Ｃａｌｃｕｌｕｓ　１　Ｌｅｃｔｕｒｅ　０．１

Ｌｉｎｅｓ， Ａｎｇｌｅ ｏｆ Ｉｎｃｌｉｎａｔｉｏｎ， ａｎｄ ｔｈｅ Ｄｉｓｔａｎｃｅ Ｆｏｒｍｕｌａ

Ｉｎｔｒｏｄｕｃｔｉｏｎ

• [0:01] Introduction
  • Review of Math 2 concepts and basic algebra needed for Calculus.
  • Topics to cover:
    • Basic lines.
    • Families of curves.
    • Trigonometric functions.
    • Introduction to Calculus.

Ｌｉｎｅｓ

• [0:34] Lines – [image]
  • What is special about a line?
    • You need at least two points to define a line.
    • Lines do not curve.
    • Lines do not end.
    • Lines have slope.
  • [1:03] What do we need to graph a line?
    • Two points or . . .
    • . . . One point and the slope
  • [1:10] Slope of a Line
    • The slope describes how a line rises or falls.
    • Deriving the slope formula:
      • Take two generic points on a line: $(𝒙_1,𝒚_1)$ and $(𝒙_2,𝒚_2)$.
      • Coordinates of a point $(𝒙,𝒚)$.
      • Distinguishing two points using subscripts.
      • Calculating the “run” (change in 𝒙): $𝒙_2-𝒙_1$.
      • Calculating the “rise” (change in 𝒚): $𝒚_2-𝒚_1$.
      • [3:30] Defining the slope $(𝒎)$ as rise over run:
        • $\boxed{\Large 𝒎=\dfrac{𝒚_2-𝒚_1}{𝒙_2-𝒙_1}}$
          • The formula works for any pair of points on the line.
  • [5:39] Equation of a Line from the Slope Formula
    • Manipulating the slope formula to get the equation of a line.
    • Fix one point $(𝒙_1,𝒚_1)$ and let the other point be variable $(𝒙,𝒚)$.
      • Substituting into the slope formula:
        • $\boxed{\Large 𝒎=\dfrac{𝒚-𝒚_1}{𝒙-𝒙_1}}$
      • The resulting equation allows solving for 𝒚 when a value for 𝒙 is provided.
      • Isolating $(𝒚-𝒚_1)$
        • $\boxed{\Large 𝒚-𝒚_1=𝒎(𝒙-𝒙_1)}$
      • Called point-slope form because you need one point $(𝒙_1,𝒚_1)$ and the slope $(𝒎)$ to define it.
  • [9:40] 🧩 Example – Finding a Line’s Equation Given Two Points: $(-2,-3)$ and $(8,2)$ – [image]
    • Necessary steps:
      • We need one point (we have two).
      • We need the slope (we must calculate it).
    • Calculating the slope using the two given points:
      • $𝒎=\dfrac{2-(-3)}{8-(-2)}=\dfrac{5}{10}=\dfrac{1}{2}$
    • Using the point-slope form to find the equation:
      • $𝒚-(-3)=\dfrac{1}{2}(𝒙-(-2))$
      • $𝒚+3=\dfrac{1}{2}(𝒙+2)$
      • $𝒚+3=\dfrac{1}{2}𝒙+1$
      • $𝒚=\dfrac{1}{2}𝒙-2$
  • [15:10] Slope-Intercept Form of a Line’s Equation
    • Transforming the point-slope form to slope-intercept form
      • $\displaystyle \boxed{\Large 𝒚=𝒎𝒙+𝒃}$
      • Identifying the slope $(𝒎)$ and the 𝒚-intercept $(𝒃)$.
      • 🧩 Example – Graphing using slope-intercept form.
      • Interpretation of positive or negative slope.
      • Two points define a unique line.
        • $𝒚=\dfrac{1}{2}𝒙-2$
          • $𝒃=-2$ point $(0,-2)$
          • $𝒎=\dfrac{1}{2}$; from $(0,2)$ one unit up (rise) ⇡ and two units to the right ⇢ (run)
            • $(-1,2)$ is the second point
  • [17:24] Special Types of Lines – [image]
    • [17:46] Horizontal lines: $𝒚=𝒄$ (where 𝒄 is a constant).
      • Intersects the 𝒚-axis at 𝒄.
      • Slope is zero.
    • [18:45] Vertical lines: $𝒙=𝒄$ (where 𝒄 is a constant).
      • Intersects the 𝒙-axis at 𝒄.
      • Slope is undefined.
  • [19:50] Standard Form of a Line’s Equation – [image]
    • [21:52] 🧩 Example – Converting from standard form to slope-intercept form by isolating 𝒚: $4𝒙+2𝒚-3=0 \;\to\; 𝒚=𝒎𝒙+𝒃$
      • Start from the standard form:
        • $4𝒙+2𝒚-3=0$
      • Move the constant term to the other side:
        • $4𝒙+2𝒚=3$
      • Divide each term to isolate 𝒚:
        • $\dfrac{4𝒙-3}{-2}=-2𝒙+\dfrac{3}{2}=𝒚$
      • Final slope–intercept form:
        • $𝒚=-2𝒙+\dfrac{3}{2}$
          • Here, slope $𝒎=-2$ and intercept $𝒃=\dfrac{3}{2}$
  • [22:02] Cover-Up Method for Graphing from Standard Form
    • To find the 𝒙-intercept, cover the term with 𝒚.
    • To find the 𝒚-intercept, cover the term with 𝒙.
    • 🧩 Example – $3𝒙+4𝒚=4$ ($𝒙$-intercept at $\dfrac{4}{3}$, $𝒚$-intercept at $1$).
  • [23:02] Parallel Lines
    • Have the same slope.
    • Stair analogy: both rise or fall at the same rate, never intersecting.
    • Algebraic condition for parallelism:
      • Two lines $𝑦=𝒎_1𝑥+𝒃_1$ and $𝑦=𝒎_2𝑥+𝒃_2$ are parallel iff $𝒎_1=𝒎_2$ but $𝒃_1\neq𝒃_2$.
      • Different intercepts $(𝒃_1,𝒃_2)$ shift the lines vertically without changing slope.
    • Geometric interpretation: equal slope ⇒ same inclination angle with respect to the 𝒙-axis.
  • [23:39] Perpendicular Lines
    • Intersect at a right angle (90°).
    • If one slope is positive, the other is negative.
    • Their slopes are negative reciprocals:
      • $\displaystyle \boxed{\Large 𝒎_1\cdot𝒎_2=-1}$
      • Example: if $𝒎_1=2$, then $𝒎_2=-\dfrac{1}{2}$
    • Special case (vertical and horizontal lines):
      • A vertical line (undefined slope) and a horizontal line (slope $=0$) are perpendicular even though $𝒎_1\cdot𝒎_2\neq-1$ (since one slope is undefined).
    • Visual cue: perpendicular lines form a “T” or “L” shape where they intersect.
  • [25:24] 🧩 Example – Parallel to $2𝒙+3𝒚=12$ through $(6,7)$ – [image]
    • Identify the slope of the given line (in slope-intercept form):
      • $2𝒙+3𝒚=12$
      • $3𝒚=-2𝒙+12$
      • $𝒚=-\dfrac{2}{3}𝒙+4$
    • The slope of the parallel line is the same: $𝒎=-\dfrac{2}{3}$
    • Use the point-slope form with the point $(6,7)$ and the parallel slope:
      • $𝒚-𝒚_1=𝒎(𝒙-𝒙_1)$
      • $𝒚-7=-\dfrac{2}{3}(𝒙-6)$
      • $𝒚-7=-\dfrac{2}{3}𝒙+4$
      • $𝒚=-\dfrac{2}{3}𝒙+11$
  • [29:50] 🧩 Example – Perpendicular to $2𝒙+3𝒚=12$ through $(6,7)$ – [image]
    • $𝒎=-\dfrac{2}{3} \;\to\; 𝒎=\dfrac{3}{2}$ (negative reciprocals)
    • Use the point-slope form:
      • $𝒚-𝒚_1=𝒎(𝒙-𝒙_1)$
      • $𝒚-7=\dfrac{3}{2}(𝒙-6)$
      • $𝒚-7=\dfrac{3}{2}𝒙-9$
      • $𝒚=\dfrac{3}{2}𝒙-2$

