Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　１．５

Sｌｏｐｅ　Oｆ　ａ　Cｕｒｖｅ，　Vｅｌｏｃｉｔｙ，　ａｎｄ　Rａｔｅｓ　Oｆ　Cｈａｎｇｅ

Ｔａｎｇｅｎｔ ｌｉｎｅ ａｎｄ ｒａｔｅ ｏｆ ｃｈａｎｇｅ

• 0:50. Introduction.
  • Main objective of calculus:
    • Find the slope of the tangent line to a curve at a point.
      • Equivalent statement: Find the slope of a curve at a point.
  • Using limits to find the slope of the tangent line to a curve at a point.
  • Importance of understanding the mathematical concept, not just the formulas.
• 3:20. Creating the formula for the slope of a curve from the slope of a secant line (that conect two points).
  • 4:03. Definitions: – [📷image]
    • A curve
    • P: a poin to find the slope of a curve at a point P.
    • Q: a point to stablish the secant line from P to Q.
    • 𝒙₀: 𝒙 coordinate of P and 𝒙₀ + h 𝒙 coordinate of Q
    • h: distance from P to Q
    • The idea is take Q really relly really close to P that is the idea of limit.
  • 6:44. Calculating the 𝒚-values of the points: 𝒇(𝒙₀) and 𝒇(𝒙₀ + h). – [📷image]
  • 7:24. Reminder of the goal: finding the slope of the curve at point P.
  • 8:32. Calculating the slope of the secant line ($\Delta \mathit{y} / \Delta \mathit{x}$) using points P and Q. – [📷image]
    • $\Delta \mathit{x}=x_0+h-x_0=h$ and $\Delta \mathit{y}=f(x_0+h)-f(x_0)$
      • $m=\displaystyle \dfrac{f(x_0+h)-f(x_0)}{h}$
  • 14:00. Introducing the concept of a limit to bring Q closer to (infinitely smaller) P: this is (h → 0). – [📷image]
    • Transition from the slope of the secant (m) to the slope of the tangent ($m_{\text{tan}}$)
      using the limit:
      • $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
  • 17:30. Explanation of why we cannot simply set h = 0 (undefined).
• 18:33. 🧩 Example 1 – Finding the equation of the tangent line to: 𝒚 = x² at (1, 1) – [📷image-1] – [📷image-2]
  • 19:42. Steps to find the equation of the tangent line.
    • Identification of 𝒇(𝒙) = 𝒙² and x₀ = 1.
    • Calculating 𝒇(𝒙₀ + h) = 𝒇(1 + h) = $(1+h)^{2}$.
    • Calculating 𝒇(𝒙₀) = 𝒇(1) = $1$.
    • Take the limit $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
      • $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{(1+h)^{2}-1}{h}=\displaystyle \lim_{h\to 0}\dfrac{1+h+2h^{2}-1}{h} =\displaystyle \lim_{h\to 0}(2+h)=0$
    • Hence, the slope of the tangent line is $2$.
    • 28:20. Using the point-slope form to find the equation of the tangent line at (1,1).
      • $\mathit{y}-y_1=m(x-x_1)$.
      • The equation of the tangent line is $\mathit{y}=2x-1$.
• 30:55. 🧩 Example 2 – Finding the equation of the tangent line to 𝒚 = 3/x at (3, 1) – [📷image-1] – [📷image-2]
  • Steps to restating the goal and using the formula for the slope of the tangent.
    • Identification of 𝒇(𝒙) = $3/x$ and $x_0=3$.
    • Calculating 𝒇(𝒙₀ + h) = 𝒇(3 + h) = $\dfrac{3}{3+h}$.
    • Calculating 𝒇(𝒙₀) = 𝒇(3) = $1$.
    • Take the limit $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
      • $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{\dfrac{3}{3+h}-1}{h}=\displaystyle \lim_{h\to 0}\dfrac{-\,h/(3+h)}{h} =\displaystyle \lim_{h\to 0}\left(-\dfrac{1}{3+h}\right)=-\dfrac{1}{3}$
    • Hence, the slope of the tangent line is $-1/3$.
    • 41:20. Using the point-slope form to find the equation of the tangent line at (1,1).
      • $\mathit{y}-y_1=m(x-x_1)$.
      • The equation of the tangent line is $\mathit{y}=-\dfrac{1}{3}x+2$.
• 44:57. 🧩 Example 3 – Finding the slope of tangent lines to 𝒚 = √x at any point – [📷image]
  • Objective: derive a general formula for the slope of the tangent to 𝒚 = √x at any point x.
    • Identification of 𝒇(𝒙) = $\sqrt{x}$.
    • Calculating 𝒇(x + h) = $\sqrt{x+h}$.
    • Take the limit $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
      • $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$
      • $\displaystyle \lim_{h\to 0}\dfrac{(x+h)-x}{h\cdot\big(\sqrt{x+h}+\sqrt{x}\big)}$ multiplicate by de conjugat $\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
      • $\displaystyle \lim_{h\to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}}$
    • Hence, the slope of the tangent line 𝐚𝐭 𝐚𝐧𝐲 𝐩𝐨𝐢𝐧𝐭 is $\dfrac{1}{2\sqrt{x}}$
    • 55:00. Obtaining the general formula for the slope of the tangent: $m=\dfrac{1}{2\sqrt{x}}$.
      • Explanation that this formula allows computing the slope at any point on the curve.
    • 57:35. Sharp point:
      • At a sharp point on a curve, 𝘁͟𝗵͟𝗲͟ ͟𝘀͟𝗹͟𝗼͟𝗽͟𝗲͟𝘀͟ ͟𝗳͟𝗿͟𝗼͟𝗺͟ ͟𝘁͟𝗵͟𝗲͟ ͟𝗹͟𝗲͟𝗳͟𝘁͟ ͟𝗮͟𝗻͟𝗱͟ ͟𝗿͟𝗶͟𝗴͟𝗵͟𝘁͟ ͟𝘀͟𝗶͟𝗱͟𝗲͟𝘀͟ ͟𝗮͟𝗿͟𝗲͟ ͟𝗱͟𝗶͟𝗳͟𝗳͟𝗲͟𝗿͟ｅ͟𝗻͟𝘁͟. This discrepancy causes the limit defining the tangent slope ($m_{\text{tan}}$) to fail, as it does not exist uniquely. Consequently, an infinite number of tangent lines with varying slopes can be drawn through that sharp point.

