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Chapter 3: Problem 28

Find an equation of the straight line having slope \(1 / 4\) that is tangent to the curve \(y=\sqrt{x}\).

Short Answer

Expert verified
The equation of the tangent line is \( y = \frac{1}{4}x + 1 \).

Step by step solution

01

Write down the equation of the line

The equation of a line with slope \( m = \frac{1}{4} \) is \( y = \frac{1}{4}x + c \), where \( c \) is the y-intercept. Our goal is to find \( c \).
02

Differentiate the curve

To find the point of tangency, differentiate the curve \( y = \sqrt{x} \). The derivative is \( y' = \frac{d}{dx}(x^{1/2}) = \frac{1}{2\sqrt{x}} \). This gives the slope of the tangent to the curve at any point \( x \).
03

Set the derivative equal to the line's slope

Set the derived slope equation \( \frac{1}{2\sqrt{x}} \) equal to the slope of the line \( \frac{1}{4} \) to find the x-coordinate of the point of tangency. Solve \( \frac{1}{2\sqrt{x}} = \frac{1}{4} \).
04

Solve for the x-coordinate

From \( \frac{1}{2\sqrt{x}} = \frac{1}{4} \), multiply both sides by \( 2\sqrt{x} \) to get \( 1 = \frac{\sqrt{x}}{2} \). Multiply both sides by \( 2 \) to obtain \( \sqrt{x} = 2 \). Squaring both sides gives \( x = 4 \).
05

Find the y-coordinate

Substitute \( x = 4 \) into the curve equation \( y = \sqrt{x} \) to find the y-coordinate. Thus \( y = \sqrt{4} = 2 \). The point of tangency is \( (4, 2) \).
06

Find the y-intercept c of the tangent line

Substitute the x-value, y-value, and slope into the line equation \( y = \frac{1}{4}x + c \). Therefore: \( 2 = \frac{1}{4}(4) + c \). Simplifying gives \( 2 = 1 + c \), so \( c = 1 \).
07

Write the final equation of the tangent line

With slope \( \frac{1}{4} \) and y-intercept \( c = 1 \), the equation of the tangent line is \( y = \frac{1}{4}x + 1 \).

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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 is a straight line that touches a curve at exactly one point without crossing it. This special line shares the same slope as the curve at the point of contact. It provides a linear approximation of the curve near this point.
To find a tangent line, you generally follow these steps:
  • Identify the function or curve you are dealing with, such as \( y = \sqrt{x} \).
  • Calculate the derivative to find the slope of the tangent at any point on the curve.
  • Use the slope and a point on the curve to write the equation of the tangent line.
In this problem, you are given the slope of the tangent line directly. You find the point on the curve where the slope matches the given one, allowing you to write the tangent's equation easily.
Derivative
The derivative of a function measures how the function value changes as its input changes. In simple terms, it gives the slope of the function at any given point.
For the curve \( y = \sqrt{x} \), the derivative is calculated as: \[ y' = \frac{d}{dx}(x^{1/2}) = \frac{1}{2\sqrt{x}}. \]
  • This derivative tells us the rate at which \( y \) is changing with respect to \( x \).
  • To find where a line is tangent to the curve, set the derivative equal to the desired slope.
Calculating the derivative is crucial because it allows us to find the exact point on the curve that has a specified slope.
Slope
The slope of a line is a measure of its steepness, usually given as \( m \) in line equations. The slope represents how much \( y \) changes for a unit change in \( x \). In equations like \( y = mx + c \), \( m \) is the slope.
In this exercise, the slope of the tangent line is \( \frac{1}{4} \).
  • To find the tangent point on \( y = \sqrt{x} \), you set the derivative equal to this value.
  • Solve the equation \( \frac{1}{2\sqrt{x}} = \frac{1}{4} \) to find the x-coordinate where the curve's slope matches \( \frac{1}{4} \).
This step is necessary to locate the point on the curve where the tangent line just touches it.
Curve
A curve is a line that bends continuously without any sharp turns. In mathematics, curves like \( y = \sqrt{x} \) are graphical representations of a function.
  • The curve \( y = \sqrt{x} \) rises slowly as \( x \) increases, resulting in less steep curves.
  • To talk tangents, the derivative is used to analyze how the curve behaves at any specific point.
Understanding the shape and behavior of the curve helps in recognizing how and where tangent lines can be drawn with particular slopes. This case shows how a line with a small slope \( \frac{1}{4} \) just gently touches the curve at a single point.

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Most popular questions from this chapter

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt[3]{\frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}}$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I .\) Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying $$|x-a|<\delta \Rightarrow|f(x)-L(x)|<\epsilon$$ for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x 2^{x}, \quad[0,2], \quad a=1$$

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\ln x)^{\ln x}$$

If we write \(g(x)\) for \(f^{-1}(x),\) Equation (1) can be written as $$g^{\prime}(f(a))=\frac{1}{f^{\prime}(a)}, \quad \text { or } \quad g^{\prime}(f(a)) \cdot f^{\prime}(a)=1$$. If we then write \(x\) for \(a\), we get $$g^{\prime}(f(x)) \cdot f^{\prime}(x)=1$$. The latter equation may remind you of the Chain Rule, and indeed there is a connection. Assume that \(f\) and \(g\) are differentiable functions that are inverses of one another, so that \((g \circ f)(x)=x .\) Differentiate both sides of this equation with respect to \(x\), using the Chain Rule to express \((g \circ f)^{\prime}(x)\) as a product of derivatives of \(g\) and \(f\) What do you find? (This is not a proof of Theorem 3 because we assume here the theorem's conclusion that \(g=f^{-1}\) is differentiable.)

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=x^{(x+1)}$$
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