Question

Use differentials to show that, for large $n$ and small $a,\sqrt{n+a}-\sqrt{n}\cong \frac{a}{2\sqrt{n}}$ Find the approximate value of $\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}.$

Short answer

The approximate value is $2.5\times {10}^{-13}$

Step-by-step solution

- Understand the Problem

We need to show that for large values of $n$ and small values of $a$, $\sqrt{n+a}-\sqrt{n}$ is approximately $\frac{a}{2\sqrt{n}}$.

- Use Differential Approximation

For small changes in $a$, the change in the function $f(x)=\sqrt{x}$ can be approximated using differentials. The differential of $f(x)$ is given by $df=f^{\prime }(x)dx$.

- Find the Derivative

Calculate the derivative of $f(x)=\sqrt{x}$: $$f^{\prime }(x)=\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$$

- Apply the Differential

For a small change $dx=a$ in $x$, we can write: $$df=\frac{1}{2\sqrt{x}}dx$$ Since $x=n$ and $dx=a$, we have $$\sqrt{n+a}-\sqrt{n}\approx \frac{1}{2\sqrt{n}}a$$

- Apply the Formula

We have established that $\sqrt{n+a}-\sqrt{n}\approx \frac{a}{2\sqrt{n}}$. Now use this to find the approximate value of $\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}$.

- Plug in the Values

Here, $n={10}^{26}$ and $a=5$. Substitute these values into the approximation formula: $$\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}\approx \frac{5}{2\sqrt{{10}^{26}}}$$

- Simplify the Result

Simplify the expression: $$\frac{5}{2\sqrt{{10}^{26}}}=\frac{5}{2{10}^{13}}=\frac{5}{2\times {10}^{13}}=\frac{5}{2\times {10}^{13}}=2.5\times {10}^{-13}$$

Key concepts

Differential Approximation

Differential approximation is a method used in calculus to estimate the change in a function's value given a small change in its input. This method relies on using differentials, which are small increments of variables.
For example, if we have a function $f(x)$, its differential $df$ can be represented as:
$df=f^{\prime }(x)\cdot dx$
Here, $f^{\prime }(x)$ is the derivative of the function, and $dx$ is the small change in the input variable $x$.
In the given problem, we are approximating the change in the square root function, $\sqrt{n+a}-\sqrt{n}$, for large $n$ and small $a$. By using differentials, we transform the problem of finding an exact change into a simpler calculation involving the derivative and the small incremental change.

Derivative of Functions

The derivative of a function provides the rate at which the function's value changes with respect to the change in the input value. For the function $f(x)=\sqrt{x}$, its derivative $f^{\prime }(x)$ is calculated as follows:
$$f^{\prime }(x)=\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$$
This tells us how the value of $\sqrt{x}$ changes as $x$ changes. In simpler terms, the derivative shows the slope of the tangent line to the function at any point $x$.
Using this derivative in differential approximation, we can find the approximate change in the function's value for any small change in $x$. In our problem, this means for a small increment $a$ added to a large number $n$, the approximate change in $\sqrt{x}$ can be calculated easily.

Mathematical Approximation Methods

Mathematical approximation methods play a crucial role in simplifying complex problems. By replacing exact values and functions with approximate equivalents, we can perform easier calculations that are often good enough for most practical purposes.
One common method is using differentials, as seen in the problem where we approximate $\sqrt{n+a}$ instead of calculating it directly.

To apply these methods correctly, it's important to keep the following in mind:

• Ensure the change in input ($a$) is small compared to the original value ($n$).
• Use the correct derivative for the function involved.
• Clearly state and simplify the final result to check its validity.

With these steps, the approximation $\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}\approx \frac{5}{2\sqrt{{10}^{26}}}$ yields a simple result of $2.5\times {10}^{-13}$, demonstrating the power of these methods in calculus.