Ａｐｐｌｉｃａｔｉｏｎｓ of Ａｖｅｒａｇｅ ａｎｄ ｉｎｓｔａｎｔａｎｅｏｕｓ ｖｅｌｏｃｉｔｙ

• 58:54. Introduction to applications.
  • 𝗔̳𝘃̳𝗲̳𝗿̳𝗮̳𝗴̳𝗲̳ ̳𝘃̳𝗲̳𝗹̳𝗼̳𝗰̳𝗶̳𝘁̳𝘆̳ ($V_{\text{ave}}$) corresponds to the slope of the secant line between two points, calculated as:
    • $m_{\text{secant}}=\displaystyle \dfrac{f(x_0+h)-f(x_0)}{h}$
  • 𝗜̳𝗻̳𝘀̳𝘁̳𝗮̳𝗻̳𝘁̳𝗮̳𝗻̳𝗲̳𝗼̳𝘂̳𝘀̳ ̳𝘃̳𝗲̳𝗹̳𝗼̳𝗰̳𝗶̳𝘁̳𝘆̳ ($V_{\text{inst}}$) matches the slope of the tangent line at a single point, given by:
    • $m_{\text{tan}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
• 59:03. 𝗔̳𝘃̳𝗲̳𝗿̳𝗮̳𝗴̳𝗲̳ ̳𝘃̳𝗲̳𝗹̳𝗼̳𝗰̳𝗶̳𝘁̳𝘆̳
  • Distance traveled over elapsed time.
    • $V_{\text{ave}}=$ displacement / time $=\displaystyle \dfrac{f(t_0+h)-f(t_0)}{h}$
  • 1:01:26. Relationship between the average velocity formula and the slope formula.
    • Both formulas represent the slope between two points.
    • Average velocity is the slope of the secant line on a position vs. time graph.
  • 1:02:20. 🧩 Example: Finding average velocity of $s(t)=1+3t-2t^{2}$ over the interval [1,3] – [📷image]
    • Identification of $t_0=1$ and $h=2$.
    • Calculating $s(t_0+h)=s(3)=-8$.
    • Calculating $s(t_0)=s(1)=2$.
    • Computing the average velocity: $\displaystyle \dfrac{s(3)-s(1)}{2}=\dfrac{-8-2}{2}=-5$.
  • 1:08:55. Introduction to the difference between velocity and speed.
    • Average Velocity:
      • Type: Vector
      • Includes: Direction and magnitude
      • Use:Indicates the overall change in position over time, providing information about both the rate of displacement and the direction.
    • Speed:
      • Type: Scalar
      • Includes: Only magnitude
      • Use: Measures the total distance traveled over a period of time without regard to direction.
• 1:10:02. 𝗜̳𝗻̳𝘀̳𝘁̳𝗮̳𝗻̳𝘁̳𝗮̳𝗻̳𝗲̳𝗼̳𝘂̳𝘀̳ ̳𝘃̳𝗲̳𝗹̳𝗼̳𝗰̳𝗶̳𝘁̳𝘆̳ – [📷image]
  • Velocity at a specific instant in time.
    • 1:11:25. 𝙃𝙤𝙬 𝙢𝙪𝙘𝙝 𝙩𝙞𝙢𝙚 𝙚𝙡𝙖𝙥𝙨𝙚𝙨 𝙞𝙣 𝙖𝙣 𝙞𝙣𝙨𝙩𝙖𝙣𝙩?
  • 1:12:55. Instantaneous velocity is found by letting h approach zero in the average velocity formula.
    • 1:13:23. This involves using a limit.
      • How we make average velocity instant?
        • $\displaystyle \lim_{h\to 0}\dfrac{f(t_0+h)-f(t_0)}{h}$; $h=$ time
  • 1:14:20. Relationship between instantaneous velocity and the slope of a curve at a point.
    • Both represent the instantaneous rate of change.
  • 1:15:23. 🧩 Example: Finding instantaneous velocity of $s(t)=500-16t^{2}$ after 5 seconds. – [📷image-1] – [📷image-2]
    • $\displaystyle \lim_{h\to 0}\dfrac{s(t_0+h)-s(t_0)}{h}$
    • Identification of $t_0=5$.
    • Calculating $s(t_0+h)=s(5+h)=500-16(5+h)^{2}$.
    • Calculating $s(t_0)=s(5)=100$.
    • 1:27:00. Substituting into the instantaneous velocity formula and simplifying.
      • Take the limit
        • $\displaystyle \lim_{h\to 0}\dfrac{s(t_0+h)-s(t_0)}{h}=V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{500-16(25+10h+h^{2})-100}{h}$
      • Simplify the expression:
        • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{500-400-160h-16h^{2}-100}{h}$
        • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{-160h-16h^{2}}{h}$
      • Cancel the common factor of $h$:
        • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\big(-160-16h\big)$
      • Evaluate the limit as $h$ approaches $0$:
        • $V_{\text{inst}}=-160\ \text{ft/sec}$
  • 1:32:37. Computing the instantaneous velocity in general (for any time $t$ so ̵t̵₀̵ ̵=̵ ̵5̵) and obtaining the formula: $v(t)=-32t$. – [📷image]
    • Set up the formula for instantaneous velocity:
      • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{s(T+h)-s(T)}{h}$
    • Substitute the given position functions:
      • $s(T+h)=500-16(T+h)^{2}$
      • $s(T)=500-16T^{2}$
    • Substitute the expressions:
      • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{500-16(T^{2}+2Th+h^{2})-(500-16T^{2})}{h}$
    • Expand and simplify:
      • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{500-16T^{2}-32Th-16h^{2}-500+16T^{2}}{h}$
    • Cancel common terms:
      • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{-32Th-16h^{2}}{h}$
    • Factor out the common term $(h)$:
      • $V_{\text{inst}}=\displaystyle \lim_{h\to 0}\big(-32T-16h\big)$
    • Evaluate the limit as $h$ approaches $0$:
      • $V_{\text{inst}}=-32T$
    • Substitute $T=5$ seconds:
      • $V_{\text{inst}}=-32(5)$
      • $V_{\text{inst}}=-160\ \text{ft/sec}$
  • 1:38:29. Explanation of the difference between velocity and speed.
    • Speed is the absolute value of velocity.

