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Magazine Chapter 4: Problem 62
Assume that \(x\) and \(y\) are differentiable functions of \(t\). Find \(\frac{d y}{d t}\) when \(y^{2}=x^{2}-x^{4}, \frac{d x}{d t}=1\) for \(x=\frac{1}{2}\), and \(y>0\).
Short Answer
Expert verified
\(\frac{dy}{dt} = \frac{1}{\sqrt{3}}\) when \(x = \frac{1}{2}\) and \(y > 0\).
Step by step solution
01
Differentiate the Equation Implicitly
Given the equation \(y^2 = x^2 - x^4\), we need to find \( \frac{dy}{dt} \). Start by differentiating both sides of the equation with respect to \(t\). The left-hand side becomes: \(2y \frac{dy}{dt}\) The right-hand side becomes: \(2x \frac{dx}{dt} - 4x^3 \frac{dx}{dt}\). Thus, we have the equation: \[2y \frac{dy}{dt} = 2x \frac{dx}{dt} - 4x^3 \frac{dx}{dt}\]
02
Substitute Known Values for \(x\) and \(\frac{dx}{dt}\)
We know \(\frac{dx}{dt} = 1\) and \(x = \frac{1}{2}\). Substitute these values into the differentiated equation. \[2y \frac{dy}{dt} = 2\left(\frac{1}{2}\right) \cdot 1 - 4\left(\frac{1}{2}\right)^3 \cdot 1\]Simplifying, \[2y \frac{dy}{dt} = 1 - \frac{1}{2}\]\[2y \frac{dy}{dt} = \frac{1}{2}\]
03
Solve for \(\frac{dy}{dt}\)
Solve the equation \(2y \frac{dy}{dt} = \frac{1}{2}\) for \( \frac{dy}{dt} \).Divide both sides by \(2y\):\[\frac{dy}{dt} = \frac{\frac{1}{2}}{2y}\]\[\frac{dy}{dt} = \frac{1}{4y}\]
04
Find the Value of \(y\)
Given \(x = \frac{1}{2}\), substitute this value into the original equation \(y^2 = x^2 - x^4\) to find \(y\). \[y^2 = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^4\]\[y^2 = \frac{1}{4} - \frac{1}{16}\]\[y^2 = \frac{4}{16} - \frac{1}{16}\]\[y^2 = \frac{3}{16}\]Taking the square root (and using \(y > 0\)), \[y = \frac{\sqrt{3}}{4}\]
05
Calculate \(\frac{dy}{dt}\) Using \(y\)
Substitute \(y = \frac{\sqrt{3}}{4}\) into the expression for \(\frac{dy}{dt}\):\[\frac{dy}{dt} = \frac{1}{4y} = \frac{1}{4 \times \frac{\sqrt{3}}{4}} = \frac{1}{\sqrt{3}}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiable Functions
Differentiable functions are a significant concept in calculus, focusing on how a function changes with respect to variable changes. If a function is differentiable, its derivative exists at every point within its domain. This property is essential when understanding rates of change in mathematical models, such as in physics or economics.
In the context of the problem, we are dealing with two functions, \(x(t)\) and \(y(t)\), both differentiable with respect to \(t\). This means that we can apply differentiation rules to find out how these functions change as \(t\) changes.
In the context of the problem, we are dealing with two functions, \(x(t)\) and \(y(t)\), both differentiable with respect to \(t\). This means that we can apply differentiation rules to find out how these functions change as \(t\) changes.
- Existence of Derivatives: The derivative \( \frac{dy}{dt} \) indicates how \(y\) changes concerning \(t\).
- Continuous Change: Differentiation gives us a snapshot of the rate of change at any specific moment in time.
- Smoothness: Differentiable functions are smooth, without breaks or sharp changes in direction, implying a consistent gradient or slope.
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. This principle is particularly important when dealing with functions that depend on one or more intermediate variables, as is the case with implicit differentiation.
In the exercise, the equation \(y^2 = x^2 - x^4\) is differentiated with respect to \(t\). Both \(x\) and \(y\) depend on \(t\), so the use of the chain rule is implicit.
In the exercise, the equation \(y^2 = x^2 - x^4\) is differentiated with respect to \(t\). Both \(x\) and \(y\) depend on \(t\), so the use of the chain rule is implicit.
- Application of the Chain Rule: When you differentiate \(y^2\) with respect to \(t\), you apply the chain rule: \(2y \cdot \frac{dy}{dt}\). Similarly, differentiating \(x^2\) and \(x^4\) involves applying the chain rule through \(\frac{dx}{dt}\).
- Linking Derivatives: Whenever a derivative is taken for a function of a function, the chain rule provides the vital link tying these derivatives together.
- Derivatives with Multiple Variables: Recognize that implicit differentiation involves managing multiple dependent variables, all evaluated over the main independent variable \(t\).
Implicit Functions
Implicit functions are equations in which the dependent variable is not isolated on one side of the equation. Instead, the relationship between the dependent and independent variables is intermixed, often complicating direct differentiation.
The given equation, \(y^2 = x^2 - x^4\), is implicit because \(y\) is not isolated as a function of \(x\). This is where implicit differentiation comes in handy.
The given equation, \(y^2 = x^2 - x^4\), is implicit because \(y\) is not isolated as a function of \(x\). This is where implicit differentiation comes in handy.
- Utilizing Implicit Differentiation: By differentiating both sides of the implicit function concerning \(t\), you're able to find \(\frac{dy}{dt}\) even when \(y\) isn't explicitly a function of \(x\).
- Cross Relating Variables: Implicit functions involve mutual relationships between variables that require simultaneous changes to one affecting the other.
- Solving for Unseen Relationships: Implicit differentiation helps uncover relationships between variables that might not be immediately apparent in straightforward equations.
