Problem 63 - Thomas Calculus

Question

Assuming that the equations in Exercises \(63-66\) define \(x\) and \(y\) implicitly as differentiable functions \(x=f(t), y=g(t),\) find the slope of the curve \(x=f(t), y=g(t)\) at the given value of \(t .\) $$ x^{2}-2 t x+2 t^{2}=4, \quad 2 y^{3}-3 t^{2}=4, \quad t=2 $$

Step-by-Step Solution

Answer

The slope is undefined at \( t = 2 \), indicating a vertical tangent.

Step 1: Differentiate the first equation with respect to t

The first equation is \( x^2 - 2tx + 2t^2 = 4 \). Differentiate both sides with respect to \( t \). Recall that \( x \) is a function of \( t \), so use the chain rule where necessary: \[ \frac{d}{dt}(x^2) - \frac{d}{dt}(2tx) + \frac{d}{dt}(2t^2) = \frac{d}{dt}(4) \]. This simplifies to: \[ 2x \frac{dx}{dt} - (2x + 2t \frac{dx}{dt}) + 4t = 0 \].Distribute the terms to get: \[ 2x \frac{dx}{dt} - 2x - 2t \frac{dx}{dt} + 4t = 0 \].Combine like terms: \[ (2x - 2t) \frac{dx}{dt} = 2x - 4t \].Which simplifies to: \[ \frac{dx}{dt} = \frac{2x - 4t}{2x - 2t} \].

Step 2: Differentiate the second equation with respect to t

The second equation is \( 2y^3 - 3t^2 = 4 \). Differentiate both sides with respect to \( t \), again, using the chain rule:\[ \frac{d}{dt}(2y^3) - \frac{d}{dt}(3t^2) = \frac{d}{dt}(4) \]. Which becomes: \[ 6y^2 \frac{dy}{dt} - 6t = 0 \].Solve for \( \frac{dy}{dt} \): \[ 6y^2 \frac{dy}{dt} = 6t \].Therefore, \[ \frac{dy}{dt} = \frac{t}{y^2} \].

Step 3: Evaluate the derivatives at t=2

We need the values of \( x \) and \( y \) at \( t = 2 \) from the original equations. 1. Substitute \( t = 2 \) into \( x^2 - 2tx + 2t^2 = 4 \): \[ x^2 - 4x + 8 = 4 \]. Simplifying gives: \[ x^2 - 4x + 4 = 0 \]. Factoring, \( (x-2)^2 = 0 \), so \( x = 2 \).2. Substitute \( t = 2 \) into \( 2y^3 - 3t^2 = 4 \): \[ 2y^3 - 12 = 4 \]. Simplifying gives: \[ 2y^3 = 16 \], so \( y^3 = 8 \), and \( y = 2 \).Now substitute these values into the derivatives:- From Step 1, \( \frac{dx}{dt} = \frac{2x - 4t}{2x - 2t} = \frac{4 - 8}{4 - 4} = \text{undefined} \).- From Step 2, \( \frac{dy}{dt} = \frac{t}{y^2} = \frac{2}{4} = \frac{1}{2} \).

Step 4: Calculate the slope of the curve

The slope of the curve is given by \( \frac{dy}{dx} \). Since \( \frac{dx}{dt} \) is undefined at \( t = 2 \), the slope cannot be expressed as a finite number. Thus the curve is vertical at \( t = 2 \), if it exists.

Key Concepts

Differentiable Functions

Differentiable functions are essential when it comes to calculus and understanding how variables change. These functions are smooth, meaning they don't have sharp corners or breaks. This smoothness allows us to find derivatives, which tell us how much a function changes at any given point. In the context of implicit differentiation, we often deal with equations where both variables are functions of another variable, like time \( t \). For example, in our exercise, both \( x \) and \( y \) are differentiable functions of \( t \). Whenever you have equations like these, it's crucial that each variable can be differentiated with respect to \( t \), meaning they have reliable rates of change that can be calculated.

Chain Rule

The chain rule is a fundamental tool for finding derivatives of composite functions. It essentially states that to differentiate a function composition, you must take the derivative of the outer function and multiply it by the derivative of the inner function. This is especially useful in implicit differentiation where the functions you're working with are interdependent.
In our given exercise when we differentiate equations like \( x^2 - 2tx + 2t^2 = 4 \), we see the chain rule in action. Since \( x \) is a function of \( t \), differentiating \( x^2 \) involves multiplying the derivative \( 2x \) by \( \frac{dx}{dt} \). This step is crucial because it accounts for how \( x \) changes with respect to \( t \), ensuring that we're considering all relationships within the function.

Slope of a Curve

The slope of a curve at a given point tells us how steep the curve is. It's the equivalent of finding "rise over run" but for a function that isn't necessarily a straight line. In calculus, the slope is found using the derivative. When a problem involves implicit equations, finding the slope means taking the derivative of every part involved.
In the step-by-step solution, the slope is expressed as \( \frac{dy}{dx} \), which involves calculating both \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). Evaluating these at \( t = 2 \), we find some unique cases, such as \( \frac{dx}{dt} \) being undefined, which implies the slope at that point is vertical, suggesting a sharp change in direction for the function.

Implicit Equations

Implicit equations describe a relationship between two variables in a way that isn’t explicitly solved for one variable in terms of the other. They often pose a challenge because unlike explicit equations, they require taking derivatives indirectly. For instance, in our exercise, we have equations like \( x^2 - 2tx + 2t^2 = 4 \) and \( 2y^3 - 3t^2 = 4 \). These implicit equations require thoughtful manipulation and use of implicit differentiation techniques to uncover more insights.
By differentiating implicitly, you solve for \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), which are the rates of change with respect to \( t \), reflecting how both \( x \) and \( y \) evolve. This insight is necessary to fully appreciate the relationships imposed by the equations.