Geology: Calculus Video #2
3Blue1Brown Intro To Calculus Series Summary
Preface ✨
Hello everyone! My intentions for writing these articles are:
- Explain technical knowledge about geology in simple terms (to the public)
- Document my journey to getting an AD-T Geology degree in one year
- Store information and habits for my future self and others (in <7 minutes)
Coolio? Sweet. Enjoy the series :-)
notes as i watched the video 📹
p.s. — the video will be linked at the bottom of the article. cheers.
many problems in calculus will ask about velocity.
big whoop.
but how does velocity of an object depend on the distance vs. time functions (graphs, to be specific)?
but wtf does velocity actually mean?
velocity is a distance traveled over a given amount of time.
to find velocity, one must have a change in time.
and to do that,
u need two points in one time:
- two for different distances
- two for different times
let’s say the points of distance is 20.21 (added distance) meters & 20 (original distance) meters.
and the times are 3.01 seconds (for added time) & 3 seconds (for original time).
velocity is: (change in distance) / (change in time)
the change in distance is “ds”.
the change in time is “dt”.
the standard equation is:
ds / dt ≈ “rise / run” slope (aka “y / x” slope)
using the numbers:
(20.21 - 20) meters / (3.01 - 3) seconds ≈ …
not so fast.
because we made a change, aka we added an extra 0.01 second…
this also means we added a smidge more distance covered by the object.
so the true equation for a change in the velocity is this:
ds / dt (t) = [ s(t + dt) - s(t) ] / dt
imma explain what all these parts mean in english, not symbols lol:
this is meant to find the slope of a point on a curve.
the slope is the “y divided by x” value, kinda (just kinda) like in algebra, where u find its steepness. also known as the “rise over run” terminology.
in reality, the slope is a line that is tangent (or barely touches) the curve of a line/graph at a specific point.
ds / dt (t) → how much did the distance change due to the extra amount of seconds (time) we added?
p.s. - math geeks will say “with respect to time” or “with respect to ‘x’”
s(t + dt) → it means: “seconds already traveled [s(t…] plus the extra time we added […+ dt)]”
s(t) → subtract the original time traveled
d(t) → the added time
ok, now let’s do grant’s example, but explained in my words, ok?
so, here it is.
using the definitions of above,
some of the pieces are given to us.
s(t) = t³
t = 2
now, to find out about how much change we will add to the graph…
ok, it will be that much.
problem?
we don’t know, it’s not given to us.
but we can start with what we got:
ds / dt (t) …
(t) will be 2.
so using the equation from above, (and thx grant for the picture too)
this is how it will look like for us:
ds / dt (2) = [ (2 + dt)³ - (2)³ ] / dt*
* let’s get “dt” to as close to zero as possible
y?
better approximation of the slope.
for now, we’ll ignore “ds / dt (2)”, that won’t change for a while.
for ([ (2 + dt)³ — (2)³ ] / dt) however, this will occur (step-by-step):
- distribute exponent & expand the equation
result: [ 2³ + {3(2)² * dt} + {3(2)(dt)²} + (dt)³ ] - 2³/ dt
2. cancel out all equal factors (like, 2³)
result: [ 3(2)² * dt} + {3(2)(dt)²} + (dt)³ ] / dt
3. cancel out “dt” in the denominator by dividing it with other “dt” in the numerator
result: [ 3(2)² + 3(2)(dt) + (dt)² ]
but remember, as “dt” gets closer to zero (which we are doing rn),
we can ignore any part of the equation that has a “dt” in it.
y?
as we push the slope to zero (or change, “dt” to zero)…
we’re basically saying, nothing changed. we didn’t add anything to the original curve/area.
so zero change, or “dt”, exists.
anyway, once this happens…
4. plug in “0” wherver u see a “dt”
result: [ 3 * (2)² + 0 + 0 ]
now, we bring back the “ds / dt (2)”, we’re almost done.
5. simplify and solve.
result: ds / dt (2) = 12
the slope (aka the derivative, or the change) of the graph where the time was equal to two seconds is 12 meters per second (distance traveled divided by the seconds passed).
that’s how steep it is (aka how the rate of change* for “ds” will be affected when “dt” is 2)
*the words “rate of change for” can cautiously be synonymous with “how much”
now here’s a plot twist.
look at the result from part #4.
result: [ 3 * (2)² + 0 + 0 ]
it’s the same as: 3(2)²
remember how “t = 2”?
plug in the big (base) “2” in “3(2)²” for “t”
now u have: “3(t)²”
that’s our new function after all that math.
and our original function was: s(t) = t³
and it went from t³ to 3t².
that’s what all that algebra is for.
to go from the original function to a specific point in time, the derivative.
one image example:
the exponent pulls down to the front.
and the exponent number goes down by 1.
some more basic examples:
won’t be that easy, definitely can go many levels up.
but that’s what a derivative is in its simplicity.
that’s as simple as i can deliver.
no more simpler tho, or it won’t make sense.
and thx grant lol
CJ
© 2023–2100 by Carlos Manuel Jarquín Sánchez. All Rights Reserved.
