The Second Derivitive

By this point, you should be very familiar with the concept of derivitives. Previously, we discussed how the first derivitive of a function can be expressed as \(f'(x)\) or \(\cfrac{dy}{dx}\).

A function can be derived multiple times, each time this process reduces the functions polynomial degree (e.g. quartic becomes cubic, which becomes a quadratic, which becomes linear). The notation for the second derivitive is given as:

\(f''(x)\) or \(\cfrac{d^{2}y}{dx^{2}}\).

Makes sense, right?

The second derivitive will produce the gradient for the first derivitive of a function. This can be applied to various topics, such as determining if a point on a curve is a local max or min, linear motion, and optimisation.

2nd DE and Curves, finding local maxima, local minima, point of inflection

Let's look at this curve, \(y = x^2(x+3)\), for this topic:

First, we can find the coordinates for stationary points (local min/max) by solving for x when the first derivitive equals 0 (aka, when the graph is flat).

\(\begin{aligned} \frac{dy}{dx} &= x^2 + 2x^2 + 6x \\[5pt] &= 3x^2 + 6x \\[5pt] 3x^2 + 6x &= 0 \\[5pt] x^2 +2x &= 0 \\[5pt] (x+1)^2 - 1 &= 0 \\[5pt] (x+1)^2 &= 1 \\[5pt] x &= -1 \pm 1 \end{aligned} \)

Thus, these coordinates are either local minima or maxima: (0, 0), (-2, 4)

To discrimate the cordinates, we use the second derivative test:

If \(f''(x) < 0 \), the co-ordinate at x is a local maximum

If \(f''(x) > 0 \), the co-ordinate at x is a local minimum

Applying this to the problem:

\(\cfrac{dy}{dx} = 3x^2 + 6x\)

\(\cfrac{d^{2}y}{dx^{2}} = 6x + 6\)

When \(x = 0\):

\(\begin{aligned} \cfrac{d^{2}y}{dx^{2}} &= 6(0) + 6 \\ &= 6 \\ &> 0 \\ \end{aligned} \)

Hence (0, 0) is local minimum

When \(x = -2\):

\(\begin{aligned} \cfrac{d^{2}y}{dx^{2}} &= 6(-2) + 6 \\ &= -6 \\ &< 0 \\ \end{aligned} \)

Hence (-2, 4) is local maximum

To find the point of inflection:

Solve \(x\) for \(f''(x) = 0\):

\(\begin{aligned} f''(x) &= 6x+6 \\[5pt] 6x+6 &= 0 \\[5pt] x &= -1 \\[5pt] f(-1) &= 2 \end{aligned} \)

Hence, the point of inflection lies at (-1, 2)

Rectillinear Motion #1

Motion of an object travelling in a single dimension can be described as a function of time, where:

• \(s(t)\) determines the displacement from the origin (e.g. (0, 0)) (sometimes the function is expressed as \(x(t)\))
• \(v(t)\) determines the velocity of the body
• \(a(t)\) determines the acceleration of the body

These functions are related to eachother in this way:

Acceleration is thereby the second derivitive of \(x(t)\):

\(x''(t) = a(t)\)

Questions about Rectillinear Motion at this level will be pretty much just algebra with differentiation involved.

1) Given that a particles displacement can be expressed as \(s(x) = 3x^3 - ax^2\):

a. Solve for \(a\) when the particle is instantaneously accelerating at 14 \(cms^-2\) after 2 seconds of motion.

\(\begin{aligned} a(t) &= s''(x) \\[5pt] &= (9x^2 - 2ax)' \\[5pt] &= 18x - 2a \\[5pt] 14 &= 18(2) - 2a \\[5pt] -22 &= -2a \\[5pt] a &= 11 \end{aligned}\)

b. Find the velocity of the particle after 7 seconds

\(\begin{aligned} v(t) &= s'(x) \\[5pt] &= 9x^2 -22x \\[5pt] v(7) &= 9(7)^2 - 22(7) \\[5pt] &= 287 \ \text{cms}^{-1} \end{aligned} \)

Optimisation Problems

The topic of optimisation tackles the same problem: what is the least value of \(x\) to yield the largest value of \(y\), or vice versa.

This is usually as simple as finding the appropriate stationary point, the trickyness of optimisation comes from the broad variety of optimisation questions. Despite this, there is a static procedure to follow always:

1. Draw a picture if plausible (yes, marks may be assosiated to this)
2. Produce an equation/function that defines what metrics you are optimising (e.g. f(x))
3. Differentiatie and solve for (e.g.) f'(x) = 0 to locate the stationary points
4. Use the second derivitive test to justify the nature (maximum or minimum) of stationary points
5. Produce an answer that exists within any mentioned domain in the question
6. Use words to state your answer

Although you may originally be dealing with multiple variables, the question would in this case reveal a hint to cancel an additional variable: by making it equal some factor of another.

With optimisation especially, the more questions you practice the better you will be at the procedure. Don't learn the procedure for that question, but practice how to apply this procedure to a wide array of optimisation problems.

1) A closed soda can has height \(h\) and radius \(r\), given the can must have a volume of \(25 \pi \ cm^3\):

a. What is the formula for the surface area of the can?

b. Find a value of \(r\) that produces the lowest surface area.

\(\begin{aligned} A &= 50 \pi r^{-1} + 2 \pi r^2 \\[5pt] \cfrac{dA}{dr} &= -50 \pi r^{-2} + 4 \pi r \\[5pt] 0 &= -50 \pi r^{-2} + 4 \pi r \\[5pt] 50\pi r^{-2} &= 4\pi r \\[5pt] 12.5 \pi &= r^3 \\[5pt] r &= \sqrt[3]{12.5\pi} \\[5pt] &\approx 3.399 \ \text{cm} \end{aligned} \)

Lowest Surface Area Justification:

\(\begin{aligned} \frac{d^2 A}{dr^2} &= 100\pi r^{-3}+4\pi \\[5pt] &= 100\pi (-3.399)^{-3} + 4\pi \\[5pt] &= 20.5665 \\[5pt] &> 0 \end{aligned} \)

Therefore, this value of r produces the lowest surface area for the largest volume.

c. What is the smallest surface area?

\(\begin{aligned} A &= \cfrac{50\pi}{3.399} + 2\pi (3.399)^2 \\[5pt] &= 118.8044 \\[5pt] &= 118.8 \ \text{cm}^2 \end{aligned} \)