Question

Find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ x=\cos \theta, y=3 \sin \theta \quad \theta=0 $$

Short answer

The first derivative is \(dy / dx = -3 \cot{\theta}\), and the second derivative is \(d^{2} y / dx^{2} = 3 / \sin^2{\theta}\). At \(\theta = 0\), the slope is undefined, and the second derivative is 3.

Step-by-step solution

Find \(dx / d\theta\) and \(dy / d\theta\)

Start by finding the derivatives of \(x\) and \(y\) with respect to \(\theta\), using basic differentiation rules for trigonometric functions. \(dx / d\theta = -\sin{\theta}\) and \(dy / d\theta = 3\cos{\theta}\).

Find \(dy / dx\)

Next, using the chain rule of calculus, \(dy / dx = (dy / d\theta) / (dx / d\theta) = -3\cos{\theta} / \sin{\theta}\). By simplifying the fraction, we find the first derivative, \(dy / dx = -3 \cot{\theta}\).

Find \(d^{2} y / dx^{2}\)

The second derivative is computed by differentiating \(dy / dx\) with respect to \(x\). Using the chain rule again, we can write \(d^{2} y / dx^{2} = (d / d\theta) [-3 \cot{\theta}] / (dx / d\theta)\), which simplifies to \(d^{2} y / dx^{2} = 3 / \sin^2{\theta}\).

Substitute \(\theta = 0\)

Finally, substitute \(\theta = 0\) into the equations for \(dy / dx\) and \(d^{2} y / dx^{2}\). Doing so yields \(dy / dx |_{{\theta =0}} = defined\) as \(\cot{(0)}\) is undefined, and \(d^{2} y / dx^{2} |_{{\theta =0}} = 3\). As the slope is undefined, we cannot determine the concavity.

Key concepts

Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions is crucial when dealing with calculus problems involving angles and periodicity. These functions include sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), cotangent (\(\cot\)), secant (\(\sec\)), and cosecant (\(\csc\)). Their derivatives are often remembered using mnemonics or derivative tables. For instance, the derivative of \(\sin\) is \(\cos\), and the derivative of \(\cos\) is \(-\sin\).

In the given exercise, the process begins with finding \(dx / d\theta\) and \(dy / d\theta\), which involves taking the derivatives of \(x = \cos \theta\) and \(y = 3 \sin \theta\) with respect to \(\theta\). The derivatives of these trigonometric functions form the groundwork of the problem and are computed using their standard derivatives: for \(x\) we have \(dx / d\theta = -\sin{\theta}\) and for \(y\) we get \(dy / d\theta = 3\cos{\theta}\).

Understanding these derivatives allows us to handle more complex exercises involving parametric equations where trigonometric functions describe the relationship between variables.

Chain Rule of Calculus

The chain rule is a fundamental tenet in calculus, permitting the differentiation of composite functions. It asserts that to find the derivative of a composed function, one multiplies the derivative of the outer function by the derivative of the inner function.

In parametric equations, where \(x\) and \(y\) are both described in terms of a third variable (in this case, \(\theta\)), the chain rule becomes an indispensable tool. To find \(dy / dx\), the chain rule dictates that we divide \(dy / d\theta\) by \(dx / d\theta\). Applying this to our exercise, we achieve \(dy / dx = -3\cos{\theta} / \sin{\theta} = -3 \cot{\theta}\).

This application of the chain rule simplifies the process of finding derivatives in parametric equations and is also used to find higher-order derivatives, as seen in the exercise when we apply it to calculate the second derivative of \(y\).

Second Derivative Test for Concavity

The second derivative test is a handy method for determining the concavity of a graph at a particular point. If the second derivative of a function at a point is positive, the graph is concave up at that point. Conversely, if it's negative, the graph is concave down. If the second derivative is zero, the test is inconclusive.

In our exercise, finding \(d^{2} y / dx^{2}\) requires us to differentiate the first derivative with respect to \(\theta\) using the chain rule and then again relate it back to \(x\) by dividing by \(dx / d\theta\). Our result, \(d^{2} y / dx^{2} = 3 / \sin^{2}{\theta}\), indicates the concavity of the function, which, if \(\theta\) were not zero, would tell us the function is concave up wherever \(\sin^{2}{\theta}\) is positive.

In this scenario, however, we face a complication: the first derivative is undefined at \(\theta = 0\), making the application of the second derivative test for concavity impossible at this specific value of \(\theta\). Nonetheless, when applicable, this test is a powerful technique for verifying the curvature of a graph.