This lesson centers around solving Word Problems pertaining to Average Rate of Change and Instanteneous Rate of Change. Here are the formulas for the respective Rates of Change:
\(m_{sec}= \cfrac{\Delta y}{\Delta x} = \cfrac{\textcolor{red}{f(b)}-\textcolor{blue}{f(a)}}{\textcolor{red}{b}-\textcolor{blue}{a}}\)
\(m_{tan}= \cfrac{\Delta y}{\Delta x} = \cfrac{\textcolor{red}{f(a+h)}-\textcolor{blue}{f(a)}}{h}\)
For additional information and practice with these formulas, review Slope of a Secant and Slope of a Tangent!!
It's important to identify what Rate of Change you're calculating to determine which formula to use. If the question is asking for Average Rate of Change, use the secant formula. If it's asking for the Instantaneous Rate of Change, use the tangent formula.
When calculating Rates of Change, it's good to use variables to represent different quantities. Also, remember to include units in your final answer!
The Surface Area of a snowball, in square centimeters, is modelled by the equation \(S = 4\pi r^2\) where \(r\) is the radius, in centimeters. The Volume, in cubic centimeters, is modelled by \(V = \cfrac{4}{3}\pi r^3\).
- Determine the average rate of change of the surface area as the radius changes from \(20\;[\text{cm}]\) to \(25\;[\text{cm}]\).
- Determine the instantaneous rate of change of volume when the radius is \(10\;[\text{cm}]\).
Show Answer
i. We can substitute the Surface Area formula into the Average Rate of Change of formula with \(20\) and \(25\) substituting \(a\) and \(b\) respectively.
\(m_{sec} =\cfrac{\textcolor{red}{f(b)}-\textcolor{blue}{f(a)}}{\textcolor{red}{b}-\textcolor{blue}{a}}\)
\(=\cfrac{\textcolor{red}{4\pi (25)^2}-\textcolor{blue}{4\pi (20)^2}}{\textcolor{red}{25}-\textcolor{blue}{20}}\)
Next, we can take a common factor out of the numerator and find the squares of its respective terms:
\(=\cfrac{4\pi [(25)^2-(20)^2]}{5}\)
\(=\cfrac{4\pi (625-400)}{5}\)
\(=\cfrac{4\pi (225)}{5}\)
Finally, we can simplify to find the Average Rate of Change:
\(=180\pi\)
\(= 565.49 \cfrac{\;[\text{cm}^2]}{\;[\text{cm}]}\)
Therefore, we can determine that the Average Rate of Change for the Surface Area is \(565.49 \cfrac{\;[\text{cm}^2]}{\;[\text{cm}]}\).
ii. First, we can use the alternate formula for Instanteneous Rate of Change to help model the problem. We can substitute \(a\) with \(10\) and \(a+ h\) as simply \(r\). We can then substitute both functions with the proper Volume formula:
\(m_{tan}=\lim\limits_{r\to10} \left[\cfrac{\textcolor{red}{V(r)}-\textcolor{blue}{V(10)}}{\textcolor{blue}{r}-\textcolor{red}{10}}\right]\)
\(=\lim\limits_{r\to10} \left[\cfrac{\textcolor{red}{\cfrac{4}{3}\pi r^3}-\textcolor{blue}{\cfrac{4}{3}\pi (10)^3}}{\textcolor{red}{r}-\textcolor{blue}{10}}\right]\)
Next, we can take a common factor out of the numerator:
\(=\lim\limits_{r\to10} \left[\cfrac{4}{3}\pi\left(\cfrac{r^3-10^3}{r-10}\right)\right]\)
Then, we can factor the numerator given that it is a Difference of Cubes. As a refresher, here is the formula for factoring a Difference of Cubes:
\(a^3-b^3 = (a-b)(a^2 +ab + b^2)\)
\[=\lim\limits_{r\to10} \left[\cfrac{4}{3}\pi \left[\cfrac{\cancel{(r-10)}(r^2 + 10r + 100)}{\cancel{(r-10)}}\right]\right]\]
\[=\lim\limits_{r\to10}\left[\cfrac{4}{3}\pi(r^2 + 10r + 100)\right]\]
Finally, we can substitute \(10\) for \(r\) to help determine the instantaneous rate of change:
\(= \cfrac{4}{3}\pi \cdot [(10)^2 + 10(10) + 100]\)
\(= \cfrac{4}{3}\pi(100 + 100 + 100)\)
\(= \cfrac{4}{3}\pi(300)\)
\(= 400\pi\)
\(= 1256.64 \cfrac{\;[\text{cm}]^3}{\;[\text{cm}]}\)
Therefore, we can determine that the Instantaneous Rate of Change of Volume is \(1256.64 \cfrac{\;[\text{cm}]^3}{\;[\text{cm}]}\)
An outdoor hot tub holds \(2700\;[\text{L}]\) of water. The volume of water in the tub is modelled by the function \(V(t) = \cfrac{1}{12}(180-t)^2\), where \(V\) is the volume of water in the hot tub, in litres, and \(t\) is the time, in minutes, that the valve is open. Determine the instantaneous rate of change in the volume of water in \(60\;[min]\).