Ａｎｇｌｅ ｏｆ Ｉｎｃｌｉｎａｔｉｏｎ

• [32:10] Angle of Inclination – [image]
  • Definition: The angle a line forms with the positive 𝒙-axis.
  • Relationship with change in 𝒙 $(\Delta 𝒙)$ and change in 𝒚 $(\Delta 𝒚)$.
  • Using basic trigonometry on a right triangle formed by the line, $\Delta 𝒙$, and $\Delta 𝒚$.
  • The tangent of the angle of inclination $(𝜃)$ equals the slope $(𝒎)$:
    • $\displaystyle \boxed{\Large \tan(𝜃)=\dfrac{\Delta 𝒚}{\Delta 𝒙}=𝒎}$
  • If the angle is known, the slope can be found; if the slope is known, the angle can be found.
  • [35:50] 🧩 Example – Given the Angle of Inclination: $30^\circ$ or $\dfrac{\pi}{6}$
    • Using the formula $𝒎=\tan(𝜃)$.
    • $\tan\!\big(\dfrac{\pi}{6}\big)=\dfrac{\sin(\pi/6)}{\cos(\pi/6)}=\dfrac{1/2}{\sqrt{3}/2}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$
  • [39:00] 🧩 Example – Given the Slope: $𝒎=-1$
    • Using $\tan(𝜃)=𝒎 \;\to\; -1=\tan(𝜃)$
    • Applying the inverse tangent: $𝜃=\tan^{-1}(-1)$
    • Interpreting the inverse tangent: finding the angle whose tangent is $-1$.
    • Using the unit circle to find angles where sine and cosine have equal magnitudes but opposite signs:
      • $135^\circ$ → Second quadrant
      • $315^\circ$ → Fourth quadrant
  • Importance of practicing these trigonometric concepts.
  • Trigonometry will be used extensively in Calculus.

Ｄｉｓｔａｎｃｅ Ｆｏｒｍｕｌａ

• [44:20] Distance Formula – [image]
  • Derived using two random points $(𝒙_1,𝒚_1)$ and $(𝒙_2,𝒚_2)$.
  • Forming a right triangle with sides $|𝒙_2-𝒙_1|$ and $|𝒚_2-𝒚_1|$.
  • Applying the Pythagorean Theorem: $𝒅^2=(𝒙_2-𝒙_1)^2+(𝒚_2-𝒚_1)^2$
  • Solving for distance 𝒅:
    • $\displaystyle \boxed{\Large 𝒅=\sqrt{(𝒙_2-𝒙_1)^2+(𝒚_2-𝒚_1)^2}}$
  • We omit the negative root because distance cannot be negative.