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Tangents to a Curve

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Introduction to Derivatives

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:

m_{sec}= \frac {y_1-y_0}{x_1-x_0} = \frac {f(x_1)-f(x_0)}{x_1-x_0}

As we bring the points closer together, x1 approaches x2 so we an write this as a limit statement:

m_{sec} = \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{x_1-x_0}

To simplify our notation, if we let − , then and x → x becomes equivalent to h → 0. This means that the equation becomes:

m_{sec}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h}

The Slope of a Tangent Line

If the point ) is on the curve , then the tangent line at has a slope that is given by
m_{tan}=\lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h}

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 point-slope form of a line: y − y0 = mtan ( x − x0 ).

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Practice

Definitions
  1. The line connecting two points on a curve is the ________________.
  2. The line that represents the slope of a single point on a curve is the ________________.
  3. The derivative is another name for the _________________.
  4. When calculating the slope of a tangent, what value is assumed to go to 0 as the two chosen points get closer and closer?
  5. Why do we use a limit statement to calculate the slope of the tangent line?
Write the equation of the tangent line when:
  1. x = 3, g(3) = 5, g'(3) = 7
  2. x = 9, g(9) = 27, g'(9) = 1/2
  3. x = -12, g(-12) = 4, g'(-12) = -3
Find the equation of the tangent line:
  1. Find the equation of the tangent line to the graph of m(x) = 3x^3 + 3x^2 + 4x + 4  at  x = 1

  2. Find the equation of the tangent line to the graph of m(x) = x  at  x = 0

  3. Find the equation of the tangent line to the graph of b(x) = -5x^4 + 3x^3 - x^2 + 5x - 3 at  x = -1

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 ' ( is read as "f prime of x", and is one way to denote the derivative of function ( x)

Alternate derivative notations include:

f'(x) \frac {dy}{dx} y' \frac{df}{dx} \frac {df(x)}{dx}


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
The function ' is defined by the formula
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}
where ' is called the derivative of with respect to . The domain of consists of all the values of for which the limit exists.

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 ∆ divided by the time interval ∆ during which the displacement occurs:

\text{Average speed }= v=\frac {\triangle x}{\triangle t}=\frac {x_1-x_0}{t_1-t_0}

Notice that the points( ) and ( ) 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
  1. AROC of the function  f(x) = x^2 - 4x + 2  over the interval x = 12 to x = 30
  2. AROC of the function  f(x) = x^2 - 5x + 201  over the interval x = 10 to x = 11
  3. AROC of the function f(x) = -5x^2 - 3x - 4  over the interval x = 29 to x = 48
Hint: Find the slope of the line that connects the two endpoints
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Find the Instantaneous Rate of Change (with respect to x)
  1. f(x) = 3x^2 - x - 5 at x = 15
  2. f(x) = 4x^2 + 195  at x = 10
  3. f(x) = -x^2 + x - 3  at x = 20
Hint: Use the Definition of the Derivative 
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Solve Rate of Change Problems
  1. 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?
  2. 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. f(x) = 2x^2 + x + 1 Find the average rate of change of J with respect to x when days is changed from x = 5 to x = 34
  3. 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). f(x) = 4x^2 + 15x + 196 Find the instantaneous rate of change of H with respect to x when x = 15.
Click here for guidance. 

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