Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(21-30\), find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=1-\cos t, y=1+\sin t ; t \neq n \pi $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \cot t \) and \( \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 t} \).

Step 1: Find the First Derivatives

To find \( \frac{dy}{dx} \) using parametric equations, we first find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Differentiating, we get:\[x = 1 - \cos t \implies \frac{dx}{dt} = \sin t\]\[y = 1 + \sin t \implies \frac{dy}{dt} = \cos t\]

Step 2: Use Chain Rule for \(\frac{dy}{dx}\)

We use the chain rule for parametric derivatives:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{\sin t} = \cot t\]

Step 3: Differentiate Again for Second Derivative

We need to find \( \frac{d}{dt}(\cot t) \) for \( \frac{d^2y}{dx^2} \). The derivative of \( \cot t \) is \( -\csc^2 t \):\[\frac{d}{dt}(\cot t) = -\csc^2 t\]

Step 4: Use Second Derivative Formula

The second derivative in parametric terms is given by:\[\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{-\csc^2 t}{\sin t} = -\frac{1}{\sin^3 t}\]

Step 5: Conclusion

Thus, the derivatives are:1. \( \frac{dy}{dx} = \cot t \)2. \( \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 t} \)

Key Concepts

Parametric Equations

In calculus, parametric equations come in handy when describing curves that cannot be easily expressed in a standard Cartesian equation like \( y = f(x) \). Students often encounter this in calculus problems where two separate equations are given in terms of a third parameter, usually \( t \). For example:

• \( x = 1 - \cos t \)
• \( y = 1 + \sin t \)

These equations represent the x and y coordinates of a point on a curve, allowing us to study motion more effectively by observing how each coordinate changes with the parameter \( t \). By not eliminating the parameter, we keep the analysis within the context of motion, often easing the computations, especially when calculating derivatives.

Chain Rule

The Chain Rule is a crucial concept when dealing with derivatives in parametric equations. It allows us to relate the derivatives of different variables through a common parameter. In simpler terms, if we have:

• A function describing \( y \) in terms of \( t \): \( \frac{dy}{dt} \)
• Another function for \( x \) in terms of \( t \): \( \frac{dx}{dt} \)

We can find \( \frac{dy}{dx} \) as the ratio of these two derivatives: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]This application showcases how changes in \( t \) produce changes in \( x \) and \( y \), and consequently, how they affect each other. For example, in the previous problem, this rule allowed us to find \( \frac{dy}{dx} = \cot t \), reflecting how changes in \( y \) compared to changes in \( x \) at each point.

Derivatives

Derivatives are fundamental in understanding the behavior of functions, especially relating to change and motion. In terms of parametrics, the derivative \( \frac{dy}{dx} \) is related to how \( y \) changes as \( x \) changes, which is crucial for tangent calculations and analyzing directions of curves.

• The first derivative calculated, \( \frac{dy}{dx} = \cot t \), indicates the slope of the tangent to the curve at any point defined by the parameter \( t \).
• This derivative can signal various properties of the curve, such as increasing or decreasing behavior and potential stationary points if \( \cot t = 0 \).

Therefore, learning how to compute derivatives of parametric forms opens doors to deeper insights into the dynamics of curves on the plane.

Second Derivative

While the first derivative gives us the slope, the second derivative provides insight into the curve's concavity and acceleration. It answers the question, "How is the slope itself changing?" Calculating \( \frac{d^2y}{dx^2} \) in parametric equations helps us determine where a curve is concave up or down.

• Our problem showed that \( \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 t} \).
• A negative second derivative, as seen here, specifies that the curve is concave down at points corresponding to \( t \).
• Conversely, if the second derivative were positive, the curve would be concave up, signaling a different dynamic.

Such insights are invaluable in fields requiring motion analysis, like physics and engineering, providing a clearer picture of how shapes progress or regress in space over time.