Problem 33 - AP Calculus with 8 Practice Tests 15th

Question

If \(x=e^{\theta} \cos \theta\) and \(y=e^{\theta} \sin \theta,\) then, when \(\theta=\frac{\pi}{2}, \frac{d y}{d x}\) is (A) 1 (B) 0 (C) \(e^{\pi / 2}\) (D) -1

Step-by-Step Solution

Answer

(D) -1

Step 1: Differentiate x with respect to θ

Given the expression for \(x\) as \(x = e^{\theta} \cos \theta\), differentiate this with respect to \(\theta\), resulting in \(\frac{dx}{d\theta} = e^{\theta} \cos \theta - e^{\theta} \sin \theta = e^{\theta}(\cos \theta - \sin \theta)\).

Step 2: Differentiate y with respect to θ

Given the expression for \(y\) as \(y = e^{\theta} \sin \theta\), differentiate this with respect to \(\theta\), resulting in \(\frac{dy}{d\theta} = e^{\theta} \sin \theta + e^{\theta} \cos \theta = e^{\theta}(\sin \theta + \cos \theta)\).

Step 3: Compute dy/dx using the chain rule

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\). Substitute the expressions from Steps 1 and 2 to get \(\frac{dy}{dx} = \frac{e^{\theta}(\sin \theta + \cos \theta)}{e^{\theta}(\cos \theta - \sin \theta)} = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}\).

Step 4: Evaluate dy/dx at θ = π/2

Substitute \(\theta = \frac{\pi}{2}\) into \(\frac{dy}{dx} = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}\). We have \(\sin \frac{\pi}{2} = 1\) and \(\cos \frac{\pi}{2} = 0\). Substituting gives \(\frac{1 + 0}{0 - 1} = -1\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that's essential for understanding how to differentiate composite functions. It's like following a pathway where each step depends on the previous one. Imagine if you have a function of a function, say, \[y = f(u)\] and \[u = g(x)\]. Using the chain rule, if you want to differentiate \(y\) with respect to \(x\), you first differentiate \(y\) with respect to \(u\), and then multiply by the derivative of \(u\) with respect to \(x\): \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \] In the context of the exercise, computing \(\frac{dy}{dx}\) required differentiating \(x\) and \(y\) with respect to \(\theta\), then using their derivatives to find the derivative of one variable with respect to another. This process is crucial to solve the problem correctly, allowing you to account for the dependence of \(x\) and \(y\) on \(\theta\). It is the step that connects the rates of change of both variables efficiently.

Trigonometric Functions

Trigonometric functions like \(\sin \theta\) and \(\cos \theta\) are the building blocks of periodic behavior in mathematics. They are essential for solutions involving angles, rotations, and many physical phenomena. In the given exercise, \(x\) is expressed as \(e^{\theta} \cos \theta\) and \(y\) as \(e^{\theta} \sin \theta\). This blend of trigonometric functions with exponential terms shows their versatility in calculus.

To differentiate such terms, you often utilize rules specific to trigonometric derivatives:

• The derivative of \(\sin \theta\) is \(\cos \theta\).
• The derivative of \(\cos \theta\) is \(-\sin \theta\).

When performing these differentiations in the exercise, note how we treat them as part of a product, necessitating the use of the product rule: \((uv)' = u'v + uv'\), which was just as crucial as the chain rule in reaching a final result.

Exponential Functions

Exponential functions involve expressions of the form \(e^{x}\), where \(e\) is a mathematical constant approximately equivalent to 2.71828. They grow rapidly and appear frequently in calculus due to their neat and predictable differentiation properties. In the problem, \(e^{\theta}\) multiplies the trigonometric components, representing a combination of exponential growth and periodic oscillation. The key property of exponential functions is that they are their own derivatives, meaning \[ \frac{d}{d\theta}(e^{\theta}) = e^{\theta}. \] This makes calculations involving these functions straightforward.
When differentiating as \(\theta\) varies, \(e^{\theta}\) remained untouched except for its multiplication role, highlighting this property. Their predictable growth behavior makes tasks involving them accessible and efficient, directly supporting the outcome of the calculation in the original problem.