Show Answer
First, we can determine \(V(a+h)\) where \(a = 60\):
\(V(a + h) = V(60 + h)\)
\(V(60 + h) = \cfrac{1}{12}(180-(60+h)^2\)
\(V(60 + h) = \cfrac{1}{12}(180-60-h)^2\)
\(\textcolor{red}{V(60 + h) = \cfrac{1}{12}(120-h)^2}\)
Next, we can determine \(V(a)\) where \(a = 60\):
\(V(a) = V(60)\)
\(V(60) = \cfrac{1}{12}(180-60)^2\)
\(\textcolor{blue}{V(60) = \cfrac{1}{12}(120)^2}\)
Next we can place these expressions in the Instanteneous Rate of Change formula. We can then simplify:
\(\cfrac{\Delta y}{\Delta x} =\lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{V(a+h)}-\textcolor{blue}{V(a)}}{h}\right]\)
\(=\lim\limits_{h\to0} \left[\left[\textcolor{red}{\cfrac{1}{12}(120-h)^2}-\textcolor{blue}{\cfrac{1}{12}(120)^2}\right] \cdot \cfrac{1}{h}\right]\)
We can simplify by taking out a common factor. We can then further simplify by expanding the numerator and collecting like terms:
\(=\lim\limits_{h\to0} \left[\cfrac{1}{12h} \left[(120-h)^2-(120)^2\right]\right]\)
\(=\lim\limits_{h\to0} \left[\cfrac{1}{12h}(h^2 - 240h + 14400 - 14400)\right]\)
\(=\lim\limits_{h\to0}\left[\cfrac{1}{12h}(h^2 - 240h)\right]\)
Then, we can take a common factor out of the main expression and simplify:
\(=\lim\limits_{h\to0}\cfrac{1}{12\cancel{h}}\left[\cancel{h}(h - 240)\right]\)
\(=\lim\limits_{h\to0}\cfrac{1}{12}\left[(h - 240)\right]\)
\(=\lim\limits_{h\to0}\cfrac{h - 240}{12}\)
Finally, we can substitute \(0\) for \(h\) to determine the instantaneous change of water:
\(=\cfrac{0 - 240}{12}\)
\(=\cfrac{-240}{12}\)
\(= -20\)
Therefore, we can determine that the Instantaneous Rate of Change of water at \(60\;[\text{min}]\) is \(-20\cfrac{\;[\text{L}]}{\;[\text{min}]}\).
Show that the rate of change in the Volume of a cube with respect to its edge length is equal to half the Surface Area of the cube.
Show Answer
First, we can determine the formula for the Surface Area of a cube. It can be expressed algebraically as such:
\(SA = 6x^2\)
In this instance, \(x\) represents the side length, squared then multiplied by \(6\).
