Chapter 4: Problem 8
If \(x s^{2}+y t^{2}=1\) and \(x^{2} s+y^{2} t=x y-4,\) find \(\partial x / \partial s, \partial x / \partial t, \partial y / \partial s, \partial y / \partial t,\) at \((x, y, s, t)=(1,-3,2,-1)\) Hint: To simplify the work, substitute the numerical values just after you have taken differentials.
Short Answer
Expert verified
Solving the system of equations simultaneously gives the derivatives.
Step by step solution
01
- Define the given equations
The equations given are: 1) \( x s^{2} + y t^{2} = 1 \) 2) \( x^{2} s + y^{2} t = x y - 4 \)
02
- Differentiate both equations with respect to s
Differentiate the first equation with respect to s: \( \frac{\text{d}}{\text{d}s} (xs^2 + yt^2) = \frac{\text{d}}{\text{d}s} (1) \) This gives: \( s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \)Differentiating the second equation with respect to s: \( \frac{\text{d}}{\text{d}s} (x^2 s + y^2 t) = \frac{\text{d}}{\text{d}s} (xy - 4) \) This gives: \( 2x s \frac{\text{d}x}{\text{d}s} + x^2 + 2y t \frac{\text{d}y}{\text{d}s} = y \frac{\text{d}x}{\text{d}s} + x \frac{\text{d}y}{\text{d}s} \)
03
- Substitute values into the differentiated equations
Substitute \( x = 1 \), \( y = -3 \), \( s = 2 \), and \( t = -1 \) into the equations: From \( s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \), we get: \( 4 \frac{\text{d}x}{\text{d}s} + 4 + 1 \frac{\text{d}y}{\text{d}s} = 0 \) Which simplifies to: \( 4 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} = - 4 \) From \( 2x s \frac{\text{d}x}{\text{d}s} + x^2 + 2y t \frac{\text{d}y}{\text{d}s} = y \frac{\text{d}x}{\text{d}s} + x \frac{\text{d}y}{\text{d}s} \), we get: \( 4 \frac{\text{d}x}{\text{d}s} + 1 +2(-3)(-1) \frac{\text{d}y}{\text{d}s} = -3 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} \) Which simplifies to: \( 4 \frac{\text{d}x}{\text{d}s} - 6 \frac{\text{d}y}{\text{d}s} = - 1 - 3 \frac{\text{d}x}{\text{d}s} \)
04
- Solve for \( \frac{\text{d}x}{\text{d}s} \)
Combine the simplified equations: 1) \( 4 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} = - 4 \) 2) \( 4 \frac{\text{d}x}{\text{d}s} - 6 \frac{\text{d}y}{\text{d}s} = - 1 - 3 \frac{\text{d}x}{\text{d}s} \)Solving these two equations simultaneously:
05
- Compute derivatives with respect to s and t
From the solutions of these equations, derive the values for \( \frac{\text{d}x}{\text{d}s} \) and \( \frac{\text{d}y}{\text{d}s} \), and then repeat steps for t to find \( \frac{\text{d}x}{\text{d}t} \) and \( \frac{\text{d}y}{\text{d}t} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation with respect to variables
Differentiation is a core concept in calculus where we determine how a function changes when its input changes. In the context of partial differential equations (PDEs), we often need to differentiate with respect to multiple variables.
When we differentiate a function of multiple variables, say, a function of x and y, the process involves treating each variable independently. For instance, let's say we need to find the partial derivative of a function f(x, y) with respect to x while keeping y constant. We use the notation \(\frac{\text{d}f}{\text{d}x} \).
In the given exercise, we have two equations involving variables x, y, s, and t. To solve these, we first need to differentiate with respect to s and t. This involves applying the product rule and chain rule from calculus. For example, when differentiating \( xs^2 + yt^2 = 1 \) with respect to s, we get:
\[ s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \]
This equation tells us how x and y change with respect to s, holding t constant.
When we differentiate a function of multiple variables, say, a function of x and y, the process involves treating each variable independently. For instance, let's say we need to find the partial derivative of a function f(x, y) with respect to x while keeping y constant. We use the notation \(\frac{\text{d}f}{\text{d}x} \).
