Slope of Tangents

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 \(\lim\limits_{h\to 0}\).


Graph with tangent line moving through function at point a and secant line moving through points a and a+h.

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 \(x=5\)

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