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Magazine Chapter 10: Problem 45
Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter. $$x=\sqrt{t}, y=2 t+4 ; t=1$$
Short Answer
Expert verified
dy/dx at t=1 is 4; d虏y/dx虏 at t=1 is 4.
Step by step solution
01
Find dx/dt
Given the parametrization for x, which is \(x = \sqrt{t}\), we need to differentiate x with respect to t. The derivative is \(\frac{dx}{dt} = \frac{1}{2\sqrt{t}}\).
02
Find dy/dt
Differentiating the given equation for y with respect to t, where \(y = 2t + 4\), we obtain \(\frac{dy}{dt} = 2\).
03
Calculate dy/dx
To find \(\frac{dy}{dx}\), use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\). Substituting the derivatives found: \(\frac{dy}{dx} = \frac{2}{\frac{1}{2\sqrt{t}}} = 4\sqrt{t}\).
04
Evaluate dy/dx at t = 1
Substitute \(t = 1\) into \(\frac{dy}{dx} = 4\sqrt{t}\): \(\frac{dy}{dx}\rvert_{t=1} = 4\sqrt{1} = 4\).
05
Calculate d虏y/dx虏
First, differentiate \(\frac{dy}{dx} = 4\sqrt{t}\) with respect to t, which gives \(\frac{d}{dt}\left(4\sqrt{t}\right) = \frac{2}{\sqrt{t}}\). Then, use the formula \(\frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\frac{dy}{dx}) \div \frac{dx}{dt} = \frac{\frac{2}{\sqrt{t}}}{\frac{1}{2\sqrt{t}}} = 4\).
06
Evaluate d虏y/dx虏 at t = 1
Substitute \(t = 1\) into \(\frac{d^{2}y}{dx^{2}} = 4\): \(\frac{d^{2}y}{dx^{2}}\rvert_{t=1} = 4\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Chain Rule in Parametric Differentiation
The chain rule is a core concept in calculus, particularly useful when dealing with parametric equations. In the context of our exercise, it helps us to find the derivative \( \frac{dy}{dx} \), even though our variables x and y are functions of a third parameter, t. By using the chain rule, we can efficiently manage these dependencies without resorting to complicated elimination processes.
Here's a simple breakdown:
Here's a simple breakdown:
- First, find the derivatives of x and y with respect to t, which are \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
- The chain rule allows us to connect these derivatives to find \( \frac{dy}{dx} \) by using the formula: \[ \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} \].
Second Derivative for Parametric Equations
The second derivative, \[ \frac{d^2y}{dx^2} \], can be critical in understanding the concavity and behavior of a curve described by parametric equations. Once we have the first derivative, finding the second derivative involves taking its derivative with respect to the parameter t and then using the chain rule again.
Follow these steps:
Follow these steps:
- First, differentiate the expression for \[ \frac{dy}{dx} \] with respect to t, which yields \[ \frac{d}{dt}(\frac{dy}{dx}) \].
- Then, divide this result by \[ \frac{dx}{dt} \] to find the second derivative: \[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\frac{dy}{dx}) \div \frac{dx}{dt} \].
The Role of Parametrization in Calculus
Parametrization plays a crucial role in calculus, serving as a bridge to deal with complex equations by expressing them in terms of a single parameter. This technique is especially useful when dealing with curves that can't be easily described by a single function y=f(x).
In our example, we express both x and y as functions of t.
By differentiating these functions with respect to t, we can explore how both variables change and interact.
The benefits of this approach include:
In our example, we express both x and y as functions of t.
By differentiating these functions with respect to t, we can explore how both variables change and interact.
The benefits of this approach include:
- Simplifying differentiation by breaking down functions into more manageable pieces.
- Easing the process of finding derivatives even for non-standard curves, without needing to solve the equations for one variable.
- Facilitating the computation of additional properties like the second derivative, through stepwise, parametric methods.
