Problem 1 - Calculus: An Intuitive and Physical Approach

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2}\) for each of the following parametric representations of the functions relating \(x\) and \(y\) : (a) \(x=2-t, y=t^{2}\). Ans. \(d y / d x=-2 t, d^{2} y / d x^{2}=2\). (b) \(x=3 t^{2}, y=4 t-5\). (c) \(x=t e^{-t}, y=e^{t}\). Ans. \(d y / d x=e^{2 t} /(1-t), d^{2} y / d x^{2}=e^{3 t}(3-2 t)(1\) \(-t)^{3}\).

Step-by-Step Solution

Answer

Part (b): \(dy/dx = \frac{2}{3t}\), \(d^{2} y / d x^{2} = -1/(9t^3)\).

Step 1 Title - Derivative of y with respect to t

For part (b), calculate the derivative of y with respect to t, denoted as \(dy/dt\). Differentiate the function for y: \(y = 4t - 5\). The derivative is \(dy/dt = 4\), since the derivative of a constant is zero, and the derivative of \(t\) with respect to \(t\) is 1.

Step 2 Title - Derivative of x with respect to t

Next, calculate the derivative of x with respect to t, denoted as \(dx/dt\). Differentiate the function for x: \(x = 3t^2\). Using the power rule, the derivative is \(dx/dt = 6t\).

Step 3 Title - Derivative of y with respect to x

Now, calculate \(dy/dx\), which is \(dy/dx = \frac{dy/dt}{dx/dt}\). Substitute the values found in Steps 1 and 2 to get \(dy/dx = \frac{4}{6t} = \frac{2}{3t}\).

Step 4 Title - Second derivative of y with respect to t (Optional for part b)

Although not required for the answer, for a complete understanding, one might consider calculating the second derivative of y with respect to t, denoted as \(d^2y/dt^2\), which is zero in this case, as the first derivative of y with respect to t is a constant.

Step 5 Title - Second derivative of x with respect to t

Calculate the second derivative of x with respect to t, denoted as \(d^2x/dt^2\), which is the derivative of \(dx/dt\). Differentiating the result from Step 2 gives \(d^2x/dt^2 = 6\).

Step 6 Title - Second derivative of y with respect to x using chain rule

To calculate the second derivative \(d^2y/dx^2\), use the following formula: \(d^2y/dx^2 = \frac{d}{dx}(dy/dx) = \frac{d}{dt}(dy/dx) / (dx/dt)\). Since \(dy/dx\) is \(2/(3t)\), and \(dx/dt = 6t\), \(d^2y/dx^2\) will be \(\frac{d}{dt}(2/(3t)) / 6t\).

Step 7 Title - Find the derivative of dy/dx with respect to t

Differentiate \(dy/dx = 2/(3t)\) with respect to t to find \(d/dt ( dy/dx )\). The derivative is \(d/dt (2/(3t)) = -2/(3t^2)\) by the power rule.

Step 8 Title - Finalize the second derivative calculation

Now substitute into the second derivative formula to get \(d^2y/dx^2 = (-2/(3t^2)) / (6t) = -1/(9t^3)\). This is the second derivative of y with respect to x for part (b).

Key Concepts

Derivative of Parametric Functions

When dealing with parametric equations, where a third variable, typically represented as 't', defines both x and y, the derivative of y with respect to x, denoted as \( \frac{dy}{dx} \), isn't directly obtained as in standard functions. Instead, we rely on indirect differentiation, where initially, both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are found separately.

Applying the concept to example (b) given as \( x = 3t^2, y = 4t - 5 \), the derivatives with respect to 't' are \( \frac{dx}{dt} = 6t \) and \( \frac{dy}{dt} = 4 \), as found in steps 1 and 2 of the solution. To get \( \frac{dy}{dx} \) we then divide the derivative of y by the derivative of x with respect to t, which leads us to \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{6t} \).

The process of finding this derivative through parametric equations can be thought of as a two-step chain reaction. The rate of change of y with respect to t is being adjusted by the rate of change of t with respect to x. This is why understanding the behavior of a function in parametric form requires an additional step of calculus.

Second Derivative Parametric Equation

Proceeding to the second derivative in parametric equations, denoted as \( \frac{d^2y}{dx^2} \), one must consider the acceleration or the curvature of the function y with respect to x. It is the rate at which \( \frac{dy}{dx} \) itself changes as x changes.

In step 6 of the solution for example (b), we transition from the first to the second derivative through the implementation of the chain rule, which again showcases the interconnectedness of x, y, and t. It's calculated by finding the derivative of \( \frac{dy}{dx} \) with respect to t, divided by \( \frac{dx}{dt} \), simplifying it further to \( \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} \).

Finally, step 8 shows that for example (b), the second derivative is \( \frac{d^2y}{dx^2} = \frac{-2/(3t^2)}{6t} = -\frac{1}{9t^3} \). This reveals the nature of the curve and how y's change rate accelerates for varying x, which can be particularly insightful for understanding motion in physics or rate of growth or decay in other applications.

Chain Rule in Parametrics

The chain rule is a fundamental tool in calculus used to take derivatives of compositions of functions. In parametric calculus, the chain rule is utilized to relate the derivatives with respect to the parameter 't' to those with respect to 'x'.

It plays an integral role when we seek the first and second derivatives of parametric equations. The chain rule's essence is evident in step 6, where the second derivative of y with respect to x is deduced. We differentiate \( \frac{dy}{dx} \) with respect to t, and then divide by \( \frac{dx}{dt} \), essentially chaining the derivative with respect to t to that with respect to x.

This process, laid out in steps 6 to 8, forms the backbone of finding parametric derivatives and is a prime example of how the chain rule is applied outside the realm of standard functions. The elegance of the chain rule in parametrics allows students to seamlessly move between rates of change with respect to different variables, making sense of complex relationships in dynamic systems.