2.1 Notes on The Tangent Problem and The Velocity Problem

The Tangent Problem

  • The tangent is linked to the Latin word tangens, meaning “touching.” A tangent to a curve is a line that touches the curve and follows the same direction as the curve at the point of contact.
  • Circle intuition: a tangent to a circle is a line that intersects the circle once and only once. This simple definition works for circles but not for more general curves.
  • For more complicated curves, a line that looks tangent but intersects the curve twice (as in Figure 1(b)) is not a tangent in the precise sense.
  • The precise definition comes from the limiting process: a tangent line is the limit of secant lines as the second point Q on the curve approaches the contact point P.
  • Secant line: a line through two points P and Q on the curve (Q ≠ P). The slope of the secant is
    m<em>PQ=y</em>Q−y<em>Px</em>Q−xP.m<em>{PQ} = \frac{y</em>Q - y<em>P}{x</em>Q - x_P}.
  • The slope of the tangent is the limit of these secant slopes as Q approaches P along the curve. Symbolically:
    m = oxed{ rac{dy}{dx}igg|{x=xP} = rac{ ext{d}y}{ ext{d}x}igg|{x=xP} =
    frac{dy}{dx}{x0} = ext{lim}{Q o P} rac{yQ - yP}{xQ - xP} } or, in the function form, for a function $y=f(x)$ near $x0$:
    m=extlim<em>xox</em>0f(x)−f(x<em>0)x−x</em>0.m = \boxed{ ext{lim}<em>{x o x</em>0} \frac{f(x) - f(x<em>0)}{x - x</em>0} } .
  • Example 1: Tangent to the parabola $y = x^2$ at $P(1,1)$
    • We don’t know the slope directly; instead, we approximate the slope by choosing a nearby point $Q$ on the parabola and computing the secant slope $m_{PQ}$.
    • If $Q=(x, x^2)$, then
      mPQ=x2−1x−1=x+1.m_{PQ} = \frac{x^2 - 1}{x - 1} = x + 1.
    • For $Q=(1.5, 2.25)$:
      mPQ=2.25−11.5−1=1.250.5=2.5.m_{PQ} = \frac{2.25 - 1}{1.5 - 1} = \frac{1.25}{0.5} = 2.5.
    • Tables show values of $m_{PQ}$ for $x$ approaching 1:
    • as $x o 1$, $m_{PQ} o 2$ (e.g., 2.5, 2.1, 2.01, 2.001, etc.).
    • Thus the slope of the tangent is m=extlimxo1x2−1x−1=2.m = ext{lim}_{x o 1} \frac{x^2 - 1}{x - 1} = 2.
    • The equation of the tangent line through $P(1,1)$ with slope $m=2$ (point-slope form):
      y−y<em>1=m(x−x</em>1) extsoy−1=2(x−1) extwhichgivesy=2x−1.y - y<em>1 = m (x - x</em>1) \ ext{so } y - 1 = 2(x - 1) \ ext{which gives } y = 2x - 1.
  • Figure interpretation: Figure 3 shows $Q$ approaching $P$ from the right and from the left; as $Q$ moves along the parabola toward $P$, the secant lines rotate about $P$ and converge to the tangent line $oldsymbol{ell}$.
  • Summary/Key takeaways:
    • Tangent line at a point is the limit of secant slopes as the second point on the curve approaches the contact point.
    • The slope of the tangent is given by the derivative at that point, i.e., $m = f'(x_0)$ if $y=f(x)$.
    • The tangent line can be constructed using $y - y1 = m (x - x1)$ with the appropriate $m$ and point $(x1,y1)$.

The Velocity Problem

  • Historical note: Galileo observed that the distance fallen by a freely falling body is proportional to the square of the time elapsed (neglecting air resistance):
    s(t)extisproportionaltot2ext,i.e.s(t)=frac12gt2extwithgexttheaccelerationduetogravity.s(t) ext{ is proportional to } t^2 ext{, i.e. } s(t) = frac{1}{2} g t^2 ext{ with } g ext{ the acceleration due to gravity.}
  • Problem setup (Example 3): A ball is dropped from a height (e.g., 450 m above ground). Find the velocity after 5 seconds.
    • The instantaneous velocity at a single time $t$ cannot be read directly from a single time value; we approximate it by the average velocity over a short time interval starting at $t$.
  • Solution approach:
    • Use the model $s(t) = frac{1}{2} g t^2$ with $g oxed{rac{9.8}{ ext{m}}{ ext{s}^2}}$ (neglecting air resistance).
    • The average velocity on the interval $[t, t+ riangle t]$ is
      vextavg=s(t+rianglet)−s(t)rianglet.v_{ ext{avg}} = \frac{s(t+ riangle t) - s(t)}{ riangle t}.
    • For $t=5$ s and $ riangle t = 0.1$ s:
    • $s(5) = frac{1}{2} g (5)^2 = 122.5$ m,
    • $s(5.1) = frac{1}{2} g (5.1)^2 = 127.449$ m,
    • vextavg=127.449−122.50.1=49.49extm/s.v_{ ext{avg}} = \frac{127.449 - 122.5}{0.1} = 49.49 ext{ m/s}.
    • Similar calculations with shorter intervals yield:
    • $5 o 5.05$: $v_{ ext{avg}} = 49.245$ m/s
    • $5 o 5.01$: $v_{ ext{avg}} = 49.049$ m/s
    • $5 o 5.001$: $v_{ ext{avg}} = 49.0049$ m/s
    • Observing that as the time interval shortens, the average velocity approaches a limiting value. The instantaneous velocity at $t=5$ is defined as the limit of these averages:
      v(5)=extlim<em>riangleto0v</em>extavg=dsdt=gt.v(5) = \boxed{ ext{lim}<em>{ riangle t o 0} v</em>{ ext{avg}} = \frac{ds}{dt} } = g t.
    • Therefore, at $t=5$ seconds,
      v(5)=gimes5 v(5)=9.8imes5=49extm/s.v(5) = g imes 5 \ \boxed{v(5) = 9.8 imes 5 = 49 ext{ m/s}}.
  • Conceptual takeaway:
    • The instantaneous velocity is the derivative of position with respect to time, obtained as a limit of average velocities over shrinking time intervals.
    • The velocity function for free fall (neglecting air resistance) is $v(t) = rac{ds}{dt} = g t$, giving $v(5) = 49$ m/s when $g \, ext{≈}\ 9.8$ m/s$^2$.
  • Real-world relevance and connections:
    • This exemplifies the fundamental idea of derivatives as instantaneous rates of change.
    • It links a physical measurement (velocity) to a mathematical limit process (limit of averages).
  • Notation recap:
    • Position: $s(t)$, with $s(0)$ as the starting height fallen.
    • Velocity: $v(t) = rac{ds}{dt}$.
    • Acceleration in this model: $a = rac{dv}{dt} = g$ (constant).