Problem 16 - 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? \(x=\cos 2 t\), \(y=\cos t\), \(0<\)t\(<\pi\)

Step-by-Step Solution

Answer

The curve is concave upward when \(\frac{\pi}{2} < t < \pi\).

Step 1: Differentiate x with Respect to t

Given that \(x = \cos 2t\), we need to find \(\frac{dx}{dt}\). Using the chain rule, the derivative of \(\cos(2t)\) is \(-\sin(2t)\) times the derivative of \(2t\), which is \(2\), so \(\frac{dx}{dt} = -2\sin(2t)\).

Step 2: Differentiate y with Respect to t

Given that \(y = \cos t\), we need to find \(\frac{dy}{dt}\). The derivative of \(\cos(t)\) is \(-\sin(t)\). Therefore, \(\frac{dy}{dt} = -\sin(t)\).

Step 3: Find dy/dx

We find \(\frac{dy}{dx}\) using the formula \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Substituting the previously found derivatives gives:\[\frac{dy}{dx} = \frac{-\sin(t)}{-2\sin(2t)} = \frac{\sin(t)}{2\sin(2t)}.\]Using the double angle identity, \(\sin(2t) = 2\sin(t)\cos(t)\), we simplify:\[\frac{dy}{dx} = \frac{1}{4\cos(t)}.\]

Step 4: Differentiate dy/dx with Respect to t

The second derivative is found by differentiating \(\frac{dy}{dx}\) with respect to \(t\). From the previous step, we differentiate \(\frac{1}{4\cos(t)}\) to get \(\frac{d^2y}{dx^2}\).The derivative is:\[\frac{d}{dt}\left( \frac{1}{4\cos(t)} \right) = \frac{-1}{4\cos^2(t)}\cdot (-\sin(t)),\]which simplifies to:\[\frac{d^2y}{dt^2} = \frac{\tan(t)}{4\cos(t)}.\]

Step 5: Find d²y/dx²

We know \(\frac{d²y}{dx²} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt}\). From the previous step and \(\frac{dx}{dt} = -2\sin(2t)\), we have:\[\frac{d²y}{dx²} = \frac{\frac{\tan(t)}{4\cos(t)}}{-2\sin(2t)} = \frac{-\tan(t)}{8\cos^2(t)\sin(t)}.\]

Step 6: Find t Where the Curve is Concave Upward

The curve is concave upward when \(\frac{d²y}{dx²} > 0\). Since \(\frac{-\tan(t)}{8\cos^2(t)\sin(t)} > 0\), it follows that \(-\tan(t)\) must be positive, implying \(\tan(t) < 0\). For \(0 < t < \pi\), this condition holds true when \(\frac{\pi}{2} < t < \pi\).

Key Concepts

Understanding the First Derivative

The first derivative, often symbolized as \(\frac{dy}{dx}\), represents the rate of change of the curve with respect to the x-axis. In the context of parametric equations where you have an independent parameter \(t\), the first derivative is found by differentiating both \(x\) and \(y\) with respect to \(t\), then applying the chain rule.

• First, calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
• Then use the formula \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

For the problem, the steps are as follows:Given \(x = \cos(2t)\), the derivative \(\frac{dx}{dt} = -2\sin(2t)\). For \(y = \cos(t)\), \(\frac{dy}{dt} = -\sin(t)\).Hence, \(\frac{dy}{dx} = \frac{-\sin(t)}{-2\sin(2t)} = \frac{1}{4\cos(t)}\). This expression tells us how y changes as x changes, in terms of the parameter \(t\).

Grasping the Second Derivative

The second derivative, noted as \(\frac{d^2y}{dx^2}\), reveals how the rate of change itself is changing. In terms of curves, it tells us about the curvature or how the slope itself is evolving as \(x\) changes.

• Differentiate \(\frac{dy}{dx}\) with respect to \(t\).
• Then divide by \(\frac{dx}{dt}\) to translate it back to \(x\)-terms.

From the previous section, we know \(\frac{dy}{dx} = \frac{1}{4\cos(t)}\). Differentiating this with respect to \(t\), we have \(\frac{d}{dt}\left(\frac{1}{4\cos(t)}\right) = \frac{-\sin(t)}{4\cos^2(t)}\).Dividing this by \(\frac{dx}{dt} = -2\sin(2t)\) gives \(\frac{d^2y}{dx^2} = \frac{-\tan(t)}{8\cos^2(t)\sin(t)}\). This second derivative informs us about the concavity of the curve.

Exploring Concavity

Concavity signifies whether a curve bends upwards or downwards. A curve is said to be "concave upward" when \(\frac{d^2y}{dx^2} > 0\), indicating that the slope of the tangent line is increasing.In our exercise, we determine the concavity by evaluating \(\frac{d^2y}{dx^2} = \frac{-\tan(t)}{8\cos^2(t)\sin(t)}\). The condition \(\frac{-\tan(t)}{8\cos^2(t)\sin(t)} > 0\) simplifies to \(-\tan(t) > 0\), implying \(\tan(t) < 0\).For the interval \(0 < t < \pi\), the tangent function is negative in the second quadrant. Thus, the condition \(\tan(t) < 0\) is satisfied when \(\frac{\pi}{2} < t < \pi\).This means that within this range of \(t\), the curve is concave upward, representing a specific behavior of the curve shape.