Question

For the following exercises, find \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter.\(x=\sqrt{t}, \quad y=2 t+4, \quad t=1\)

Short answer

The second derivative \(d^2 y / d x^2\) at \(t = 1\) is 4.

Step-by-step solution

Differentiate x with respect to t

First, we need to find the derivative of \(x = \sqrt{t} \) with respect to \(t\). This can be expressed as \(\frac{dx}{dt}\). Using the power rule: \[\frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}\]

Differentiate y with respect to t

Next, calculate the derivative of \(y = 2t + 4\) with respect to \(t\). This is written as \(\frac{dy}{dt}\): \[\frac{dy}{dt} = \frac{d}{dt}(2t + 4) = 2\]

Find the first derivative \(dy/dx\) using parametric equations

Let's find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\): \[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{\frac{1}{2\sqrt{t}}} = 4\sqrt{t} \]

Differentiate \(dy/dx\) to get \(d^2y/dx^2\)

Now we must find the second derivative. Since \(\frac{dy}{dx} = 4\sqrt{t} \), we differentiate \(4\sqrt{t}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\):First, find \(\frac{d}{dt}(4\sqrt{t})\):\[\frac{d}{dt}(4t^{1/2}) = 4 \times \frac{1}{2}t^{-1/2} = 2t^{-1/2} = \frac{2}{\sqrt{t}}\]Then, the second derivative \(\frac{d^2y}{dx^2}\) is given by:\[\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{2}{\sqrt{t}}}{\frac{1}{2\sqrt{t}}} = 4\]

Evaluate \(d^2y/dx^2\) at \(t = 1\)

Substitute \(t = 1\) into \(\frac{d^2y}{dx^2} = 4\): \[ \frac{d^2y}{dx^2} = 4 \]The second derivative evaluated at \(t = 1\) remains \(4\).

Key concepts

Understanding Derivatives

Derivatives are at the heart of calculus and serve as a primary tool to investigate how one quantity changes with respect to another. Think of a derivative as a rate of change or slope of a function's graph. If you consider a parameterized path on a curve defined by a pair of functions,

• we find derivatives to calculate the slope of the tangent to the curve at any point.
• The notation \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\).

Calculating this for parametric equations involves two steps:

• Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) by differentiating each function with respect to the parameter \(t\).
• Then determine \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\).

This process allows us to analyze more complicated curves where directly solving for \(y\) in terms of \(x\) might not be feasible or convenient.

Exploring the Second Derivative

When we need to understand how the rate of change itself is changing, we look at the second derivative. The second derivative, denoted as \(\frac{d^2y}{dx^2}\), tells us about the concavity of the function and points to whether a curve is bending upwards or downwards.

• A positive second derivative means the curve is concave up, indicating a local minimum or a bowl-like shape.
• A negative second derivative means the curve is concave down, suggesting a local maximum or a hill-like shape.

For parametric equations, the process involves differentiating \(\frac{dy}{dx}\) with respect to \(t\) again and dividing by \(\frac{dx}{dt}\). This step informs us about the function's "steepness" change rate along the curve. At a given point on the curve, evaluating \(\frac{d^2y}{dx^2}\) helps judge the exact nature of that curve's behavior.

The Role of Calculus in Analyzing Curves

Calculus provides tools such as derivatives and integrals, crucial for analyzing various physical, biological, and mathematical systems. By applying these techniques to parametric equations:

• we can understand complex motion and geometric shapes without directly eliminating the parameter \(t\).
• Calculus allows us to determine vital features like curvature, inflection points, and overall behavior of a path traced by a point.

In this exercise, we demonstrated how derivatives can be harnessed to analyze and interpret the characteristics of curves. Parametric equations offer a unique approach to studying multi-dimensional paths, and calculus equips us with a complete framework for exploration. Understanding calculus' concepts and applications makes complex mathematical expressions much more digestible and applicable in both theoretical and real-world scenarios.