typos

quarto/integrals/arc_length.qmd

@@ -761,7 +761,7 @@ This comes from solving the projectile motion equations with a drag force *propo
 
 ```{julia}
 #| hold: true
-@syms gₑ::postive, k::postive, v₀::positive, θ::postive, x::postive
+@syms gₑ::positive, k::positive, v₀::positive, θ::positive, x::positive
 ex = (gₑ/(k*v₀*cos(θ)) + tan(θ))*x + gₑ/k^2 * log(1 - k/(v₀*cos(θ))*x)
 diff(ex, x, x), diff(ex, x, x, x,)
 ```

quarto/integrals/area_between_curves.qmd

@@ -824,11 +824,11 @@ Doubling the answer above gives a value that Galileo had struggled with for many
 #| echo: false
 imgfile="figures/companion-curve-bisects-rectangle.png"
 caption = """
-Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is `2pi`.
+Roberval, avoiding a trigonometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is `2pi`.
 """
 # ImageFile(:integrals, imgfile, caption)
 nothing
 ```
 
-![Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is $2\pi$.
+![Roberval, avoiding a trignoometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is $2\pi$.
 ](./figures/companion-curve-bisects-rectangle.png)

quarto/integrals/center_of_mass.qmd

@@ -63,7 +63,7 @@ $$
 Writing $w_i = m_i / (m_1 + m_2 + \cdots + m_n)$, we get the center of mass is just a weighted sum: $w_1 x_1 + \cdots + w_n x_n$, where the $w_i$ are the relative weights.
 
 
-With some rearrangment, we can see that the center of mass satisfies the equation:
+With some rearrangement, we can see that the center of mass satisfies the equation:
 
 
 $$

quarto/integrals/mean_value_theorem.qmd

@@ -29,7 +29,7 @@ $$
 \frac{1}{b-a} \int_a^b f(x) dx.
 $$
 
-If $f$ is a constant, this is just the contant value, as would be expected. If $f$ is *piecewise* linear, then this is the weighted average of these constants.
+If $f$ is a constant, this is just the constant value, as would be expected. If $f$ is *piecewise* linear, then this is the weighted average of these constants.
 
 
 #### Examples

quarto/integrals/surface_area.qmd

@@ -85,7 +85,7 @@ To see why this formula is as it is, we look at the parameterized case, the firs
 Let a partition of $[a,b]$ be given by $a = t_0 < t_1 < t_2 < \cdots < t_n =b$. This breaks the curve into a collection of line segments. Consider the line segment connecting $(g(t_{i-1}), f(t_{i-1}))$ to $(g(t_i), f(t_i))$. Rotating this around the $x$ axis will generate something approximating a disc, but in reality will be the frustum of a cone. What will be the surface area?
 
 
-Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and layed out flat, it would form a sector of a circle, whose area would be $R^2\gamma/2$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
+Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and laid out flat, it would form a sector of a circle, whose area would be $R^2\gamma/2$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
 
 
 Relating this to our values in terms of $f$ and $g$, we have $r_1=f(t_i)$, $r_0 = f(t_{i-1})$, and $\sin(\theta) = \Delta f / \sqrt{(\Delta g)^2 + (\Delta f)^2}$, where $\Delta f = f(t_i) - f(t_{i-1})$ and similarly for $\Delta g$.

quarto/integrals/volumes_slice.qmd

@@ -242,17 +242,17 @@ cone = integrate(pi*R^2, (y, 0, h))
 It is not unusual to parameterize a cone by the angle $\theta$ it makes and the height. Since $r/h=\tan\theta$, this gives the formula $V = \pi/3\cdot h^3\tan(\theta)^2$.
 
 
-The frustum of a cone is simply viewed as a cone with its top cut off. If the original height would have been $h_0$ and the actual height $h_1$, then the volume remaining is just $\int_0^{h_1}  \pi r(y)^2 dy$. We can see the formula for the frustrum of the right cone:
+The frustum of a cone is simply viewed as a cone with its top cut off. If the original height would have been $h_0$ and the actual height $h_1$, then the volume remaining is just $\int_0^{h_1}  \pi r(y)^2 dy$. We can see the formula for the frustum of the right cone:
 
 ```{julia}
 @syms h0
-frustrum = integrate(pi*R^2, (y, 0, h0))
+frustum = integrate(pi*R^2, (y, 0, h0))
 ```
 
 Simplifying, we can see this volume can be expressed using the ratio $(h_0/h_1)$:
 
 ```{julia}
-frustrum - cone * ( 3h0/h - 3(h0/h)^2 + (h0/h)^3) |> simplify
+frustum - cone * ( 3h0/h - 3(h0/h)^2 + (h0/h)^3) |> simplify
 ```