Problem 34 - AP Calculus with 8 Practice Tests 15th

Question

If \(x=\cos t\) and \(y=\cos 2 t,\) then \(\frac{d^{2} y}{d x^{2}}(\sin t \neq 0)\) is (A) \(4 \cos t\) (B) 4 (C) -4 (D) \(-4 \cot t\)

Step-by-Step Solution

Answer

B) 4

Step 1: Express y in terms of t

Given that \(y = \cos 2t\), we can use the double angle identity for cosine: \(y = \cos 2t = 2 \cos^2 t - 1\). Since \(x = \cos t\), we can express \(y\) as \(y = 2x^2 - 1\).

Step 2: First derivative dy/dx

Find \(\frac{dy}{dx}\) by differentiating \(y = 2x^2 - 1\) with respect to \(x\). Using the derivative formula for \(a x^n\), we have \(\frac{dy}{dx} = 4x\).

Step 3: Second derivative d²y/dx²

Differentiate \(\frac{dy}{dx} = 4x\) with respect to \(x\) to find \(\frac{d^2y}{dx^2}\). The derivative of \(4x\) with respect to \(x\) is 4. Therefore, \(\frac{d^2y}{dx^2} = 4\).

Step 4: Verify condition and finalize

Since the condition \(\sin t eq 0\) implies \(t eq n\pi\) where \(n\) is an integer, there is no impact on \(\frac{d^2y}{dx^2} = 4\). The answer is unaffected by \(\sin t eq 0\) in this case.

Key Concepts

Second Derivative

The second derivative of a function gives us insight into the curvature of the function's graph. It shows how the rate of change of the slope, or the first derivative, behaves. If the second derivative is positive, it means the function is concave up, resembling a smile. When it's negative, the function is concave down, like a frown. In simple terms, it tells us about the acceleration of the function's behavior.In the problem at hand, we started with the function given by \( y = 2x^2 - 1 \). The first step was to find the first derivative, \( \frac{dy}{dx} = 4x \). Once we have the first derivative, we differentiate it once more to find the second derivative. Thus, \( \frac{d^2y}{dx^2} = 4 \).

• Positive second derivative means the graph is concave up.
• Negative second derivative indicates a concave down graph.
• The second derivative helps in determining points of inflection, where the graph changes concavity.

Finding the second derivative is crucial for understanding the behavior and shape of the function beyond just its slope.

Chain Rule

The chain rule in calculus is a fundamental tool for finding the derivative of composite functions. It's handy when dealing with functions nested inside each other, like an onion with multiple layers.To apply the chain rule, we differentiate the outer function and then multiply it by the derivative of the inner function. For example, when you have a function like \( y = f(g(x)) \), the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).In the context of our problem, we initially have \( x = \cos t \) and \( y = \cos 2t \). Using the chain rule helps us differentiate these expressions as t changes:

• Start by identifying the outer and inner functions.
• Differentiate the outer function.
• Multiply by the derivative of the inner function.

The chain rule is an essential part of calculating derivatives, especially when functions become complex or are given in parametric forms.

Parametric Equations

Parametric equations allow us to represent curves using a set of equations, often involving a third variable called the parameter. This method is particularly useful when describing more complex geometric shapes that are difficult to encapsulate with a single function of x or y.In our exercise, \( x = \cos t \) and \( y = \cos 2t \) are parametric equations with 't' being the parameter. This setup beautifully illustrates how both x and y vary as 't' changes.

• Parametric equations provide flexibility in modeling curves with multiple variables.
• Helpful in plotting trajectories and motions in physics and engineering.
• Enable conversion back to a single equation by eliminating the parameter, if possible.

Parametric forms are particularly beneficial because they allow us to consider the motion along a path, taking into account the time or angle, as seen with 't' in our problem.