Problem 15 - Calculus Early Transcendentals

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2} .\) For which values of \(t\) is the curve concave upward? \(x=2 \sin t$$y=3 \cos t$$0<\)t\(<2 \pi\)

Step-by-Step Solution

Answer

The curve is never concave upward for any \(t\) in the interval \((0, 2\pi)\).

Step 1: Differentiate x and y with respect to t

To find \(\frac{dy}{dx}\), we first need \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Differentiate each parametric equation.\(x = 2\sin t \rightarrow \frac{dx}{dt} = 2 \cos t\)\(y = 3 \cos t \rightarrow \frac{dy}{dt} = -3 \sin t\)

Step 2: Find dy/dx using the Chain Rule

Using the chain rule, where \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\), calculate \(\frac{dy}{dx}\).\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3 \sin t}{2 \cos t} = -\frac{3}{2} \tan t\)

Step 3: Differentiate dy/dx with respect to t to find d²y/dx²

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = -\frac{3}{2} \tan t\) with respect to \(t\) and divide by \(\frac{dx}{dt}\).First, differentiate \(-\frac{3}{2} \tan t\):\(\frac{d}{dt}(-\frac{3}{2} \tan t) = -\frac{3}{2} \sec^2 t\).Then divide by \(\frac{dx}{dt} = 2 \cos t\):\(\frac{d^2 y}{dx^2} = \frac{-\frac{3}{2} \sec^2 t}{2 \cos t} = -\frac{3}{4} \frac{1}{\cos^3 t}\).

Step 4: Determine when the curve is concave upward

The curve is concave upward when \(\frac{d^2y}{dx^2} > 0\). Since \(\frac{d^2y}{dx^2} = -\frac{3}{4} \sec^3 t\) is negative unless \(-\sec^3 t\) is zero or undefined, there are no values of \(t\) in \((0, 2\pi)\) where the curve is concave upward.

Key Concepts

Understanding Derivatives in Parametric Curves

When working with parametric curves, we often want to find the derivative, which tells us how the position of the curve changes. For a curve defined by two parametric equations, such as \(x = 2\sin t\) and \(y = 3\cos t\), the derivative \(\frac{dy}{dx}\) represents the slope of the curve at any point.

To find this, we need the derivatives of \(x\) and \(y\) with respect to the parameter \(t\). These are known as \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). From our equations:

• \(\frac{dx}{dt} = 2\cos t\)
• \(\frac{dy}{dt} = -3\sin t\)

Using these, we apply the chain rule, which will help us find \(\frac{dy}{dx}\) by considering it as the division of \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\). The result is \(\frac{dy}{dx} = -\frac{3}{2}\tan t\).

This expression tells us how steep the curve is at any point, depending on the value of \(t\). This information is crucial for analyzing the behavior and shape of the parametric curve.

Examining Concavity of Parametric Curves

Concavity involves understanding how a curve bends. It tells us whether the curve is curving upwards (concave up) or downwards (concave down). In the context of parametric curves, we determine concavity by calculating the second derivative \(\frac{d^2y}{dx^2}\).

We start from the first derivative \(\frac{dy}{dx} = -\frac{3}{2}\tan t\). To find the second derivative, we differentiate \(\frac{dy}{dx}\) with respect to \(t\), resulting in \(\frac{d}{dt}(-\frac{3}{2}\tan t) = -\frac{3}{2}\sec^2 t\).

Next, we divide this by \(\frac{dx}{dt} = 2\cos t\), giving \(\frac{d^2 y}{dx^2} = -\frac{3}{4} \frac{1}{\cos^3 t}\).

A curve is concave upward when \(\frac{d^2y}{dx^2} > 0\). However, in this case, \(\frac{d^2y}{dx^2}\) is negative for all \(t\) within \((0, 2\pi)\), which indicates no concavity upwards within this range. This insight is pivotal for understanding the nature of the curve's bend or shape as defined by the parameter \(t\).

The Role of the Chain Rule in Calculus

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It is particularly useful in dealing with parametric curves because these curves are defined by separate parametric equations for \(x\) and \(y\).

The key idea of the chain rule is that the derivative of a composite function can be calculated as the product of the derivative of the outer function and the derivative of the inner function.

For instance, to find \(\frac{dy}{dx}\) in the context of parametric curves, we need \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). Applying the chain rule means setting up the ratio \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), which breaks down a potentially complex problem into simpler parts.

In our example, this leads to \(\frac{dy}{dx} = -\frac{3}{2}\tan t\), giving the slope of the curve. This ability to seamlessly navigate between derivatives is what makes the chain rule such a powerful tool in calculus and understanding parametric equations.