A Preview of Calculus: Fresh Take

  • Recognize a tangent to a curve at a point as the limit of secant lines
  • Explain the integral through the area problem

The Tangent Problem and Differential Calculus

The Main Idea 

  • Variable Rates of Change:
    • Linear functions have constant rates of change (slope)
    • Nonlinear functions have varying rates of change
  • Secant Lines:
    • Approximate the rate of change between two points
    • Slope: [latex]m_{\sec} = \frac{f(x) - f(a)}{x - a}[/latex]
  • Tangent Lines:
    • Represent instantaneous rate of change at a point
    • Limit of secant lines as they approach the point
  • Average Velocity:
    • [latex]v_{\text{avg}} = \frac{s(t) - s(a)}{t - a}[/latex]
    • Analogous to slope of secant line
  • Instantaneous Velocity:
    • Limit of average velocities as time interval approaches zero
    • Analogous to slope of tangent line
  • Connection to Calculus:
    • Differential calculus focuses on instantaneous rates of change
    • Foundation for solving the tangent problem

Estimate the slope of the tangent line (rate of change) to [latex]f(x)=x^2[/latex] at [latex]x=1[/latex] by finding slopes of secant lines through [latex](1,1)[/latex] and the point [latex]\left(\frac{5}{4},\frac{25}{16}\right)[/latex] on the graph of [latex]f(x)=x^2[/latex].

An object moves along a coordinate axis so that its position at time [latex]t[/latex] is given by [latex]s(t)=t^3[/latex]. Estimate its instantaneous velocity at time [latex]t=2[/latex] by computing its average velocity over the time interval [latex][2,2.001][/latex].

A ball is thrown upward with an initial velocity of [latex]64[/latex]ft/s from a height of [latex]6[/latex] ft. Its height (in feet) after [latex]t[/latex] seconds is given by:

[latex]h(t) = -16t^2 + 64t + 6[/latex]

  1. Find the average velocity between [latex]t = 1[/latex]and [latex]t = 1.5[/latex] seconds.
  2. Estimate the instantaneous velocity at [latex]t = 1[/latex]second.

The Area Problem and Integral Calculus

The Main Idea 

  • The Area Problem:
    • Finding the area between a curve and the [latex]x[/latex]-axis
    • Basis for integral calculus
  • Approximating Area:
    • Divide interval into smaller rectangles
    • Sum the areas of rectangles
    • As rectangle width approaches zero, sum approaches true area
  • Definite Integral:
    • Limit of sums of rectangular areas
    • Represents exact area under a curve

Estimate the area between the [latex]x[/latex]-axis and the graph of [latex]f(x)=x^2+1[/latex] over the interval [latex][0,3][/latex] by using the three rectangles shown here:

A graph of the same parabola f(x) = x^2 + 1, but with a different shading strategy over the interval [0,3]. This time, the shaded rectangles are given the height of the taller corner that could intersect with the function. As such, the rectangles go higher than the height of the function.
Figure 10.

 

Approximate the area under the curve [latex]f(x) = \sqrt{x}[/latex] from [latex]x = 1[/latex] to [latex]x = 4[/latex] using four rectangles with equal width.