Envelope of Trajectories: Parabola of Safety

Hello! Welcome to our next lesson in kinematics.

In our last session, we explored the deep geometric connection between a projectile's path and the focus-directrix definition of a parabola. We saw that the directrix represents a line of constant energy and used this insight to solve complex problems with geometric elegance. You correctly identified that for a given launch speed $v_{0}$ , the focus of any possible trajectory must lie on a circle of radius $v_{0}^{2} /2 g$ around the launch point.

Today, we address the natural follow-up question: What is the boundary of all possible points a projectile can reach? This boundary is a fascinating concept known as the envelope of trajectories, or more dramatically, the parabola of safety. Our goal is to derive its equation from first principles, and as requested by the learning outcome, we will use the power of calculus to do so. This will perfectly suit your preference for rigorous, proof-based derivations.

1. Conceptualizing the Envelope

Imagine launching projectiles from the origin with a fixed speed $v_{0}$ but at all possible angles $θ$ . Each launch creates a parabolic path. The envelope is the curve that "wraps around" this entire family of parabolas, touching each one at a single point and defining the outermost boundary of the region that can be hit.

Envelope of Projectile Trajectories (Parabola of Safety) — This image illustrates the concept of the envelope of trajectories, also known as the 'parabola of safety'. The individual blue curves are trajectories for different launch angles, all with the same initial speed. The red parabola is the envelope, representing the boundary of all reachable points.

Before we derive its equation, it's crucial to understand the mathematical properties of an envelope. This will provide the foundation for our calculus-based methods.

The Parabola of Safety — Unveiling the Firework's Outer ...

The article 'The Parabola of Safety' provides an excellent conceptual introduction to envelope curves. It explains the key properties we will use in our derivation.

Please read Section 2, 'Mathematical Toolkit: Understanding Envelope Curves'. Focus on the 'Tangency', 'Extremal Property', and especially the 'Derivative Condition'. This formal condition is the key to our first derivation.

As you've just read, the envelope can be seen as the curve representing the maximum height $y$ achievable for any given horizontal distance $x$ . This extremal property is precisely what allows us to use calculus to find it.

2. Setting up the Mathematical Framework

We begin with the familiar trajectory equation for a projectile launched from the origin with speed $v_{0}$ and angle $θ$ :

y = x tan θ - \frac{g x ^{2}}{2 v _{0}^{2} cos ^{2} θ}

This equation describes a family of parabolas, with the angle $θ$ acting as the parameter that distinguishes each curve in the family. To make the mathematics more tractable, we will use two standard substitutions:

Use the identity $sec^{2} θ = 1 + tan^{2} θ$ .
Let $m = tan θ$ . The parameter $m$ can take any real value, covering all launch angles from $- π /2$ to $π /2$ .

Applying these, the trajectory equation becomes:

y (x, m) = m x - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + m^{2})

This equation, $y (x, m)$ , defines our family of curves. Our task is to find the envelope of this family. We will explore three distinct, yet convergent, methods to achieve this.

3. Derivation Method 1: The Formal Derivative Method

This is the most direct calculus-based method for finding an envelope. As stated in the resource you read, for a family of curves defined by an equation like $F (x, y, m) = y - y (x, m) = 0$ , the envelope is the set of points $(x, y)$ that satisfy both the original equation and the partial derivative of that equation with respect to the parameter $m$ being zero.

⎩ ⎨ ⎧ y - (m x - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + m^{2})) = 0 \frac{\partial}{\partial m} [y - (m x - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + m^{2}))] = 0

Since $y$ and $x$ are treated as constants when differentiating with respect to $m$ , the second equation simplifies to finding $\frac{\partial y ( x , m )}{\partial m} = 0$ . This is equivalent to finding the value of $m$ that maximizes the height $y$ for a given $x$ .

