We have previously discussed how to find a curve's secant in order to determine its slope.
Another way of calculating a curve's slope is by finding its tangent.
This is also referred to as finding its Instantaneous Rate of Change at a single point. This can be expressed algebraically as:
\(m_{tan}= \cfrac{\Delta y}{\Delta x} = \lim\limits_{h\to0}\cfrac{\textcolor{red}{f(a+h)}-\textcolor{blue}{f(a)}}{h}\)
We can start the calculation by using the equation of a secant by using 2 points, \(a\) and \(a + h\) and making \(h\) as small as possible. We can achieve the second point by using limits such as \(\lim\limits_{h\to 0}\).
Example
Determine the slope of the tangent for the rational function \(f(x) = \cfrac{5+x}{x^2}\) at \(x=5\).
First, we can evaluate \(f(a+h)\) by substituting \(5+h\) for \(x\):
\(f(a+h) = f(5+h)\)
\(f(5+h) = \cfrac{5+(5+h)}{(5+h)^2}\)
\(f(5+h) = \cfrac{10+h}{(5+h)(5+h)}\)
\(\textcolor{red}{f(5+h) = \cfrac{10+h}{h^2 + 10h + 25}}\)
Then, we can evaluate \(f(a)\) by substituting \(5\) for \(x\):
\(f(a) = f(5)\)
\(f(5) = \cfrac{5+5}{(5)^2}\)
\(f(5) = \cfrac{10}{25}\)
\(\textcolor{blue}{f(5) = \cfrac{2}{5}}\)
Next, we can substitute the values we found for \(f(5+h)\) and \(f(5)\) into the numerator for the tangent formula:
\(\cfrac{\Delta y}{\Delta x} = \lim\limits_{h\to0}\cfrac{\textcolor{red}{f(5+h)}-\textcolor{blue}{f(5)}}{h}\)
\(= \lim\limits_{h\to0} \left[\textcolor{red}{\cfrac{10+h}{h^2 + 10h + 25}} - \textcolor{blue}{\cfrac{2}{5}} \right]\cdot \left[\cfrac{1}{h} \right]\)
We can't plug the limit into the given expression as that would cause the entire denominator to become \(0\). As a result, we can combine the terms by giving the expression a common denominator:
\(= \lim\limits_{h\to0} \left[\left(\cfrac{\textcolor{green}{5}}{\textcolor{green}{5}}\right)\left(\cfrac{10+h}{h^2 + 10h + 25}\right) - \left(\cfrac{\textcolor{magenta}{h^2 + 10h + 25}}{\textcolor{magenta}{h^2 + 10h + 25}}\right)\left(\cfrac{2}{5}\right)\right]\cdot \left[\cfrac{1}{h}\right]\)
\(= \lim\limits_{h\to0} \left[\cfrac{(10+h)(\textcolor{green}{5})}{(h^2 + 10h + 25)(\textcolor{green}{5})} - \cfrac{(2)(\textcolor{magenta}{h^2 + 10h + 25})}{(5)(\textcolor{magenta}{h^2 + 10h + 25})} \right]\cdot \left[\cfrac{1}{h}\right]\)
\(= \lim\limits_{h\to0} \left[\cfrac{5(10+h) - 2(h^2 + 10h + 25)}{5(h^2 + 10h + 25)}\right]\cdot \left[\cfrac{1}{h}\right]\)
Now that both terms in the main expression have the same denominator, we can combine all terms in the numerator under the same denominator:
\(= \lim\limits_{h\to0}\cfrac{5(10+h) - 2(h^2 + 10h + 25)}{5h(h^2 + 10h + 25)}\)
Now we can expand the numerator and collect like terms:
\(= \lim\limits_{h\to0}\cfrac{5(10) + 5(h) - 2(h^2) -2(10h) -2(25)}{5h(h^2 + 10h + 25)}\)
\(= \lim\limits_{h\to0}\cfrac{50 + 5h - 2h^2 - 20h - 50}{5h(h^2 + 10h + 25)}\)
Now that the numerator is expanded, we can simplify by collecting and cancelling like terms:
\(= \lim\limits_{h\to0}\cfrac{50 + 5h - 2h^2 - 20h - 50}{5h(h^2 + 10h + 25)}\)
\(= \lim\limits_{h\to0}\cfrac{-2h^2 - 15h}{5h(h^2 + 10h + 25)}\)
\(= \lim\limits_{h\to0}\cfrac{\cancel{h}(-2h - 15)}{5\cancel{h}(h^2 + 10h + 25)}\)
\(= \lim\limits_{h\to0}\cfrac{-2h - 15}{5(h^2 + 10h + 25)}\)
Finally, we can substitute \(0\) for \(h\) to determine the slope of the tangent:
\(= \cfrac{-2(0) - 15}{5((0)^2 + 10(0) + 25)}\)
\(= \cfrac{-15}{5(25)}\)
\(= \cfrac{-15}{125} = \cfrac{-3}{25}\)
Therefore, we can determine that the slope of the tangent is \(\cfrac{-3}{25}\)
Determine the slope of the tangent for the square root function \(g(x) = \sqrt{x-4}\) at \(x=5\)
Show Answer
First, we can evaluate \(f(a+h)\) by substituting \(5+h\) for \(x\):
\(f(a+h) = f(5+h)\)
\(f(a+h) = \sqrt{5+h-4}\)
\(\textcolor{red}{f(a+h) = \sqrt{h + 1}}\)
Then, we can evaluate \(f(a)\) by substituting \(5\) for \(x\):
\(f(a) = f(5)\)
\(f(5) = \sqrt{5-4}\)
\(f(5) = \sqrt{1}\)
\(\textcolor{blue}{f(5) = 1}\)
Next, we can substitute the values we found for \(f(5+h)\) and \(f(5)\) into the numerator for the tangent formula:
\(\cfrac{\Delta y}{\Delta x} = \lim\limits_{h\to0}\cfrac{\textcolor{red}{f(a+h)}-\textcolor{blue}{f(a)}}{h}\)
\(= \lim\limits_{h\to0}\cfrac{\textcolor{red}{\sqrt{h + 1}}-\textcolor{blue}{\sqrt{1}}}{h}\)
If we plug the limit into the expression, the denominator will be \(0\), resulting in an undefined calculation. In order to prevent this, we can rationalize the expression by multiplying it by the conjugate of the numerator to eliminate its the square root:
\(= \lim\limits_{h\to0}\left(\cfrac{\sqrt{h + 1}-1}{h}\right)\left(\cfrac{\sqrt{h + 1}+1}{\sqrt{h + 1}+1}\right)\)
\(= \lim\limits_{h\to0}\cfrac{h + 1-1}{h(\sqrt{h + 1}+1)}\)
Now we can simplify by collecting and cancelling like terms:
\(= \lim\limits_{h\to0}\cfrac{h + 1-1}{h(\sqrt{h + 1}+1)}\)
\(= \lim\limits_{h\to0}\cfrac{\cancel{h}}{\cancel{h}(\sqrt{h + 1}+1)}\)
\(= \lim\limits_{h\to0}\cfrac{1}{\sqrt{h + 1}+1}\)
Finally, we can substitute \(0\) for \(h\) to determine the slope of the tangent:
\(= \cfrac{1}{\sqrt{0 + 1}+1}\)
\(= \cfrac{1}{\sqrt{1}+1}\)
\(= \cfrac{1}{1+1} = \cfrac{1}{2}\)
Therefore, we can determine that the slope of the tangent is \(\cfrac{1}{2}\).