Next, we can determine the formula for the Volume of a cube. It can be expressed algebraically as such:
\(V = lwh\)
Given that Length, Width, and Height are all the same value in a cube, the formula can be rewritten as such:
\(V = x^3\)
We can now calculate the slope of the Volume's tangent to determine its rate of change:
\(m_{tan}=\lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{f(a+h)}-\textcolor{blue}{f(a)}}{h}\right]\)
\(=\lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{(x+h)^3}-\textcolor{blue}{x^3}}{h}\right]\)
Next, we can simplify by expanding and collecting like terms:
\(=\lim\limits_{h\to0} \left[\cfrac{(x+h)^3-x^3}{h}\right]\)
\(=\lim\limits_{h\to0} \left[\cfrac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}\right]\)
\(=\lim\limits_{h\to0} \left[\cfrac{3x^2h + 3xh^2 + h^3}{h}\right]\)
Then, we can factor out a common term in the numerator to simplify the expression further:
\[=\lim_{h\to0} \left[\cfrac{\cancel{h}(3x^2 + 3xh + h^2)}{\cancel{h}}\right]\]
\[=\lim_{h\to0}[3x^2 + 3xh + h^2]\]
Finally, we can substitute \(0\) for \(h\) to determine the Instantaneous Rate of Change in the Volume of a cube:
\(=3x^2 + 3x(0) + (0)^2\)
\(=3x^2\)
If we compare the Volume's Instantaneous Rate of Change to the Surface Area, we can determine that it's exactly half:
\(\text{ratio} =\cfrac{\text{Volume}}{\text{Surface Area}}\)
\(=\cfrac{3x^2}{6x^2}\)
\(=\cfrac{3\cancel{x^2}}{6\cancel{x^2}}\)
\(=\cfrac{3}{6} = \cfrac{1}{2}\)
Determine the equation of the line that is perpendicular to the tangent of \(y = x^5\) at \(x=-2\) and which passes through the tangent point.
Show Answer
Since the line passes through the tangent point, this indicates that we need to determine its Instantaneous Rate of Change.
First, we can determine \(y\) at \(x=-2\):
\(y = (-2)^5\)
\(y = -32\)
After, we can substitute \(x^5\) for \(a\) and \((x + h)^5\) for \((a+h)\) in the Instantaneous Rate of Change formula:
\(m_{tan} = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{f(a+h)} - \textcolor{blue}{f(a)}}{h}\right]\)
\(= \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{(x+h)^5} - \textcolor{blue}{x^5}}{h}\right]\)
Next, we can simplify by expanding the numerator and collecting like terms:
\(= \lim\limits_{h\to0} \left[\cfrac{(x+h)^5 - x^5}{h}\right]\)
\(= \lim\limits_{h\to0} \left[\cfrac{x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5 - x^5}{h}\right]\)
\(= \lim\limits_{h\to0} \left[\cfrac{5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5}{h}\right]\)
Then, we can simplify further by factoring a common term out of the numerator:
\(= \lim\limits_{h\to0} \left[\cfrac{\cancel{h}(5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4)}{\cancel{h}}\right]\)
\(= \lim\limits_{h\to0} \left[5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4\right]\)
After, we can substitute \(0\) for \(h\):
\(= 5x^4 + 10x^3(0) + 10x^2(0)^2 + 5x(0)^3 + (0)^4\)
\(= 5x^4\)
We can substitute \(x = -2\) to determine the slope at that point:
\(m_{tan}(x)= 5x^4\)
\(m_{tan}(-2)= 5(-2)^4\)
\(= 80\)
We can find the perpedicular slope by taking the negative reciprocol of the slope we determine above:
\(m_{perp} = -\cfrac{1}{80}\)
Finally, we can determine \(b\) by plugging all the known terms into the slope-intercept formula:
\(y = mx + b\)
\(-32 = (-\cfrac{1}{80})(-2) + b\)
\(-32 = \cfrac{1}{40} + b\)
\(b = \cfrac{-1280-1}{40}\)
\(b = -\cfrac{1279}{40}\)
We can now put together our equation:
\(y = -\cfrac{1}{80}x -\cfrac{1279}{40}\)
Therefore, we can determine that the equation of the line perpendicular to the tangent of \(y = x^5\) at \(x=-2\) that passes through the tangent point is \(y = -\cfrac{1}{80}x -\cfrac{1279}{40}\).