∂w/∂ρ = (8x) * (sinφ cosθ) + (8y) * (sinφ sinθ) + (2z) * (cosφ)
Now, we substitute x = ρ sinφ cosθ, y = ρ sinφ sinθ, and z = ρ cosφ into the equation:
∂w/∂ρ = 8(ρ sinφ cosθ)(sinφ cosθ) + 8(ρ sinφ sinθ)(sinφ sinθ) + 2(ρ cosφ)(cosφ)
Simplifying the equation, we have:
∂w/∂ρ = 8ρ sin²φ cos²θ + 8ρ sin²φ sin²θ + 2ρ cos²φ
We can factor out 8ρ sin²φ from the first two terms:
∂w/∂ρ = 8ρ sin²φ (cos²θ + sin²θ) + 2ρ cos²φ
Since cos²θ + sin²θ = 1, the equation simplifies to:
∂w/∂ρ = 8ρ sin²φ + 2ρ cos²φ
This gives us the partial derivative of w with respect to ρ.
Conclusion for Problem 29
We have successfully found the partial derivative ∂w/∂ρ by methodically applying the chain rule. This involved differentiating w with respect to x, y, and z, then differentiating x, y, and z with respect to ρ, and finally combining these derivatives. This problem illustrates the importance of understanding coordinate transformations and the chain rule in multivariable calculus.
Conclusion
Through the detailed solutions of these problems, we have explored the application of partial derivatives in multivariable calculus. Each problem required a careful application of the chain rule, highlighting the importance of breaking down complex functions into simpler components. Whether dealing with logarithmic functions, nested functions, or coordinate transformations, a methodical approach is key to successfully computing partial derivatives. Mastering these techniques is essential for further studies in calculus and its applications in various fields such as physics, engineering, and economics.
