Problem 21
Question
In Problems \(21-30\), find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=3 \tau^{2}, y=4 \tau^{3} ; \tau \neq 0 $$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = 2\tau \) and \( \frac{d^2y}{dx^2} = \frac{1}{3\tau} \).
1Step 1: Differentiate x with respect to τ
Begin by differentiating the expression for x with respect to the parameter \( \tau \). Given \( x = 3\tau^2 \), the derivative is \( \frac{dx}{d\tau} = 6\tau \).
2Step 2: Differentiate y with respect to τ
Now, differentiate the expression for y with respect to \( \tau \). Given \( y = 4\tau^3 \), the derivative is \( \frac{dy}{d\tau} = 12\tau^2 \).
3Step 3: Find dy/dx
Use the chain rule to find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{d\tau} \) by \( \frac{dx}{d\tau} \). Thus, \( \frac{dy}{dx} = \frac{12\tau^2}{6\tau} = 2\tau \).
4Step 4: Differentiate dy/dx to find d²y/dx²
To find the second derivative, differentiate \( \frac{dy}{dx} = 2\tau \) with respect to \( \tau \), yielding \( \frac{d}{d\tau}(2\tau) = 2 \).
5Step 5: Use the chain rule to find d²y/dx²
Apply the chain rule: \( \frac{d^2y}{dx^2} = \frac{d}{d\tau}\left( \frac{dy}{dx} \right) \div \frac{dx}{d\tau} = \frac{2}{6\tau} = \frac{1}{3\tau} \).
Key Concepts
DifferentiationChain RuleSecond Derivative
Differentiation
Differentiation is a core concept in calculus. It's the process of finding the derivative of a function, which represents the rate at which the function's value changes. In simpler terms, it's how we measure change.
For parametric equations, where both x and y are dependent on a separate parameter—in this case, \( \tau \)—we differentiate each one individually with respect to \( \tau \). This approach simplifies the process of finding derivatives when the relationship between x and y isn't straightforward.
For instance, in the original problem, we start with \( x = 3\tau^2 \) and \( y = 4\tau^3 \). We find \( \frac{dx}{d\tau} = 6\tau \) and \( \frac{dy}{d\tau} = 12\tau^2 \). These derivatives allow us to link changes in x and y with changes in \( \tau \), the parameter managing both equations, setting up the need for further derivative calculations like \( \frac{dy}{dx} \).
Differentiating with respect to a parameter can sometimes provide insights or simplify calculations for complex relationships that may otherwise seem difficult to handle directly.
For parametric equations, where both x and y are dependent on a separate parameter—in this case, \( \tau \)—we differentiate each one individually with respect to \( \tau \). This approach simplifies the process of finding derivatives when the relationship between x and y isn't straightforward.
For instance, in the original problem, we start with \( x = 3\tau^2 \) and \( y = 4\tau^3 \). We find \( \frac{dx}{d\tau} = 6\tau \) and \( \frac{dy}{d\tau} = 12\tau^2 \). These derivatives allow us to link changes in x and y with changes in \( \tau \), the parameter managing both equations, setting up the need for further derivative calculations like \( \frac{dy}{dx} \).
Differentiating with respect to a parameter can sometimes provide insights or simplify calculations for complex relationships that may otherwise seem difficult to handle directly.
Chain Rule
The chain rule is a fundamental tool in calculus used to differentiate composite functions. It's particularly useful in situations like parametric differentiation, where functions depend on an intermediary variable, \( \tau \) in this case.
When we want to find \( \frac{dy}{dx} \), we're interested in how changes in x and y relate to one another, through their mutual dependence on \( \tau \). We achieve this by applying the chain rule, which helps break down the differentiation of these composed structures.
Illustratively, in our example, once we've differentiated x and y with respect to \( \tau \), we use the chain rule to find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{d\tau} \) by \( \frac{dx}{d\tau} \). This yields \( \frac{dy}{dx} = \frac{12\tau^2}{6\tau} = 2\tau \). The chain rule here acts as an analytical bridge, showing us how movements in \( \tau \) affect the relationship directly between x and y.
Mastering the chain rule offers powerful insights into how variables interconnect based on their respective derivatives, simplifying the investigation of more complex differential equations.
When we want to find \( \frac{dy}{dx} \), we're interested in how changes in x and y relate to one another, through their mutual dependence on \( \tau \). We achieve this by applying the chain rule, which helps break down the differentiation of these composed structures.
Illustratively, in our example, once we've differentiated x and y with respect to \( \tau \), we use the chain rule to find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{d\tau} \) by \( \frac{dx}{d\tau} \). This yields \( \frac{dy}{dx} = \frac{12\tau^2}{6\tau} = 2\tau \). The chain rule here acts as an analytical bridge, showing us how movements in \( \tau \) affect the relationship directly between x and y.
Mastering the chain rule offers powerful insights into how variables interconnect based on their respective derivatives, simplifying the investigation of more complex differential equations.
Second Derivative
The second derivative offers insight into the nature of a function's rate of change. While the first derivative, \( \frac{dy}{dx} \), tells us the slope of the tangent to the curve at any point, the second derivative, \( \frac{d^2y}{dx^2} \), informs us about the curvature of the function, or how that slope is changing.
In parametric differentiation, after finding \( \frac{dy}{dx} \) using the chain rule, we proceed to determine \( \frac{d^2y}{dx^2} \). This involves differentiating \( \frac{dy}{dx} \) again with respect to \( \tau \), followed by dividing by \( \frac{dx}{d\tau} \). In our current context, this second step gives us \( \frac{d^2y}{dx^2} = \frac{2}{6\tau} = \frac{1}{3\tau} \).
The second derivative uncovers how the graph of the function bends and can hint at transitions from concave to convex or vice versa. Being able to interpret and calculate second derivatives is essential for understanding the broader shapes and potential turning points of functions, particularly in physics, engineering, and higher-level mathematics.
In parametric differentiation, after finding \( \frac{dy}{dx} \) using the chain rule, we proceed to determine \( \frac{d^2y}{dx^2} \). This involves differentiating \( \frac{dy}{dx} \) again with respect to \( \tau \), followed by dividing by \( \frac{dx}{d\tau} \). In our current context, this second step gives us \( \frac{d^2y}{dx^2} = \frac{2}{6\tau} = \frac{1}{3\tau} \).
The second derivative uncovers how the graph of the function bends and can hint at transitions from concave to convex or vice versa. Being able to interpret and calculate second derivatives is essential for understanding the broader shapes and potential turning points of functions, particularly in physics, engineering, and higher-level mathematics.
Other exercises in this chapter
Problem 20
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Sketch the graph of the given equation. $$ (y-1)^{2}=16 $$
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Sketch the region in the first quadrant that is inside the cardioid \(r=3+3 \cos \theta\) and outside the cardioid \(r=3+3 \sin \theta\), and find its area.
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