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(a) The curve \(y=|x| / \sqrt{2-x^{2}}\) is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point \((1,1)\) . (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

Short Answer

Expert verified
The tangent line equation at \((1,1)\) is \( y = 2x - 1 \).

Step by step solution

01

Differentiate the Function

The given function is \( y = \frac{|x|}{\sqrt{2-x^2}} \). We need to find its derivative, particularly at \( x = 1 \). Since the function involves \(|x|\), we need to consider the positive definition for \(x > 0\), which simplifies to \( y = \frac{x}{\sqrt{2-x^2}} \). Use the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \) for differentiation, where \( u = x \) and \( v = \sqrt{2-x^2} \).
02

Apply the Quotient Rule

First, compute \( u' = 1 \) and \( v' = \frac{-x}{\sqrt{2-x^2}} \) using the chain rule on \( v \). Now apply the quotient rule to find \( y' = \frac{1 \cdot \sqrt{2-x^2} - x \cdot \left(\frac{-x}{\sqrt{2-x^2}}\right)}{(\sqrt{2-x^2})^2} \). This simplifies to \( y' = \frac{2}{(2-x^2)^{3/2}} \).
03

Evaluate the Derivative at x = 1

Substitute \( x = 1 \) into the derivative \( y' = \frac{2}{(2-x^2)^{3/2}} \), resulting in \( y' = \frac{2}{(2-1^2)^{3/2}} = 2 \). This indicates the slope of the tangent line at \( (1, 1) \) is 2.
04

Write the Equation of the Tangent Line

Using the point-slope form \( y - y_1 = m(x - x_1) \) where \((x_1, y_1) = (1, 1)\) and \(m = 2\), the tangent line equation becomes \( y - 1 = 2(x - 1) \). Simplifying, we get \( y = 2x - 1 \).
05

Graph the Curve and Tangent Line

To visualize the solution, plot the curve \( y = \frac{|x|}{\sqrt{2-x^2}} \) and the tangent line \( y = 2x - 1 \) on the same graph. The tangent should intersect the curve precisely at the point \((1, 1)\).

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Key Concepts

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

Derivative
The concept of a derivative is essential in calculus as it represents the rate of change or the slope of a curve at a given point. In other words, it tells us how a function is changing at any particular point. When calculating derivatives, particularly for finding tangent lines, it helps us compute the slope of the curve.

In our bullet-nose curve problem, we were asked to find the tangent's slope at the point \(1,1\). The function given is \(y = \frac{|x|}{\sqrt{2-x^2}}\). To find the derivative, we simplify the absolute value to \(x\) since \(x > 0\). We then use rules of differentiation to find how the y-value changes as x changes. Understanding this change gives us a clear idea of the curve's slope at any point, which is crucial for identifying the equation of the tangent line.
Quotient Rule
The quotient rule is a technique for differentiating functions that are expressed as the ratio of two other functions. When we have a function in the form of \(\frac{u}{v}\), the derivative can be found using:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]The essence of this rule is in knowing how each part of the ratio contributes to the rate of change of the entire function.

For the bullet-nose curve \(y = \frac{x}{\sqrt{2-x^2}}\), we identify \(u = x\) and \(v = \sqrt{2-x^2}\). Using the quotient rule here, we first differentiate \(u\) and \(v\) separately before applying the formula. This step is necessary to find the derivative correctly. The quotient rule helps simplify the overall complexity when faced with divisions in functions, making it easier to isolate individual changing rates.
Chain Rule
The chain rule is another fundamental concept used in calculus for differentiating composites of functions. It is essential when the function in question is made up of other functions working within each other, like nesting dolls. The chain rule states:\[( f(g(x)) )' = f'(g(x)) \cdot g'(x)\]In simple terms, you "chain" the derivatives of the inner function and the outer function.

In our exercise, the chain rule is used when differentiating \(v = \sqrt{2-x^2}\). The inner function \(g(x)\) is \(2-x^2\), and we can see it nested inside the square root. Differentiating the outer function (the square root) gives us part of the derivative, while differentiating \(g(x)\) gives us the rest. The chain rule helps us navigate through this nesting smoothly by linking the derivative routes, ensuring we understand the rate at which each part of the function changes. By mastering the chain rule, you are equipped to tackle more advanced calculus problems involving layered functions.

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Most popular questions from this chapter

One side of a right triangle is known to be 20 \(\mathrm{cm}\) long and the opposite angle is measured as \(30^{\circ},\) with a possible error of \(\pm 1^{\circ} .\) (a) Use differentials to estimate the error in computing the length of the hypotenuse. (b) What is the percentage error?

Show, using implicit differentiation, that any tangent line at a point \(P\) to a circle with center \(O\) is perpendicular to the radius \(O P .\)

(a) The van der Waals equation for moles of a gas is $$\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T$$ where \(P\) is the pressure, \(V\) is the volume, and \(T\) is the temperature of the gas. The constant \(R\) is the universal gas constant and \(a\) and \(b\) are positive constants that are characteristic of a particular gas. If \(T\) remains constant, use implicit differentiation to find \(d V / d P .\) (b) Find the rate of change of volume with respect to pressure of 1 mole of carbon dioxide at a volume of \(V=10 \mathrm{L}\) and a pressure of \(P=2.5\) atm. Use \(a=3.592 \mathrm{L}^{2}-\mathrm{atm} / \mathrm{mole}^{2}\) and \(b=0.04267 \mathrm{L} / \mathrm{mole}\)

If a current \(I\) passes through a resistor with resistance \(R\) Ohm's Law states that the voltage drop is \(V=R I\) . If \(V\) is constant and \(R\) is measured with a certain error, use differ- entials to show that the relative error in calculating \(I\) is approximately the same (in magnitude) as the relative error in \(R .\)

Find the parabola with equation \(y=a x^{2}+b x\) whose tangent line at \((1,1)\) has equation \(y=3 x-2\)

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