Ａｖｅｒａｇｅ ａｎｄ Ｉｎｓｔａｎｔａｎｅｏｕｓ ｒａｔｅ ｏｆ ｃｈａｎｇｅ

• 1:39:23. Connection between slope and rate of change.
  • The slope is a rate of change.
• 1:39:59. The average rate of change is the slope of the secant line between two points.
• 1:41:58. The instantaneous rate of change is the slope of the curve at one point.
• 1:43:10. Rate of change:
  • Average rate of change:
    • $r_{\text{av}}=\displaystyle \dfrac{f(x_0+h)-f(x_0)}{h}$
  • instantaneous rate of change:
    • $r_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
• 1:46:37. 🧩 Example: Finding the average and instantaneous rate of change of a curve 𝒇(𝒙) = 3𝒙² - 4, r_ave:[2,5] and r_inst:-2
  • Average rate of change: $r_{\text{av}}=\displaystyle \dfrac{f(x_0+h)-f(x_0)}{h}$
    • $f(x_0)=f(2)$
    • $f(x_0+h)=f\big(2+(5-2)\big)=f(5)$
    • $r_{\text{av}}=\displaystyle \dfrac{f(x_0+h)-f(x_0)}{h}=\dfrac{f(5)-f(2)}{3}=\dfrac{63}{3}=21$
  • instantaneous rate of change: $r_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$
    • $f(x_0)=3x_0^{2}-4$
    • $f(x_0+h)=f\big(3(x_0+h)^{2}-4\big)=3x_0^{2}+6x_0h+h^{2}-4$
    • $r_{\text{inst}}=\displaystyle \lim_{h\to 0}\dfrac{\big(3x_0^{2}+6x_0h+h^{2}-4\big)-\big(3x_0^{2}-4\big)}{h}=\displaystyle \lim_{h\to 0}\dfrac{6x_0h+h^{2}}{h} =\displaystyle \lim_{h\to 0}\big(6x_0+h\big)=6x_0$
    • $r_{\text{inst}\_\!x_0:-2}=-12$