Question

$$\text { If } u=x^{2} y^{3} z \text { and } x=\sin (s+t), y=\cos (s+t), z=e^{s t}, \text { find } \partial u / \partial s \text { and } \partial u / \partial t$$.

Short answer

\( \partial u / \partial s \) and \( \partial u / \partial t \) can be found using the chain rule and substituting the given functions.

Step-by-step solution

Write down the given functions

The functions given are: \(u = x^2 y^3 z\)\(x = \sin(s+t)\)\(y = \cos(s+t)\) \(z = e^{s t}\).

Apply the chain rule for partial derivatives

To find \( \partial u / \partial s \) and \( \partial u / \partial t \), use the chain rule: \[ \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial s} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial s} \] \[ \frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial t} \].

Compute the partial derivatives of \(u\) with respect to \(x\), \(y\), and \(z\)

Calculate the partial derivatives: \( \frac{\partial u}{\partial x} = 2x y^3 z\) \( \frac{\partial u}{\partial y} = 3x^2 y^2 z\) \( \frac{\partial u}{\partial z} = x^2 y^3\).

Compute the partial derivatives of \(x\), \(y\), and \(z\) with respect to \(s\) and \(t\)

Calculate: \( \frac{\partial x}{\partial s} = \cos(s+t)\) \( \frac{\partial x}{\partial t} = \cos(s+t)\) \( \frac{\partial y}{\partial s} = -\sin(s+t)\) \( \frac{\partial y}{\partial t} = -\sin(s+t)\) \( \frac{\partial z}{\partial s} = t e^{st}\) \( \frac{\partial z}{\partial t} = s e^{st}\).

Substitute and simplify for \( \partial u / \partial s \)

Using the chain rule, substitute all partial derivatives into: \[ \frac{\partial u}{\partial s} = 2x y^3 z \cos(s+t) + 3x^2 y^2 z (-\sin(s+t)) + x^2 y^3 (t e^{st}) \] Substituting \(x = \sin(s+t)\), \(y = \cos(s+t)\), and \(z = e^{st}\), we get: \[ \frac{\partial u}{\partial s} = 2 \sin^1(s+t) \cos^3(s+t) e^{st} \cos(s+t) - 3 \sin^2(s+t) \cos^2(s+t) e^{st} \-\sin(s+t) + \sin^2(s+t) \cos^3(s+t)t e^{st} \].

Substitute and simplify for \( \partial u / \partial t \)

Similarly, use the chain rule for: \[ \frac{\partial u}{\partial t} = 2x y^3 z \cos(s+t) + 3x^2 y^2 z (-\sin(s+t)) + x^2 y^3 (s e^{st}) \] Substituting \(x = \sin(s+t)\), \(y = \cos(s+t)\), and \(z = e^{st}\), we get: \[ \frac{\partial u}{\partial t} = 2 \sin^1(s+t) \cos^3(s+t) e^{st} \cos(s+t) - 3 \sin^2(s+t) \cos^2(s+t) e^{st} \-\sin(s+t) + \sin^2(s+t) \cos^3(s+t)s e^{st} \].

Key concepts

the importance of differentiation

Differentiation is a core concept in calculus that's all about finding the rate at which a function changes. For functions of a single variable, differentiation tells us how the function's value changes as its input changes.

In the context of multivariable functions, differentiation involves partial derivatives, where we examine how the function changes in response to changes in each of its variables independently.

Let's dive into an example. Consider a function **u = x^2 y^3 z**:

• To find **∂u/∂x**, we treat **y** and **z** as constants and differentiate with respect to **x**:
  \( \frac{\text{∂}u}{\text{∂}x} = 2xy^3z \).
  
• For **∂u/∂y**, **x** and **z** are constants:
  \( \frac{\text{∂}u}{\text{∂}y} = 3x^2y^2z \).
  
• And for **∂u/∂z**, **x** and **y** are held constant:
  \( \frac{\text{∂}u}{\text{∂}z} = x^2y^3 \).
  

Through these partial derivatives, we can dissect the behavior of **u** in various dimensions. This approach is fundamental in optimization problems, where we need to find maximum or minimum values of multivariable functions. It's also essential in understanding how complex systems respond to changes in their parameters.

Ultimately, mastering differentiation in the multivariable context enables you to solve a wide range of practical and theoretical problems.