Problem 66 - 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 \(t .\) $$ x \sin t+2 x=t, \quad t \sin t-2 t=y, \quad t=\pi $$

Step-by-Step Solution

Answer

The slope of the curve at \(t = \pi\) is \(-4\).

Step 1: Calculate Partial Derivatives of Equation 1

The first equation given is \(x \sin t + 2x = t\). We need to find \(\frac{dx}{dt}\) by differentiating partially with respect to \(t\). Differentiating, we get: \(\sin t \frac{dx}{dt} + x \cos t = 1 - 2\frac{dx}{dt}\). Solve for \(\frac{dx}{dt}\) to get: \[\frac{dx}{dt} = \frac{1 - x \cos t}{\sin t - 2}\].

Step 2: Calculate Partial Derivatives of Equation 2

The second equation is \(t \sin t - 2t = y\). Differentiate with respect to \(t\) to find \(\frac{dy}{dt}\). This gives us: \(\sin t + t \cos t - 2 = \frac{dy}{dt}\). So, \(\frac{dy}{dt} = \sin t + t \cos t - 2\).

Step 3: Substitute the Given Value of \(t\)

We know \(t = \pi\). Substitute \(t = \pi\) in both differentiated expressions. For \(\frac{dx}{dt}\): \(\frac{dx}{dt} = \frac{1 - x \cos(\pi)}{\sin(\pi) - 2}\ = \frac{1 + x}{-2}\). For \(\frac{dy}{dt}\): \(\frac{dy}{dt} = \sin(\pi) + \pi \cos(\pi) - 2 = \pi(-1) - 2 = -\pi - 2\).

Step 4: Find the Slope at \(t = \pi\)

The slope of the curve is given by \(\frac{dy/dt}{dx/dt}\). Substitute the values from the previous steps: \[\frac{dy/dt}{dx/dt} = \frac{-\pi - 2}{-\frac{1 + x}{2}} = \frac{-2(\pi + 2)}{1 + x}\].

Step 5: Solve for \(x\) at \(t = \pi\)

Substitute \(t = \pi\) into the original equation for \(x\): \[x \sin(\pi) + 2x = \pi\] simplifies to \[2x = \pi\]. Solving for \(x\) gives \(x = \frac{\pi}{2}\).

Step 6: Compute the Final Slope

Substitute \(x = \frac{\pi}{2}\) into the slope expression: \[\frac{-2(\pi + 2)}{1 + \frac{\pi}{2}} = \frac{-2(\pi + 2)}{1 + \frac{\pi}{2}} = \frac{-4(\pi + 2)}{2 + \pi}\]. Simplifying further gives us \(-4\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental tool in calculus for breaking down functions that have more than one variable. In the context of implicit differentiation, they help us understand how a particular variable changes with respect to time, holding others constant. For instance, when you look at a function like \(x \sin t + 2x = t\), both \(x\) and \(t\) can change. Applying partial differentiation involves treating one variable, such as \(x\), as a constant while differentiating with respect to another variable, like \(t\). This allows us to pinpoint the contribution of a single element to the overall change.

To calculate \( \frac{dx}{dt} \) for the first equation, we differentiate each term concerning \(t\). Terms without a \(t\) are treated as constants, thus ignoring them in differentiation. Here, the sine term reflects how \(x\) dynamically changes with \(t\). Solving for \( \frac{dx}{dt} \) involves algebraic manipulation to isolate it, indicating how fast \(x\) itself shifts as \(t\) progresses. This step-by-step breakdown eases the complexity often encountered with functions involving multiple variables. As you continue with partial derivatives, try to identify each variable’s role and influence in the equation.

Differentiable Functions

Differentiable functions are the backbone of calculus, serving as the basis for both explicit and implicit differentiation. A function is differentiable at a point if it has a defined derivative at that point, implying it is smooth and has no abrupt edges or breaks. This property is crucial since it assures us we can obtain meaningful information about the function's behavior—such as its slope—at any point within its domain.

In the given exercise, because the functions \(x = f(t)\) and \(y = g(t)\) are stated as differentiable, it allows us to confidently compute derivatives of these implicit functions with respect to \(t\). Without differentiability, our calculated derivatives wouldn’t provide reliable or interpretable insights into how \(x\) and \(y\) evolve as \(t\) changes. Differentiable functions appear smooth on the graph and can be dissected mathematically to understand intricate behaviors, paving the way to analyze slopes effectively using derivatives.

Slope of a Curve

The slope of a curve at a particular point describes the steepness or direction of the curve at that point. It's computed as the ratio of the change in the \(y\)-coordinate to the change in the \(x\)-coordinate, often expressed as \( \frac{dy}{dx} \). In implicit differentiation, where functions for \(x\) and \(y\) are defined indirectly with respect to \(t\), finding the slope involves calculating both \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), and then dividing these derivatives to obtain the slope \( \frac{dy/dt}{dx/dt} \).

In our example, the slope at \(t = \pi\) was determined by first solving for \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) when \(t = \pi\). By substituting these back into the slope formula, it reveals how the \(y\)-value changes compared to the \(x\)-value. In essence, this signifies the tangent's inclination concerning the underlying curve—whether it rises, falls, or stays flat. The precision in calculating these derivatives ensures that the slope encapsulates all the instantaneous changes occurring to the curve at exactly that point of interest, providing a snapshot of the curve's behavior.