Concavity is determined by the second derivative of \(y\) with respect to \(x\text{,}\) \(\frac{d^2y}{dx^2}\text{,}\) so we compute that here following Key Idea 9.3.9.

In Example 9.3.3, we found \(\ds\frac{dy}{dx} = \frac{2t+6}{10t-6}\) and \(\fp(t) = 10t-6\text{.}\) So: \begin{align*} \frac{d^2y}{dx^2} \amp = \frac{d}{dt}\left[\frac{2t+6}{10t-6}\right]\Bigg/(10t-6)\\ \amp = -\frac{72}{(10t-6)^2}\Bigg/(10t-6)\\ \amp = -\frac{72}{(10t-6)^3}\\ \amp = -\frac{9}{(5t-3)^3} \end{align*}

The graph of the parametric functions is concave up when \(\frac{d^2y}{dx^2} > 0\) and concave down when \(\frac{d^2y}{dx^2} \lt 0\text{.}\) We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.

As the numerator of \(\ds -\frac{9}{(5t-3)^3}\) is never 0, \(\frac{d^2y}{dx^2} \neq 0\) for all \(t\text{.}\) It is undefined when \(5t-3=0\text{;}\) that is, when \(t= 3/5\text{.}\) Following the work established in Section 3.4, we look at values of \(t\) greater/less than \(3/5\) on a number line:

Reviewing Example 9.3.3, we see that when \(t=3/5=0.6\text{,}\) the graph of the parametric equations has a vertical tangent line. This point is also a point of inflection for the graph, illustrated in Figure 9.3.11.

We need to compute \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\text{.}\) \begin{equation*} \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{\cos(t) }{1/(2\sqrt{t})} = 2\sqrt{t}\cos(t) . \end{equation*} \begin{equation*} \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{x'(t)} = \frac{\cos(t) /\sqrt{t}-2\sqrt{t}\sin(t) }{1/(2\sqrt{t})}=2\cos(t) -4t\sin(t) . \end{equation*}

The points of inflection are found by setting \(\frac{d^2y}{dx^2}=0\text{.}\) This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have “nice” solutions.

In Figure 9.3.13(a) we see a plot of the second derivative. It shows that it has zeros at approximately \(t=0.5,\ 3.5,\ 6.5,\ 9.5,\ 12.5\) and \(16\text{.}\) These approximations are not very good, made only by looking at the graph. Newton's Method provides more accurate approximations. Accurate to 2 decimal places, we have: \begin{equation*} t=0.65,\ 3.29,\ 6.36,\ 9.48,\ 12.61\ \text{ and } \ 15.74. \end{equation*}

The corresponding points have been plotted on the graph of the parametric equations in Figure 9.3.13(b). Note how most occur near the \(x\)-axis, but not exactly on the axis.

By direct application of Theorem 9.3.16, we have \begin{align*} L \amp = \int_0^{3\pi/2} \sqrt{(-3\sin(t) )^2 +(3\cos(t) )^2} \ dt.\\ \end{align*} Apply the Pythagorean Theorem. \begin{align*} \amp = \int_0^{3\pi/2} 3 \ dt\\ \amp = 3t\Big|_0^{3\pi/2} = 9\pi/2. \end{align*}

This should make sense; we know from geometry that the circumference of a circle with radius 3 is \(6\pi\text{;}\) since we are finding the arc length of \(3/4\) of a circle, the arc length is \(3/4\cdot 6\pi = 9\pi/2\text{.}\)

We can see by the parametrizations of \(x\) and \(y\) that when \(t=\pm 1\text{,}\) \(x=0\) and \(y=0\text{.}\) This means we'll integrate from \(t=-1\) to \(t=1\text{.}\) Applying Theorem 9.3.16, we have \begin{align*} L \amp = \int_{-1}^1\sqrt{(3t^2-1)^2+(2t)^2}\ dt\\ \amp = \int_{-1}^1 \sqrt{9t^4-2t^2+1} \ dt. \end{align*}

Unfortunately, the integrand does not have an antiderivative expressible by elementary functions. We turn to numerical integration to approximate its value. Using 4 subintervals, Simpson's Rule approximates the value of the integral as \(2.65051\text{.}\) Using a computer, more subintervals are easy to employ, and \(n=20\) gives a value of \(2.71559\text{.}\) Increasing \(n\) shows that this value is stable and a good approximation of the actual value.

The teardrop shape is formed between \(t=-1\) and \(t=1\text{.}\) Using Key Idea 9.3.20, we see we need for \(g(t)\geq 0\) on \([-1,1]\text{,}\) and this is not the case. To fix this, we simplify replace \(g(t)\) with \(-g(t)\text{,}\) which flips the whole graph about the \(x\)-axis (and does not change the surface area of the resulting solid). The surface area is: \begin{align*} \text{ Area } \ S \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{(3t^2-1)^2+(2t)^2}\ dt\\ \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{9t^4-2t^2+1} \ dt. \end{align*}

Once again we arrive at an integral that we cannot compute in terms of elementary functions. Using Simpson's Rule with \(n=20\text{,}\) we find the area to be \(S=9.44\text{.}\) Using larger values of \(n\) shows this is accurate to 2 places after the decimal.