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Chapter 3: Problem 64

Normal lines \(A\) line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point \(P\) $$y=\sqrt{x} ; P(4,2)$$

Short Answer

Expert verified
Answer: The equation of the normal line to the curve \(y = \sqrt{x}\) at point \(P(4, 2)\) is \(y - 2 = -4(x - 4)\).

Step by step solution

01

Find the derivative of the curve

Given the curve \(\displaystyle y=\sqrt{x}\), first, find its derivative with respect to \(x\). As \(\displaystyle y=x^{\frac{1}{2}}\), we can use the power rule to find the derivative. $$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$$
02

Evaluate the derivative at point P

Evaluate the derivative at point \(\displaystyle P( 4,2)\): $$\frac{dy}{dx}|_{x=4} = \frac{1}{2}(4)^{-\frac{1}{2}} = \frac{1}{4}$$ The slope of the tangent line at point \(\displaystyle P( 4,2)\) is \(\displaystyle \frac{1}{4}\).
03

Find the equation of the tangent line

Using the point-slope form of a line, we can find the equation of the tangent line: $$y - y_{1} = m(x - x_{1})$$ Inserting point \(\displaystyle P\displaystyle \left( 4,2\right) \) and the slope \(\displaystyle \frac{1}{4}\), we get: $$y - 2 = \frac{1}{4}(x - 4)$$
04

Determine the slope of the normal line

Since the normal line is perpendicular to the tangent line, its slope is the negative inverse of the tangent's slope: $$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{1}{4}} = -4$$
05

Find the equation of the normal line

Using the point-slope form of a line again with point \(\displaystyle P( 4,2)\) and slope \(\displaystyle -4\), we can find the equation of the normal line: $$y - y_{1} = m(x - x_{1})$$ Inserting the point and the slope, we get: $$y - 2 = -4(x - 4)$$ That is the equation of the normal line.

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Key Concepts

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

Tangent Lines
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. It represents the immediate direction the curve is heading at that exact point. Imagine a car driving on a curved road, the tangent line shows the path of the car at that moment.
To find a tangent line, first, understand:
  • Tangent lines have the same slope as the curve's derivative at the point of tangency.
  • They are key in finding normal lines since normal lines are perpendicular to tangent lines.
For the curve given by the equation \(y = \sqrt{x}\), and at point \(P(4,2)\), the tangent line equation uses the point-slope formula with the derivative's value as the slope.
Derivative
Derivatives are fundamental in calculus and measure how a function changes as its input changes. They can be aptly described as the rate of change or the function's instantaneous slope at any given point.
For functions, derivatives can be computed using different rules, with one of the simplest being the power rule.
  • The power rule: If \(y = x^n\), then \( \frac{dy}{dx} = n \cdot x^{n-1} \)
In this problem, the derivative of \(y = \sqrt{x}\) or \(y = x^{1/2}\) is calculated as \( \frac{dy}{dx} = \frac{1}{2} x^{-1/2} \).
This derivative tells us the slope of the tangent line at any point on the curve \(y = \sqrt{x}\). By evaluating it at \(x = 4\), we find it is \( \frac{1}{4} \). This shows how steep the curve is at point \(P\).
Slope
The slope of a line in mathematics is a measure of its steepness or the angle at which it inclines from the horizontal. It's a vital concept when determining the characteristics of a line in algebra.
Slopes:
  • Positive slope: Line goes upwards as you move right.
  • Negative slope: Line goes downwards as you move right.
  • Zero slope: Line is horizontal.
  • Undefined slope: Line is vertical.
In the context of tangent and normal lines:
  • The slope of the tangent line at \(P(4,2)\) is \(1/4\), as computed from the derivative.
  • For normal lines, which are perpendicular to tangent lines, the slope is the negative reciprocal. Thus, the normal line at \(P\) has a slope of \(-4\).
Point-Slope Form
The point-slope form of a linear equation is an algebraic way to express lines on a coordinate plane. It is given by the formula:
  • \(y - y_1 = m(x - x_1)\)
Where:
  • \(m\) represents the slope.
  • \((x_1, y_1)\) is a specific point on the line.
In this particular exercise:
  • The tangent line at \(P(4,2)\) uses \(m = \frac{1}{4}\) and rearranges into the formula \(y - 2 = \frac{1}{4}(x - 4)\).
  • The normal line uses a slope of \(-4\) and the same point to form \(y - 2 = -4(x - 4)\).
Understanding this form helps in quickly writing the equation of a line when given a point and its slope, making it an essential tool in geometry and calculus.

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Most popular questions from this chapter

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. The period \(T\) is the time required by the mass to complete one oscillation. Show that \(T=2 \pi \sqrt{\frac{m}{k}}\) b. Assume \(k\) is constant and calculate \(\frac{d T}{d m}\) c. Give a physical explanation of why \(\frac{d T}{d m}\) is positive.

a. Determine the points where the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 60 ). b. Does the curve have any horizontal tangent lines? Explain.

a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (Exercise 61 ). b. Does the curve have any horizontal tangent lines? Explain.

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at \(9.00 ?\) (Hint: Use the Law of Cosines.)

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$\sin x y=x+y$$
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