Tangents to a Curve
Vocabulary
Secant Line: a line that cuts across a curve; a reference for the slope of a line tangent to a curve
Tangent Line: a line that "just touches" a curve at a single point and no others
Differential Calculus: a study of calculus based on finding the difference in location between two points that get closer together until the distance between them is infinitely small
Slope of the Tangent Line
Remember in Algebra 1 how you learned about finding the slope of a line? What is the formula for slope of a line?
The tangent line represents the slope of an individual point on a curve. You find it by finding the slope of two points with an infinitely small distance between them. By decreasing the distance between the two points, the slope of the secant line gets closer and closer to the slope of the tangent line.
The slope of the secant line is found by the equation:
\begin{align*}m_{sec}\end{align*}\begin{align*}= \frac {y_1y_0}{x_1x_0} = \frac {f(x_1)f(x_0)}{x_1x_0}\end{align*}
As we bring the points closer together, x_{1} approaches x_{2} so we an write this as a limit statement:
\begin{align*}m_{sec}\end{align*} \begin{align*}= \lim_{x_1 \to x_0} \frac {f(x_1)f(x_0)}{x_1x_0} \end{align*}
To simplify our notation, if we let h = x 1 − x 0 , then x 1 = x 0 + h and x 1 → x 0 becomes equivalent to h → 0. This means that the equation becomes:
\begin{align*}m_{sec}\end{align*}\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)f(x_0)}{h} \end{align*}
The Slope of a Tangent Line
provided that the limit exist. 

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Click here for help finding the slope of the tangent line.
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Once you know the slope you can find the equation of the tangent line by using pointslope form of a line: y − y_{0} = m_{tan} ( x − x_{0} ).
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Practice
Definitions
 The line connecting two points on a curve is the ________________.
 The line that represents the slope of a single point on a curve is the ________________.
 The derivative is another name for the _________________.
 When calculating the slope of a tangent, what value is assumed to go to 0 as the two chosen points get closer and closer?
 Why do we use a limit statement to calculate the slope of the tangent line?
Write the equation of the tangent line when:
 x = 3, g(3) = 5, g'(3) = 7
 x = 9, g(9) = 27, g'(9) = 1/2
 x = 12, g(12) = 4, g'(12) = 3
Find the equation of the tangent line:

Find the equation of the tangent line to the graph of \begin{align*}m(x) = 3x^3 + 3x^2 + 4x + 4\end{align*} at \begin{align*}x = 1\end{align*}

Find the equation of the tangent line to the graph of \begin{align*}m(x) = x \end{align*} at \begin{align*}x = 0\end{align*}

Find the equation of the tangent line to the graph of \begin{align*}b(x) = 5x^4 + 3x^3  x^2 + 5x  3 \end{align*}at \begin{align*}x = 1\end{align*}
Instantaneous Rates of Change
Vocabulary
Derivative: the slope of the tangent line to a single point on a graph
Instantaneous Speed: the speed an object is travelling at any given single point in time
Average Speed: the distance an object travels divided by the time interval required
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The notation f ' ( x ) is read as "f prime of x", and is one way to denote the derivative of function ( x)
Alternate derivative notations include:
\begin{align*}f'(x)\end{align*}  \begin{align*}\frac {dy}{dx}\end{align*}  \begin{align*}y'\end{align*}  \begin{align*}\frac{df}{dx}\end{align*}  \begin{align*}\frac {df(x)}{dx}\end{align*} 

Rates of Change
The slope of the tangent line is also called the derivative. The derivative of a function f ( x ) can be written as f ' ( x ).
The Derivative


Speed is calculated by taking the Average Rate of Change (AROC), or the slope of the secant line, while the derivative is the Instantaneous Rate of Change (IROC), or the slope of the tangent line. Can you think of two situations where the AROC would be useful? What about the IROC?
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Finding Average Velocity
The average speed of an object is defined as the object's displacement ∆ x divided by the time interval ∆ t during which the displacement occurs:
\begin{align*}\text{Average speed }= v=\frac {\triangle x}{\triangle t}=\frac {x_1x_0}{t_1t_0}\end{align*} 

Notice that the points( t 0 , x 0 ) and ( t 1 , x 1 ) lie on the position versus time curve, as the figure below shows.
This expression is also the expression for the slope of a secant line connecting the two points. Therefore, the average velocity is equal to the AROC of a function over an interval.
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The IROC is the object's velocity at that point, so it is equal to the IROC, or the derivative.
Practice
Find the Average Rate of Change
 AROC of the function \begin{align*}f(x) = x^2  4x + 2\end{align*} over the interval x = 12 to x = 30
 AROC of the function \begin{align*} f(x) = x^2  5x + 201\end{align*} over the interval x = 10 to x = 11
 AROC of the function \begin{align*}f(x) = 5x^2  3x  4\end{align*} over the interval x = 29 to x = 48
Find the Instantaneous Rate of Change (with respect to x)
 \begin{align*}f(x) = 3x^2  x  5\end{align*} at x = 15
 \begin{align*}f(x) = 4x^2 + 195\end{align*} at x = 10
 \begin{align*}f(x) = x^2 + x  3\end{align*} at x = 20
Solve Rate of Change Problems
 The number of people in the US affected by the common cold in the month of November is defined by N = f(x) where x is the day of the month. What is the meaning of f'(x) in this context?
 The number of households in Florida affected by hurricane season in the month of July is defined by J = f(x) where x is the day of the month. \begin{align*}f(x) = 2x^2 + x + 1\end{align*} Find the average rate of change of J with respect to x when days is changed from x = 5 to x = 34
 A cake is taken from an oven when its temperature is 196°F and placed on a cooling rack in a room where the temperature is 75°F. The temperature of the cake over (x) minutes is given by H = f(x). \begin{align*}f(x) = 4x^2 + 15x + 196\end{align*} Find the instantaneous rate of change of H with respect to x when x = 15.