Chapter 10: Problem 14
Find \(d y / d x\) and \(d^{2} y / d x^{2} .\) For which values of \(t\) is the curve concave upward? $$x=t+\ln t, \quad y=t-\ln t$$
Short Answer
Expert verified
The curve is concave upward for all \(t \neq -1\).
Step by step solution
01
Differentiate x and y with respect to t
Given that \(x = t + \ln t\) and \(y = t - \ln t\). First, we find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). \[\frac{dx}{dt} = 1 + \frac{1}{t}\]\[\frac{dy}{dt} = 1 - \frac{1}{t}\]
02
Find dy/dx
To find \(\frac{dy}{dx}\), use the chain rule by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\):\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 - \frac{1}{t}}{1 + \frac{1}{t}}\]
03
Simplify dy/dx
Simplify the expression for \(\frac{dy}{dx}\):Multiply numerator and denominator by \(t\):\[\frac{dy}{dx} = \frac{t(1 - \frac{1}{t})}{t(1 + \frac{1}{t})} = \frac{t - 1}{t + 1}\]
04
Differentiate dy/dx to find d²y/dx²
Now differentiate \(\frac{dy}{dx} = \frac{t-1}{t+1}\) with respect to \(t\) using the quotient rule, where \(u = t-1\) and \(v = t+1\):\[\frac{d^2y}{dx^2} = \frac{(v)(\frac{du}{dt}) - (u)(\frac{dv}{dt})}{v^2}\]\[\frac{du}{dt} = 1, \quad \frac{dv}{dt} = 1\]\[\frac{d^2y}{dx^2} = \frac{(t+1)(1) - (t-1)(1)}{(t+1)^2} = \frac{t + 1 - t + 1}{(t+1)^2} = \frac{2}{(t+1)^2}\]
05
Determine concavity of the curve
The curve is concave upward when \(\frac{d^2y}{dx^2} > 0\). Since \(\frac{d^2y}{dx^2} = \frac{2}{(t+1)^2}\), it is always positive since \((t+1)^2 > 0\) for all \(t eq -1\). Thus, the curve is concave upward for all \(t eq -1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Concavity
Concavity is a property of a curve that tells us about its shape or direction. When a curve is said to be concave upward, it means that the curve opens upwards, similar to a cup or a bowl. This occurs when the second derivative of the curve, denoted as \( \frac{d^2y}{dx^2} \), is positive. It indicates that the slope of the tangent lines is increasing.
To determine the concavity of a curve described by parametric equations, like the given \( x = t + \ln t \) and \( y = t - \ln t \), we first find the second derivative \( \frac{d^2y}{dx^2} \). This involves differentiating \( \frac{dy}{dx} \) using techniques like the quotient rule. If the resulting function is positive for most or all values of \( t \), the curve is concave upward. In this case, \( \frac{d^2y}{dx^2} = \frac{2}{(t+1)^2} \), which is always greater than zero for \( t eq -1 \), confirming upward concavity for most of the curve.
To determine the concavity of a curve described by parametric equations, like the given \( x = t + \ln t \) and \( y = t - \ln t \), we first find the second derivative \( \frac{d^2y}{dx^2} \). This involves differentiating \( \frac{dy}{dx} \) using techniques like the quotient rule. If the resulting function is positive for most or all values of \( t \), the curve is concave upward. In this case, \( \frac{d^2y}{dx^2} = \frac{2}{(t+1)^2} \), which is always greater than zero for \( t eq -1 \), confirming upward concavity for most of the curve.
The Chain Rule in Differentiation
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When dealing with parametric equations, like \( x = t + \ln t \) and \( y = t - \ln t \), we often apply the chain rule to connect derivatives with respect to one variable to another. This is achieved by determining \( \frac{dy}{dx} \) indirectly through \( t \).
To apply the chain rule for parametric differentiation, we find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first. Then, we use the relation: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] This converts derivatives with respect to \( t \) back to conventional derivatives with respect to \( x \), as needed to analyze the curve's behavior, like its slope at different points.
To apply the chain rule for parametric differentiation, we find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first. Then, we use the relation: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] This converts derivatives with respect to \( t \) back to conventional derivatives with respect to \( x \), as needed to analyze the curve's behavior, like its slope at different points.
Using the Quotient Rule
The quotient rule is a method in calculus used to find the derivative of a quotient of two functions. When we have \( \frac{dy}{dx} = \frac{t-1}{t+1} \), and need to compute its derivative to find the second derivative \( \frac{d^2y}{dx^2} \), the quotient rule becomes necessary. The formula for the quotient rule is:
- \( \frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \)
- \( \frac{d^2y}{dx^2} = \frac{(t+1)(1) - (t-1)(1)}{(t+1)^2} = \frac{2}{(t+1)^2} \)
Parametric Equations and their Derivatives
Parametric equations offer a way to describe curves by expressing the coordinates \( (x, y) \) as functions of an independent parameter \( t \). Instead of using an equation relating \( x \) and \( y \) directly, the curve is defined by equations like \( x = f(t) \) and \( y = g(t) \).
For these equations, we derive \( \frac{dy}{dx} \) - the rate of change of \( y \) with respect to \( x \) - by using the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This approach is quite powerful, particularly in complex shapes or curves that aren't simple functions of \( x \). With the example \( x = t + \ln t \) and \( y = t - \ln t \), we find:
For these equations, we derive \( \frac{dy}{dx} \) - the rate of change of \( y \) with respect to \( x \) - by using the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This approach is quite powerful, particularly in complex shapes or curves that aren't simple functions of \( x \). With the example \( x = t + \ln t \) and \( y = t - \ln t \), we find:
- \( \frac{dx}{dt} = 1 + \frac{1}{t} \)
- \( \frac{dy}{dt} = 1 - \frac{1}{t} \)