Let's perform the differentiation:

\frac{\partial y}{\partial m} = \frac{\partial}{\partial m} (m x - \frac{g x ^{2}}{2 v _{0}^{2}} - \frac{g x ^{2} m ^{2}}{2 v _{0}^{2}})

\frac{\partial y}{\partial m} = x - \frac{g x ^{2} m}{v _{0}^{2}}

Setting this derivative to zero gives us the condition for a point to be on the envelope:

x - \frac{g x ^{2} m}{v _{0}^{2}} = 0

Assuming $x \neq = 0$ , we can solve for $m$ :

m = \frac{v _{0}^{2}}{g x}

This expression gives us the specific launch parameter $m$ (and thus launch angle $θ$ ) required to reach the maximum possible height at a horizontal distance $x$ . To find the equation of the envelope, we substitute this value of $m$ back into our original trajectory equation $y (x, m)$ .

y_{e n v} = (\frac{v _{0}^{2}}{g x}) x - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + (\frac{v _{0}^{2}}{g x})^{2})

Now, we simplify this expression:

y_{e n v} = \frac{v _{0}^{2}}{g} - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + \frac{v _{0}^{4}}{g ^{2} x ^{2}})

y_{e n v} = \frac{v _{0}^{2}}{g} - (\frac{g x ^{2}}{2 v _{0}^{2}} + \frac{g x ^{2}}{2 v _{0}^{2}} \cdot \frac{v _{0}^{4}}{g ^{2} x ^{2}})

y_{e n v} = \frac{v _{0}^{2}}{g} - (\frac{g x ^{2}}{2 v _{0}^{2}} + \frac{v _{0}^{2}}{2 g})

y_{e n v} = \frac{v _{0}^{2}}{g} - \frac{v _{0}^{2}}{2 g} - \frac{g x ^{2}}{2 v _{0}^{2}}

Finally, we arrive at the equation for the envelope of trajectories:

y_{e n v} = \frac{v _{0}^{2}}{2 g} - \frac{g x ^{2}}{2 v _{0}^{2}}

This is the equation of a downward-opening parabola, which we will analyze further at the end. For another clear walkthrough of this exact method, you can refer to the first answer in the Stack Exchange thread "Envelope of Projectile Trajectories" (resource ID LINK, section 1).

4. Derivation Method 2: The Discriminant Method

This method is algebraic, but its conceptual foundation is deeply tied to the nature of the envelope as a boundary. It's a very powerful technique often used in competitive exams like the JEE Advanced.

Let's rearrange our trajectory equation $y = m x - \frac{g x ^{2}}{2 v _{0}^{2}} (1 + m^{2})$ into a standard quadratic form in the variable $m$ :

(\frac{g x ^{2}}{2 v _{0}^{2}}) m^{2} - x m + (y + \frac{g x ^{2}}{2 v _{0}^{2}}) = 0

This is a quadratic equation of the form $A m^{2} + B m + C = 0$ , where:

$A = \frac{g x ^{2}}{2 v _{0}^{2}}$
$B = - x$
$C = y + \frac{g x ^{2}}{2 v _{0}^{2}}$

Now, consider a point $(x, y)$ in the plane. For this point to be reachable by a projectile, there must be a real value of $m = tan θ$ that satisfies this equation.

If the discriminant $Δ = B^{2} - 4 A C > 0$ , there are two distinct real roots for $m$ . This means there are two different launch angles to hit the point $(x, y)$ . The point lies inside the envelope.
If $Δ < 0$ , there are no real roots for $m$ . The point $(x, y)$ is unreachable and lies outside the envelope.
If $Δ = 0$ , there is exactly one real root for $m$ . This means there is only one unique launch angle to hit the point $(x, y)$ . This point must lie on the boundary—that is, on the envelope itself.

Therefore, the equation of the envelope is given by the condition $Δ = 0$ .

The following video provides a clear and concise derivation using this discriminant method.

Trajectory Equation and Envelope Equation

The video 'Trajectory Equation and Envelope Equation' by Infinium clearly demonstrates the discriminant method. It starts by deriving the trajectory equation and then uses the discriminant of the quadratic in tan(alpha) to find the envelope.

Please watch from 03:30 to 06:30. The video uses z for the vertical axis instead of y, and alpha instead of theta, but the physics and mathematics are identical. Pay close attention to how setting the discriminant to zero defines the boundary condition.

