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Chapter 7: Problem 37

Let \(f(x, y)=3 x^{2}+2 x y+5 y,\) as in Example 5. Show that $$f(1+h, 4)-f(1,4)=14 h+3 h^{2}$$ Thus, the error in approximating \(f(1+h, 4)-f(1,4)\) by \(14 h\) is \(3 h^{2},\) (If \(h=.01,\) for instance, the error is only .0003 .)

Short Answer

Expert verified
The error in approximating \(f(1+h, 4) - f(1,4)\) by \(14 h\) is \(3 h^2\).

Step by step solution

01

- Substitute into the function

First, substitute \(x = 1+h\) and \(y = 4\) into the function \(f(x,y) = 3x^2 + 2xy + 5y\). This gives: \[ f(1+h, 4) = 3(1+h)^2 + 2(1+h)(4) + 5(4) \]
02

- Expand the expression

Now, expand the terms within the function: \[ f(1+h, 4) = 3(1 + 2h + h^2) + 2(4 + 4h) + 20 \]
03

- Simplify the expression

Simplify the expression by distributing and combining like terms: \[ f(1+h, 4) = 3 + 6h + 3h^2 + 8 + 8h + 20 \] \[ f(1+h, 4) = 31 + 14h + 3h^2 \]
04

- Evaluate the function at (1,4)

Now, evaluate the function \(f\) at the point (1,4): \[ f(1, 4) = 3(1)^2 + 2(1)(4) + 5(4) \] \[ f(1, 4) = 3 + 8 + 20 \] \[ f(1, 4) = 31 \]
05

- Calculate the difference

Subtract \(f(1,4)\) from \(f(1+h, 4)\): \[ f(1+h, 4) - f(1,4) = (31 + 14h + 3h^2) - 31 \] \[ f(1+h, 4) - f(1, 4) = 14h + 3h^2 \]
06

Conclusion - Error in approximation

Therefore, the error in approximating \(f(1+h, 4) - f(1,4)\) by \(14h\) is \(3h^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
Multivariable calculus involves functions with more than one variable. In our exercise, we have the function \(f(x, y) = 3x^2 + 2xy + 5y\). Here, \(x\) and \(y\) are independent variables. Multiple variables allow us to analyze how changes in one variable affect the function when the other variable is constant. This is crucial in optimizing functions and studying dynamic systems. Let's substitute and expand \(f(1+h, 4)\) to understand this better with our example.
Function Expansion
Function expansion helps to estimate a function at a certain point by expressing it in terms of simpler components. By expanding \(f(1+h, 4)\), we break it down into smaller, understandable terms. After substituting \(x = 1+h\) and \(y = 4\), we get: \[ f(1+h, 4) = 3(1+h)^2 + 2(1+h)4 + 5(4) \]Expanding this: \[ 3(1 + 2h + h^2) + 2(4 + 4h) + 20 \] Combining like terms, we simplify: \[ 3 + 6h + 3h^2 + 8 + 8h + 20 = 31 + 14h + 3h^2 \] This step-by-step breakdown helps in understanding how each term affects the function value. Evaluating the original function at points helps to establish a baseline \(f(1, 4) = 31\).
Approximation Error Analysis
In real-life applications, we often need to approximate functions. It's important to also understand the error associated with these approximations. The expanded function \(31 + 14h + 3h^2\) helps to see how close the linear approximation \(14h\) is. The difference, \(3h^2\), represents the error. For small values of \(h\), such as \(0.01\), the error \(0.0003\) is minimal but crucial for precision. Checking these errors ensures the reliability of approximations. Thus, we've shown the accuracy of \(f(1+h, 4) - f(1, 4)\) and the bounds of approximation in calculus problems.

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