Question Statement

Find the approximate value of $\sqrt{37}$ using differentials.

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Background and Explanation

These problems use linear approximation (differential approximation) to estimate function values near a known point using the tangent line. The method requires the formula $f(x+\Delta x)\approx f(x)+f^{\prime }(x)\Delta x$, where $x$ is chosen as a convenient value where the function and its derivative are easily calculated.

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Solution

Q30: Approximating $\sqrt{37}$

We express $\sqrt{37}$ as a small increment from a perfect square: $\sqrt{37}=(37{)}^{1/2}=(36+1{)}^{1/2}$

Choose $x=36$ (since $\sqrt{36}=6$ is known exactly) and $\Delta x=dx=1$.

Let $f(x)=\sqrt{x}$.

Differentiate with respect to $x$: $\begin{array}{rl}{f}^{\prime }(x)& =\frac{1}{2}{x}^{\frac{1}{2}-1}\\ & =\frac{1}{2}{x}^{-\frac{1}{2}}\\ & =\frac{1}{2\sqrt{x}}\end{array}$

Using the linear approximation formula: $f(x+\Delta x)\approx f(x)+f^{\prime }(x)\cdot dx$

This gives us: $\sqrt{x+\Delta x}\approx \sqrt{x}+\frac{1}{2\sqrt{x}}\cdot dx$

Substituting $x=36$ and $dx=1$: $\begin{array}{rl}\sqrt{36+1}& \approx \sqrt{36}+\frac{1}{2\sqrt{36}}(1)\\ \sqrt{37}& \approx 6+\frac{1}{2(6)}\\ & =6+\frac{1}{12}\\ & =6+0.0833\\ & =6.0833\end{array}$

Hence: $\sqrt{37}\approx 6.0833$

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Key Formulas or Methods Used

• Linear Approximation Formula: $f(x+\Delta x)\approx f(x)+f^{\prime }(x)\Delta x$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$ (applied to $\sqrt{x}=x^{1/2}$)
• Derivative of Sine: $\frac{d}{dx}(\sin x)=\cos x$
• Degree to Radian Conversion: $1^{\circ }=\frac{\pi }{180}$ radians $\approx 0.01745$ rad

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Summary of Steps

For $\sqrt{37}$:

1. Express as $\sqrt{36+1}$, choosing $x=36$ and $\Delta x=1$
2. Define $f(x)=\sqrt{x}$ and compute $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$
3. Apply formula: $\sqrt{37}\approx 6+\frac{1}{12}\approx 6.0833$