By setting $B^{2} - 4 A C = 0$ , you will find that it leads to the exact same equation for the envelope:

(- x)^{2} - 4 (\frac{g x ^{2}}{2 v _{0}^{2}}) (y + \frac{g x ^{2}}{2 v _{0}^{2}}) = 0

Solving this for $y$ confirms our previous result.

Test your understanding!

Show the algebraic steps to solve $(- x)^{2} - 4 (\frac{g x ^{2}}{2 v _{0}^{2}}) (y + \frac{g x ^{2}}{2 v _{0}^{2}}) = 0$ for $y$ and verify that it yields the envelope equation we found earlier.

Show answer

Starting with the equation $Δ = 0$ :

x^{2} - \frac{2 g x ^{2}}{v _{0}^{2}} (y + \frac{g x ^{2}}{2 v _{0}^{2}}) = 0

Assuming $x \neq = 0$ , we can divide the entire equation by $x^{2}$ :

1 - \frac{2 g}{v _{0}^{2}} (y + \frac{g x ^{2}}{2 v _{0}^{2}}) = 0

Now, solve for the term in the parenthesis:

1 = \frac{2 g}{v _{0}^{2}} (y + \frac{g x ^{2}}{2 v _{0}^{2}})

\frac{v _{0}^{2}}{2 g} = y + \frac{g x ^{2}}{2 v _{0}^{2}}

Finally, isolate $y$ :

y = \frac{v _{0}^{2}}{2 g} - \frac{g x ^{2}}{2 v _{0}^{2}}

This confirms the result obtained via the calculus method.

5. The Beauty of Convergent Derivations

It is remarkable and deeply satisfying that two such different mathematical approaches—one rooted in differential calculus and optimization, the other in the algebraic theory of quadratic equations—converge to the identical result.

y_{e n v} = \frac{v _{0}^{2}}{2 g} - \frac{g x ^{2}}{2 v _{0}^{2}}

This is the equation of the Parabola of Safety. Let's analyze it:

It is a parabola because $y$ is a quadratic function of $x$ .
The negative sign on the $x^{2}$ term confirms it opens downwards.
Its vertex (and y-intercept) is at $x = 0$ , $y = \frac{v _{0}^{2}}{2 g}$ . This is the maximum possible height a projectile can reach, achieved by a vertical launch ( $θ = 9 0^{\circ}$ ).
Its x-intercepts (where $y = 0$ ) define the maximum horizontal range. Setting $y = 0$ , we get $x^{2} = \frac{v _{0}^{4}}{g ^{2}}$ , so $x = \pm \frac{v _{0}^{2}}{g}$ . The maximum range on level ground is $\frac{v _{0}^{2}}{g}$ , achieved at $θ = 4 5^{\circ}$ .

The fact that multiple lines of reasoning lead to this single, elegant formula is a testament to the consistency of mathematical physics, a theme I know you appreciate. The article "The Parabola of Safety" (resource LINK) you looked at earlier actually details four different methods, including completing the square and the intersection of neighboring curves, all yielding the same result. This convergence of methods strengthens our confidence in the physical truth of the model.

Conclusion

In this lesson, we have rigorously derived the equation for the envelope of projectile trajectories. By applying calculus and algebraic reasoning, we have moved beyond analyzing single trajectories to defining the entire boundary of possible motion.

Key Takeaways:

The envelope of trajectories for a fixed launch speed $v_{0}$ is the boundary of all reachable points. It is also called the parabola of safety.
This envelope can be derived using calculus by finding the launch parameter $m = tan θ$ that maximizes height $y$ for any given $x$ , and then substituting this condition back into the trajectory equation.
An alternative and powerful method involves reformulating the trajectory equation as a quadratic in $m$ and setting its discriminant to zero, which represents the boundary case of having a single unique trajectory to a point.
All valid methods converge to the same equation: $y = \frac{v _{0}^{2}}{2 g} - \frac{g x ^{2}}{2 v _{0}^{2}}$ . This is the equation of a downward-opening parabola.

Preview of the Next Lesson:
We have successfully derived the tool. In our next lesson, we will put it to use. We will apply the envelope equation to determine the boundary of reachable points and solve problems involving safe/unsafe regions. This will involve analyzing whether specific targets can be hit and exploring more complex scenarios, solidifying the practical application of today's theoretical work.