https://file.aiquickdraw.com/mathbot/file/172840412601115dm54pb.jpg

To solve this problem, we need to show that:

\[
\frac{\partial u}{\partial t} = a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y}
\]

Given that:
- \( u = f(r, s) \)
- \( r = x + at \)
- \( s = y + bt \)

and \( x, y, t \) are independent variables.

### Step-by-Step Solution

#### 1. Expression for Total Derivative of \( u \)
Since \( u = f(r, s) \), \( u \) depends on \( r \) and \( s \), and \( r \) and \( s \) in turn depend on \( x, y, t \). Therefore, the chain rule gives us:

\[
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial t} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial t}
\]

#### 2. Derivatives of \( r \) and \( s \) with Respect to \( t \)
From the given expressions:
- \( r = x + at \) implies \( \frac{\partial r}{\partial t} = a \)
- \( s = y + bt \) implies \( \frac{\partial s}{\partial t} = b \)

Thus, the partial derivative of \( u \) with respect to \( t \) becomes:

\[
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial r} \cdot a + \frac{\partial u}{\partial s} \cdot b
\]

#### 3. Chain Rule for \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \)
Next, we need expressions for \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \). Using the chain rule again:

\[
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial x}
\]

Since \( r = x + at \) and \( s = y + bt \), we have:
- \( \frac{\partial r}{\partial x} = 1 \) and \( \frac{\partial s}{\partial x} = 0 \)

Thus:
\[
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}
\]

Similarly, for \( \frac{\partial u}{\partial y} \):

\[
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial y}
\]

Here:
- \( \frac{\partial r}{\partial y} = 0 \) and \( \frac{\partial s}{\partial y} = 1 \)

Thus:
\[
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial s}
\]

#### 4. Final Expression
Now we substitute these results into the original equation. We already know:

\[
\frac{\partial u}{\partial t} = a \frac{\partial u}{\partial r} + b \frac{\partial u}{\partial s}
\]

But from the chain rule results:

- \( \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \)
- \( \frac{\partial u}{\partial y} = \frac{\partial u}{\partial s} \)

Thus, we can rewrite the equation as:

\[
\frac{\partial u}{\partial t} = a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y}
\]

This proves the required result.

### Do you have any questions or would you like a detailed explanation on any step?

---

Here are five related questions:
1. How would this change if \( r \) and \( s \) were functions of \( x, y, z, t \)?
2. Can you derive the expression for \( \frac{\partial u}{\partial x} \) when \( r = x^2 + at \)?
3. What happens if \( u \) is a function of three variables, say \( r, s, q \), where \( q = z + ct \)?
4. How do we apply the chain rule when dealing with functions of more than two variables?
5. How do implicit functions affect the computation of partial derivatives?

**Tip:** When working with chain rule in multivariable calculus, it's important to carefully track how each variable depends on the others and apply the appropriate derivative rules.

If u = f(r, s), r = x + at, s = y + bt and x, y, t are independent variables, then show that ∂u/∂t = a ∂u/∂x + b ∂u/∂y.

This page explains how to prove the partial derivative relationship ∂u/∂t = a ∂u/∂x + b ∂u/∂y for the function u = f(r, s), where r = x + at and s = y + bt. Using the chain rule for partial derivatives, follow a step-by-step solution suitable for university-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation