Notes
Practice Questions
\(\small{\textbf{1)}}\) Find the equation of the normal line to the curve \(f(x) = x^3 – 4x + 1\) at the point \((2, 1)\).
The equation of the normal line is \(y = -\frac{1}{8}x + \frac{5}{4}\)
\(\,\,\,\,\,f(x) = x^3 – 4x + 1\)
\(\,\,\,\,\,f'(x) = 3x^2 – 4\)
\(\,\,\,\,\,f'(2) = 3(2)^2 – 4 = 8\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(2)} = -\frac{1}{8}\)
\(\,\,\,\,\,f(2) = (2)^3 – 4(2) + 1 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -\frac{1}{8}(x – 2)\)
\(\,\,\,\,\,\text{Simplified: } y = -\frac{1}{8}x + \frac{5}{4}\)
\(\small{\textbf{2)}}\) Find the equation of the normal line to the curve \(f(x) = e^x\) at the point \((0, 1)\).
The equation of the normal line is \(y = -x + 1\)
\(\,\,\,\,\,f(x) = e^x\)
\(\,\,\,\,\,f'(x) = e^x\)
\(\,\,\,\,\,f'(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(0)} = -1\)
\(\,\,\,\,\,f(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -1(x – 0)\)
\(\,\,\,\,\,\text{Simplified: } y = -x + 1\)
\(\small{\textbf{3)}}\) Find the equation of the normal line to the curve \(f(x) = \ln(x)\) at the point \((1, 0)\).
The equation of the normal line is \(y = -x +1\)
\(\,\,\,\,\,f(x) = \ln(x)\)
\(\,\,\,\,\,f'(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = -1\)
\(\,\,\,\,\,f(1) = \ln(1) = 0\)
\(\,\,\,\,\,\text{Point-slope form: } y – 0 = -1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = -x +1\)
\(\small{\textbf{4)}}\) Find the equation of the normal line to the curve \(f(x) = \sin(x)\) at the point \(\left(\frac{\pi}{2}, 1\right)\).
The equation of the normal line is \(x= \frac{\pi}{2}\)
\(\,\,\,\,\,f(x) = \sin(x)\)
\(\,\,\,\,\,f'(x) = \cos(x)\)
\(\,\,\,\,\,f’\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\)
\(\,\,\,\,\,\text{Slope of tangent line: } m = 0\)
\(\,\,\,\,\,\text{Slope of normal line: } m = \text{undefined (since } -\frac{1}{0} \text{ is undefined)}\)
\(\,\,\,\,\,\text{The normal line is vertical and passes through } x = \frac{\pi}{2}\)
\(\,\,\,\,\,\text{Equation: } x = \frac{\pi}{2}\)
\(\small{\textbf{5)}}\) Find the equation of the normal line to the curve \(f(x) = \frac{1}{x}\) at the point \((1, 1)\).
The equation of the normal line is \(y = x \)
\(\,\,\,\,\,f(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = -\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1) = -\frac{1}{1^2} = -1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = 1\)
\(\,\,\,\,\,f(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = 1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = x \)
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