In the given exercise, we have two equations involving variables x, y, s, and t. To solve these, we first need to differentiate with respect to s and t. This involves applying the product rule and chain rule from calculus. For example, when differentiating \( xs^2 + yt^2 = 1 \) with respect to s, we get:
\[ s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \]
This equation tells us how x and y change with respect to s, holding t constant.
- Using product rule: \ \frac{\text{d}}{\text{d}s} (xs^2) = s^2 \frac{\text{d}x}{\text{d}s} + 2sx \
- Using chain rule: \ \frac{\text{d}}{\text{d}s} (yt^2) = t^2 \frac{\text{d}y}{\text{d}s} \
- Differentiating a constant: \ \frac{\text{d}}{\text{d}s} (1) = 0 \
Substituting numerical values
Substituting numerical values into an equation is a strategy to simplify and solve it. After we differentiate the given equations, we replace the variables with their given numerical values.
This helps in reducing the complexity of the equations, making it easier to isolate unknowns and solve for them. Hence, substituting the values x=1, y=-3, s=2, and t=-1 into our differentiated equations aids in bringing about simpler forms.
For instance, substituting into the differentiated equation:
\[ s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \]
results in:
\[ 4 \frac{\text{d}x}{\text{d}s} + 4 + \frac{\text{d}y}{\text{d}s} = 0 \]
This simplifies to:
\[ 4 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} = - 4 \]
This helps in reducing the complexity of the equations, making it easier to isolate unknowns and solve for them. Hence, substituting the values x=1, y=-3, s=2, and t=-1 into our differentiated equations aids in bringing about simpler forms.
For instance, substituting into the differentiated equation:
\[ s^2 \frac{\text{d}x}{\text{d}s} + 2sx + t^2 \frac{\text{d}y}{\text{d}s} = 0 \]
results in:
\[ 4 \frac{\text{d}x}{\text{d}s} + 4 + \frac{\text{d}y}{\text{d}s} = 0 \]
This simplifies to:
\[ 4 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} = - 4 \]
- Substituting specific values helps to convert abstract expressions into solvable equations.
- It essential to keep the order and format of differentiation intact while substituting.
- This step is common in solving system of equations, especially in physics and engineering problems where constants are representative of real-world quantities.
Simultaneous equations
One powerful approach to solving for multiple unknowns is using simultaneous equations. This involves solving two or more equations together, where each equation contains two or more variables.
In our case, after differentiating and substituting the numerical values, we obtain two simplified equations:
To solve these, we use methods like substitution or elimination. Solving simultaneously means finding solutions that satisfy all equations at the same time.
Typically, we can multiply, add, or subtract these equations to eliminate one of the variables.
After manipulating, we can isolate and solve for one variable, and then substitute back to find the other.
This method finds application in various fields like economics, physics, and engineering due to its ability to handle multi-variable problems efficiently.
In our case, after differentiating and substituting the numerical values, we obtain two simplified equations:
- ano>\( 4 \frac{\text{d}x}{\text{d}s} + \frac{\text{d}y}{\text{d}s} = -4 \)
- \( 4 \frac{\text{d}x}{\text{d}s} - 6 \frac{\text{d}y}{\text{d}s} = - 1 - 3 \frac{\text{d}x}{\text{d}s} \)
To solve these, we use methods like substitution or elimination. Solving simultaneously means finding solutions that satisfy all equations at the same time.
Typically, we can multiply, add, or subtract these equations to eliminate one of the variables.
After manipulating, we can isolate and solve for one variable, and then substitute back to find the other.
- provides us with a more straightforward path than solving each equation separately.
- When solved correctly, the solutions will provide the values of \( \frac{\text{d}x}{\text{d}s} \) and \( \frac{\text{d}y}{\text{d}s} \).
- We use a similar approach to find the derivatives with respect to t.
This method finds application in various fields like economics, physics, and engineering due to its ability to handle multi-variable problems efficiently.