Question

For the curve \(\mathrm{y}=\left[\mathrm{a}^{3} /\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)\right]\), then length of tangent at \([\mathrm{a},(\mathrm{a} / 2)]\) is (a) \((\sqrt{5} / 2)|\mathrm{a}|\) (b) \(\mid\) a (c) \(\mid \mathrm{a} / 4\) (d) \(\mid \mathrm{a} / 2\)

Short answer

The length of the tangent at the given point is \(|a/2|\), which corresponds to option (d).

Step-by-step solution

Find the derivative of the curve

First, let's find the derivative of the curve with respect to x: \[y(a,x) = \frac{a^3}{a^2 + x^2}\] \[y'(x) = \frac{d(y(x))}{dx} = \frac{d}{dx} \left(\frac{a^3}{a^2 + x^2}\right)\] We can find the derivative by the Quotient rule: \[\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\] In our case, \(u = a^3\) and \(v = a^2 + x^2\), so: \[u' = \frac{d(a^3)}{dx} = 0\] \[v' = \frac{d(a^2 + x^2)}{dx} = 2x\] Now apply the Quotient Rule, \[y'(x) = \frac{(a^2 + x^2)(0) - a^3(2x)}{(a^2 + x^2)^2} = \frac{-2a^3x}{(a^2 + x^2)^2}\]

Evaluate the derivative at the given point

Now, we have to evaluate the derivative \(y'(x)\) at the given point \((a, a/2)\): \[y'(a) = \frac{-2a^3a}{(a^2 + a^2)^2} = \frac{-2a^4}{(2a^2)^2}\]

Find the equation of the tangent line

Now we will use the point-slope form to find the equation of the tangent line at the given point. Recall that the point-slope form of a line is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the given point. The slope of the tangent line at the point \([a, (a/2)]\) is equal to \(y'(a)\). So, we substitute the given point \((x_1 = a, y_1 = a/2)\) and the tangent slope \(y'(a) = -2a^4 / (2a^2)^2\) into the point-slope form. \[y - \frac{a}{2} = \frac{-2a^4}{(2a^2)^2}(x - a)\]

Solve for length of tangent

Since the length of the tangent is the distance from the tangent line to the x-axis, we calculate the distance from the given point \((a, a/2)\) to the tangent line: Let \(y = 0\) in the equation of the tangent line: \[0 - \frac{a}{2} = \frac{-2a^4}{(2a^2)^2}(x - a)\] \[ - \frac{a}{2} = \frac{-2a^4}{16a^4}(x - a)\] Simplifying the equation: \[ x = a - \frac{a}{2} = a\left(1 - \frac{1}{2}\right) = \frac{a}{2}\] Thus, the length of the tangent at the given point is \(|a/2|\), which corresponds to option (d).

Key concepts

Curve Analysis

Curve analysis involves examining the properties of a given curve to understand its behavior and characteristics at various points. To begin with, a curve can be described by a mathematical function, like the one provided in the exercise:

• \( y = \frac{a^3}{a^2 + x^2} \)

This function shows how the curve behaves as \(x\) changes. An important part of curve analysis is finding key features such as slopes, tangents, and any points of interest, like maxima or minima.
Understanding these aspects helps in applications such as predicting the path of moving objects or optimizing designs in engineering. For the problem, specific calculations involve evaluating the tangent at a point \([a, a/2]\), which represents a localized straight line that touches but doesn't intersect the curve.

Quotient Rule

The quotient rule is an essential tool in calculus for finding the derivative of ratios of two differentiable functions. In mathematical terms, when you have a function \(y = \frac{u}{v}\), the quotient rule states:

• \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} \]

This helps in determining the rate at which one variable changes with respect to another.
In our exercise, applying the quotient rule is crucial since the function \( y = \frac{a^3}{a^2 + x^2} \) involves a fraction. Here, \(u = a^3\) and \(v = a^2 + x^2\). The derivatives of these portions \(u' = 0\) and \(v' = 2x\) are used to apply this rule effectively. By following the rule, the derivative of the curve is found, which plays a vital role in determining the tangent line equation.

Derivatives

Derivatives represent a cornerstone concept in calculus, reflecting the rate of change of a function with respect to a variable. For our curve \(y = \frac{a^3}{a^2 + x^2}\), finding the derivative \(y'(x)\) tells us the slope of the curve at any point.

• \[ y'(x) = \frac{-2a^3x}{(a^2 + x^2)^2} \]

The derivative function provides critical information needed to understand the behavior and shape of the graph.
In the context of finding the tangent line, the derivative at the point of tangency gives the slope of the tangent.
Thus, by evaluating \(y'(x)\) at specific \(x\)-values, one can calculate how steep or flat the curve is at those points. This insight is invaluable in physics, engineering, and economics for optimization and modeling situations.

Tangent Line Equations

Tangent line equations describe lines that touch but do not cross a curve at a point. The slope of this line is the same as the slope of the curve at the touchpoint, provided by the derivative.
For a line tangent to the curve \(y = \frac{a^3}{a^2 + x^2}\) at \([a, a/2]\), we use the point-slope form:

• \[ y - y_1 = m(x - x_1) \]

where \(m\) is the derivative at \(x_1 = a\), and \(y_1\) is \(a/2\). After calculating the derivative at this point, we substitute it into the equation:

• \[ y - \frac{a}{2} = \frac{-2a^4}{(2a^2)^2}(x - a) \]

This equation captures the linear representation of the tangent to the curve. Such equations are crucial in determining the length, intersection points, and area of regions bounded by curves and their tangents in real-world applications, like designing roads or analyzing wave patterns.