daviddoji/CalculusWithJuliaNotes.jl
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Merge branch 'main' of https://github.com/jverzani/CalculusWithJuliaNotes.jl

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jverzani 2023-05-08 13:34:17 -04:00
commit c3518dbba7
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4
quarto/derivatives/implicit_differentiation.qmd
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@ -440,7 +440,7 @@ To visualize, we plot implicitly and notice that:
```{julia} ```{julia}
K(x,y) = x^3 - y^3 K(x,y) = x^3 - y^3 - 3
implicit_plot(K, xlims=(-3, 3), ylims=(-3, 3)) implicit_plot(K, xlims=(-3, 3), ylims=(-3, 3))
``` ```
@ -451,7 +451,7 @@ The same problem can be done symbolically. The steps are similar, though the las
#| hold: true #| hold: true
@syms x y u() @syms x y u()
eqn = K(x,y) - 3 eqn = K(x,y)
eqn1 = eqn(y => u(x)) eqn1 = eqn(y => u(x))
dydx = solve(diff(eqn1,x), diff(u(x), x))[1] # 1 solution dydx = solve(diff(eqn1,x), diff(u(x), x))[1] # 1 solution
d2ydx2 = solve(diff(eqn1, x, 2), diff(u(x),x, 2))[1] # 1 solution d2ydx2 = solve(diff(eqn1, x, 2), diff(u(x),x, 2))[1] # 1 solution

2
quarto/derivatives/related_rates.qmd
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@ -182,7 +182,7 @@ How can this relationship be summarized? Well, let's go back to what we know, th
diff(A(t), t) diff(A(t), t)
``` ```
This should be clear: the rate of change, $dA/dt$, is increasing linearly, hence the second derivative, $dA^2/dt^2$ would be constant, just as we saw for the average rate of change. This should be clear: the rate of change, $dA/dt$, is increasing linearly, hence the second derivative, $d^2A/dt^2$ would be constant, just as we saw for the average rate of change.
So, for this problem, a constant rate of change in width and height leads to a linear rate of change in area, put otherwise, linear growth in both width and height leads to quadratic growth in area. So, for this problem, a constant rate of change in width and height leads to a linear rate of change in area, put otherwise, linear growth in both width and height leads to quadratic growth in area.