Problem 283
Question
$$ \text { If } x=\phi(t), y=\psi(t), \text { then find } \frac{d^{2} y}{d x^{2}} \text { . } $$
Step-by-Step Solution
Verified Answer
\(\frac{d^2y}{dx^2} = \frac{(dx/dt) \cdot (d^2y/dt^2) - (dy/dt) \cdot (d^2x/dt^2)}{(dx/dt)^3}\)
1Step 1: Compute the First Derivative
First, we need to calculate \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) using the equations \(y = \psi(t)\) and \(x = \phi(t)\). Then, the first derivative of y with respect to x can be calculated as \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) using the chain rule.
2Step 2: Compute the Derivatives of \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)
These derivates are \(\frac{d^2y}{dt^2}\) and \(\frac{d^2x}{dt^2}\). These derivatives will be needed for further computations.
3Step 3: Apply the Quotient Rule
\(\frac{d^2y}{dx^2}\) is computed by applying the quotient rule to \(\frac{dy}{dx}\). The formula for the quotient rule is: \(\frac{d^2y}{dx^2} = \frac{(dx/dt) \cdot (d^2y/dt^2) - (dy/dt) \cdot (d^2x/dt^2)}{(dx/dt)^3}\). Implement these calculations to obtain the second derivative of y with respect to x.
Key Concepts
Chain RuleQuotient RuleParametric Equations
Chain Rule
The chain rule is a fundamental concept in calculus, used for finding the derivative of composite functions. Imagine you have a function within another function, like nesting two boxes. The chain rule helps us differentiate the outer function in relation to the inner one.
The chain rule formula is: \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
This process allows us to express the derivative of \(y\) in terms of \(x\), even though both are defined parametrically through \(t\).
The chain rule formula is: \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
- Where \(y\) is a function of \(u\), and \(u\) is a function of \(x\).
- Thus, \(y\) is indirectly a function of \(x\).
This process allows us to express the derivative of \(y\) in terms of \(x\), even though both are defined parametrically through \(t\).
Quotient Rule
The quotient rule is used when you need to differentiate a fraction where both the numerator and the denominator are functions. It's like if you have two animals stacked on top of each other, and you need to figure out how their shapes change. The rule provides a method to find the overall change.
The formula for the quotient rule is: \[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
By applying this rule, we manage to simplify the complex relationship of second derivatives, ensuring we respect the contribution of both the numerator and denominator over the parameter \(t\).
The formula for the quotient rule is: \[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
- Here, \(u\) is the numerator function.
- \(v\) is the denominator function.
By applying this rule, we manage to simplify the complex relationship of second derivatives, ensuring we respect the contribution of both the numerator and denominator over the parameter \(t\).
Parametric Equations
Parametric equations let you express a curve by linking coordinates to parameters. You can think of the parameter \(t\) as time, showing how \(x\) and \(y\) positions evolve. This concept is different from standard Cartesian equations and is really useful for curves not easily described by one equation.
In the exercise, parametric equations \(x=\phi(t)\) and \(y=\psi(t)\) give shapes as functions of the parameter \(t\).
This is how you connect the dots between independent parametric expressions and their interdependent derivative for a full description of the curve's behavior.
In the exercise, parametric equations \(x=\phi(t)\) and \(y=\psi(t)\) give shapes as functions of the parameter \(t\).
- These equations are brilliant for modeling real-world phenomena, like projectile paths.
- Changing \(t\) moves you along the curve, like a slider.
This is how you connect the dots between independent parametric expressions and their interdependent derivative for a full description of the curve's behavior.
Other exercises in this chapter
Problem 281
$$ \text { If } y=\tan (x+y), \text { find } \frac{d^{3} y}{d x^{3}} $$
View solution Problem 282
$$ \text { If } x^{2}+y^{2}=r^{2}, \text { find }\left(\frac{d^{3} y}{d x^{3}}\right)_{x=0} $$
View solution Problem 284
$$ \text { If } x=a t^{2}, y=2 a t, \text { find } \frac{d^{2} y}{d x^{2}} $$
View solution Problem 285
$$ \text { If } x=a t^{2}, y=b t^{3}, \text { find } \frac{d^{2} y}{d x^{2}} \text { . } $$
View solution