I'm confused about the following problem. Calculus by Spivak, 3rd Edition, Chapter 11, exercise 30:
A cannon ball is shot from the ground with velocity $v$ at and angle $\alpha$ (Figure 31) so that it has a vertical component of velocity $v\sin\alpha$ and a horizontal component $v\cos\alpha$. Its distance $s(t)$ above the ground obeys the law $s(t) = -16t^2 + (v\sin\alpha)t$, while its horizontal velocity remains constantly $v\cos\alpha$.
(a) Show that the path of the cannon ball is a parabola (find the position at each time $t$ and show that these points lie on a parabola).
So, we have $$x^\prime(t)= v\cos\alpha$$ $$x(t) = (v\cos\alpha)t + c$$
Assuming $x(0) = 0$, we have
$$x(t) = (v\cos\alpha)t$$
Now, if $\cos\alpha \neq 0$, we can write: $$t = \frac{x}{v\cos\alpha}$$
Per the text, the vertical position of the cannon ball is given by: $$y(t) = -16t^2 + (v\sin\alpha)t$$
Substituting $\frac{x}{v\cos\alpha}$ in for $t$ we get:
$$y = -16\left(\frac{x}{v\cos\alpha}\right)^2 + (v\sin\alpha)\left(\frac{x}{v\cos\alpha}\right)$$ $$y = \frac{-16x^2}{v^2\cos^2\alpha} + (\tan\alpha)x$$
For $\cos\alpha \neq 0$, this is the equation of a parabola.
However, this differs from Spivak's solution:
- (a) The position at time $t$ is $$((v\cos\alpha)t, -16t^2 + (v\sin\alpha)t).$$ If $\cos\alpha = 0$, so that the cannon ball is shot straight up, then these points all lie on a straight line. If $\cos\alpha \neq 0$, then the set of all such points is equal to the set of all points $$\left(t, -\frac{16t^2}{v\cos\alpha} + (\tan\alpha)t\right),$$ so the path of the cannon ball lies on the graph of $$f(x) = \frac{-16x^2}{v\cos\alpha} + (\tan\alpha)x,$$ which is the graph of a parabola.
Notice that my equation for $y$ in terms of $x$ differs from Spivak's: $$\frac{-16x^2}{v^2\cos^2\alpha} + (\tan\alpha)x \neq \frac{-16x^2}{v\cos\alpha} + (\tan\alpha)x$$
I think maybe he made a mistake going from here: $$((v\cos\alpha)t, -16t^2 + (v\sin\alpha)t)$$ to here: $$\left(t, -\frac{16t^2}{v\cos\alpha} + (\tan\alpha)t\right)$$
He multiplies the $x$ and $y$ coordinates by $\frac{1}{v\cos\alpha}$. This is incorrect, right?
For example, if you have:
$$f(x) = x^2$$
this would correspond with the points $(x, x^2)$, and includes the point $(a,a^2)$. This is not the same as the function associated with $(ax, ax^2)$ ($a \neq 0$), which includes the point $(a,a)$.