Problem 44
Question
(a) find the equations of the tangent line and the normal line to the curve at the given point, and (b) use a graphing utility to plot the graph of the function, the tangent line, and the normal line on the same screen. The curve \(y=2 x+(1 / \sqrt{x})\) at the point \((1,3)\).
Step-by-Step Solution
Verified Answer
The tangent line's equation at point (1, 3) is: \(y = \frac{3}{2}x + \frac{1}{2}\). The normal line's equation at the same point is: \(y = -\frac{2}{3}x + \frac{11}{3}\). Use a graphing utility to plot the given function \(y = 2x + \frac{1}{\sqrt{x}}\), the tangent line, and the normal line on the same screen, making sure they intersect at the point (1, 3).
1Step 1: 1. Derive the Function
First we need to find the derivative of the given function, y = 2x + (1/√x). We will apply the power rule to find the derivative.
Differentiate with respect to x :
y = 2x + x^(-1/2)
\(y' = 2 - \frac{1}{2}x^{-\frac{3}{2}}\)
2Step 2: 2. Evaluate the Derivative at the Given Point
Now we will evaluate the derivative at the given point (1, 3) to find the slope:
\(y'(1) = 2 - \frac{1}{2}(1)^{-\frac{3}{2}} = 2 - \frac{1}{2} \Rightarrow m_t = \frac{3}{2}\)
The slope of the tangent line, \(m_t\), is 3/2 at the point (1, 3).
3Step 3: 3. Calculate the Slope of the Normal Line
To find the slope of the normal line, we take the negative reciprocal of the tangent line's slope.
\(m_n = -\frac{1}{m_t} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}\)
The slope of the normal line, \(m_n\), is -2/3.
4Step 4: 4. Find the Equations of the Tangent and Normal Lines
Now that we have the slopes, we will use the point-slope form (y - y₁ = m(x - x₁)) to find the equations of the tangent line and the normal line.
The tangent line's equation:
y - 3 = (3/2)(x - 1)
y = (3/2)x + (1/2)
The normal line's equation:
y - 3 = (-2/3)(x - 1)
y = (-2/3)x + 11/3
5Step 5: 5. Graph the Function and the Lines
Using a graphing utility, plot the graph of the function, tangent line, and normal line on the same screen.
Function: y = 2x + (1/√x)
Tangent Line: y = (3/2)x + (1/2)
Normal Line: y = (-2/3)x + 11/3
Remember to verify that all the graphs intersect at the given point (1, 3).
Key Concepts
Derivative of a FunctionPoint-Slope FormGraphing UtilityNormal Line
Derivative of a Function
To find the derivative of a function, we differentiate it with respect to its variable, usually denoted as \(x\). The derivative essentially measures the rate at which the function changes at any given point. For the function \(y = 2x + \frac{1}{\sqrt{x}}\), we rewrite it as \(y = 2x + x^{-1/2}\) to facilitate differentiation using the power rule.
The power rule states that for \(x^n\), the derivative \(\frac{d}{dx}\) is \(nx^{n-1}\). Applying this rule, the derivative of \(2x\) is 2, and the derivative of \(x^{-1/2}\) is \(-\frac{1}{2}x^{-3/2}\).
So, the derivative of the entire function is \(y' = 2 - \frac{1}{2}x^{-3/2}\). Evaluating this derivative at \(x = 1\) gives us the slope of the tangent line at that point, \(m_t = \frac{3}{2}\).
The power rule states that for \(x^n\), the derivative \(\frac{d}{dx}\) is \(nx^{n-1}\). Applying this rule, the derivative of \(2x\) is 2, and the derivative of \(x^{-1/2}\) is \(-\frac{1}{2}x^{-3/2}\).
So, the derivative of the entire function is \(y' = 2 - \frac{1}{2}x^{-3/2}\). Evaluating this derivative at \(x = 1\) gives us the slope of the tangent line at that point, \(m_t = \frac{3}{2}\).
Point-Slope Form
The point-slope form of a linear equation is a method used to find the equation of a line given a point on the line and its slope. It is written as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point and \(m\) is the slope.
For the tangent line at the point \((1, 3)\) with a slope of \(\frac{3}{2}\), we use the point-slope form to find its equation: \(y - 3 = \frac{3}{2}(x - 1)\). Solving this yields the equation \(y = \frac{3}{2}x + \frac{1}{2}\).
Similarly, to find the equation of the normal line, which is perpendicular to the tangent line, we first determine its slope by taking the negative reciprocal of the tangent slope, which is \(-\frac{2}{3}\). Following the same process, the equation for the normal line becomes \(y - 3 = -\frac{2}{3}(x - 1)\), resulting in \(y = -\frac{2}{3}x + \frac{11}{3}\).
For the tangent line at the point \((1, 3)\) with a slope of \(\frac{3}{2}\), we use the point-slope form to find its equation: \(y - 3 = \frac{3}{2}(x - 1)\). Solving this yields the equation \(y = \frac{3}{2}x + \frac{1}{2}\).
Similarly, to find the equation of the normal line, which is perpendicular to the tangent line, we first determine its slope by taking the negative reciprocal of the tangent slope, which is \(-\frac{2}{3}\). Following the same process, the equation for the normal line becomes \(y - 3 = -\frac{2}{3}(x - 1)\), resulting in \(y = -\frac{2}{3}x + \frac{11}{3}\).
Graphing Utility
Using a graphing utility allows us to visually confirm our calculations by plotting functions and lines on a graph. It helps in verifying intersections and ensuring accuracy of our solutions. For this problem, we need to plot:
This intersection confirms that both the tangent and normal lines are correctly calculated, and aligns with the function at the point of tangency.
- The function: \(y = 2x + \frac{1}{\sqrt{x}}\)
- The tangent line: \(y = \frac{3}{2}x + \frac{1}{2}\)
- The normal line: \(y = -\frac{2}{3}x + \frac{11}{3}\)
This intersection confirms that both the tangent and normal lines are correctly calculated, and aligns with the function at the point of tangency.
Normal Line
The normal line is a line perpendicular to the tangent line at the point of tangency on a curve. To find the normal line, we first need the slope of the tangent line. The slope of the normal line is then the negative reciprocal of the tangent line's slope.
In mathematical terms, if the tangent line has a slope \(m_t\), the normal line’s slope, \(m_n\), is \(-\frac{1}{m_t}\). For this exercise, with the tangent slope \(m_t = \frac{3}{2}\), the normal line slope is \(-\frac{2}{3}\).
Using the point-slope form, the equation of the normal line given point \((1, 3)\) is \(y - 3 = -\frac{2}{3}(x - 1)\). Simplifying gives \(y = -\frac{2}{3}x + \frac{11}{3}\), confirming it’s correctly aligned perpendicularly to the tangent line.
In mathematical terms, if the tangent line has a slope \(m_t\), the normal line’s slope, \(m_n\), is \(-\frac{1}{m_t}\). For this exercise, with the tangent slope \(m_t = \frac{3}{2}\), the normal line slope is \(-\frac{2}{3}\).
Using the point-slope form, the equation of the normal line given point \((1, 3)\) is \(y - 3 = -\frac{2}{3}(x - 1)\). Simplifying gives \(y = -\frac{2}{3}x + \frac{11}{3}\), confirming it’s correctly aligned perpendicularly to the tangent line.
Other exercises in this chapter
Problem 44
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View solution Problem 45
Find the derivative of the function. $$ f(x)=\sqrt{\sin 2 x-\cos 2 x} $$
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