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Chapter 27: Problem 39

Find the slope of a line tangent to the curve of \(y=x / \tan ^{-1} x\) at \(x=0.80 .\) Verify the result by using the numerical derivative feature of a calculator.

Short Answer

Expert verified
The slope of the tangent line at \( x = 0.80 \) is approximately 0.4104.

Step by step solution

01

Differentiate the given function

The function given is \( y = \frac{x}{\tan^{-1}x} \). First, we need to find its derivative with respect to \( x \) in order to determine the slope of the tangent line. Apply the quotient rule: if \( u(x) = x \) and \( v(x) = \tan^{-1}x \), then the derivative \( y' \) is given by: \[ y' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} = \frac{\tan^{-1}x \cdot 1 - x \cdot \frac{1}{1+x^2}}{(\tan^{-1}x)^2} \] Simplifying gives: \[ y' = \frac{\tan^{-1}x - \frac{x}{1+x^2}}{(\tan^{-1}x)^2} \].
02

Substitute the specific value of x into the derivative

Now that we have the derivative, substitute \( x = 0.80 \) into the expression to find the slope of the tangent line at that point. \[ y'(0.80) = \frac{\tan^{-1}0.80 - \frac{0.80}{1+0.80^2}}{(\tan^{-1}0.80)^2} \] Calculate \( \tan^{-1}0.80 \) and substitute the value into the equation. Using a calculator, \( \tan^{-1}0.80 \approx 0.6747 \). Therefore: \[ y'(0.80) = \frac{0.6747 - \frac{0.80}{1.64}}{(0.6747)^2} \] \[ y'(0.80) \approx \frac{0.6747 - 0.4878}{0.4552} \], which simplifies to approximately \( 0.4104 \).
03

Verify using a calculator's numerical derivative feature

Use a calculator that can compute numerical derivatives to confirm the result. Enter the function \( y = \frac{x}{\tan^{-1}x} \) and use the derivative feature to differentiate at \( x = 0.80 \). The calculator should provide a numerical value for the derivative close to the analytical result we calculated, \( y'(0.80) \approx 0.4104 \).
04

Conclusion: Interpret the Result

The slope of the tangent line to the curve \( y = \frac{x}{\tan^{-1}x} \) at \( x = 0.80 \) is approximately 0.4104. This means that at \( x = 0.80 \), the curve is inclining upwards with this slope.

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Key Concepts

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

Tangent Line
A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point, without crossing it. This line represents the instantaneous direction that the curve is heading at that point.
  • Imagine driving along a winding road; the tangent line is like a brief snapshot of the road at your exact location.
  • The slope of the tangent line effectively tells you how steep the road is right where you are.
Understanding the tangent line gives us critical insight into how functions behave at specific points. In calculus, determining the tangent line involves finding its slope, which is done using derivatives.
Slope
The slope of a tangent line, or any line, indicates how steep the line is. In more technical terms, the slope is the "rate of change" of the line, and for a tangent line, this rate of change is specific to a point on the curve.
  • If you鈥檝e ever skied down a hill, the slope gives you an idea of how gentle or how steep your path will be.
  • In the context of the given problem, the slope at the point where \( x = 0.80 \) is approximately \( 0.4104 \), indicating a gentle upward incline at this spot on the curve.
Knowing the slope helps in predicting how a function behaves in the immediate neighborhood of a point.
Derivative
The derivative of a function is a fundamental concept in calculus that describes the rate at which a function's output changes as its input changes. In simpler words, it tells you how fast or slow a function is climbing or descending.
  • In our problem, we differentiated the function \( y = \frac{x}{\tan^{-1}x} \) to find the slope of the tangent.
  • Calculating the derivative is like peering into the crystal ball of the function, showing how it tweaks and turns.
To solve our problem, we used the 'quotient rule' to find the derivative of the function in question, which brings us to our next concept.
Quotient Rule
The quotient rule is a method used in calculus for finding the derivative of a ratio of two functions. When you have a function that is one function divided by another, you use the quotient rule to differentiate it.
  • Think of the quotient rule as a recipe that involves combining the derivatives of the top and bottom of a fraction to find the slope of its graph.
  • The formula for the quotient rule is: \(\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\). Here, \(u(x)\) and \(v(x)\) are the numerator and the denominator functions respectively.
In our example, it helps us find how the function \( y = \frac{x}{\tan^{-1}x} \) changes, giving us a tool to decipher the curve's subtleties at any point, especially at \( x = 0.80 \).

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Most popular questions from this chapter

Solve the given problems by finding the appropriate derivative. Assuming that force is proportional to acceleration, show that a particle moving along the \(x\) -axis, so that its displacement \(x=a e^{k t}+b e^{-k t},\) has a force acting on it which is proportional to its displacement.

Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0} x \cot x$$

Solve the given problems by finding the appropriate derivative. A package of weather instruments is propelled into the air to an altitude of about \(7 \mathrm{km} .\) A parachute then opens, and the package returns to the surface. The altitude \(y\) of the package as a function of the time \(t\) (in \(\min\) ) is given by \(y=\frac{10 t}{e^{0.4 t}+1} \cdot\) Find the vertical velocity of the package for \(t=8.0\) min.

Solve the given problems. If the force resisting the fall of an object of mass \(m\) through the atmosphere is directly proportional to the velocity \(v,\) then the velocity at time \(t\) is \(v=\frac{m g}{k}\left(1-e^{-k t / m}\right),\) where \(g\) is the acceleration due to gravity and \(k\) is a positive constant. Find \(\lim _{k \rightarrow 0^{+}} v\)

Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^{3}}$$
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