Problem 12 - Essential Calculus

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ x=\sqrt{t}, y=\sqrt{t-1} \quad t=2 $$

Step-by-Step Solution

Answer

The slope or the first derivative \(dy/dx\) at \(t=2\) is \(\sqrt{2}\) and the concavity or the second derivative \(d^2y/dx^2\) at \(t=2\) is \(0\).

Step 1: Derive \(x\) and \(y\) with respect to \(t\)

Firstly, differentiate \(x = \sqrt{t}\) and \(y = \sqrt{t-1}\).Given, \(x=\sqrt{t} = t^{1/2}\), then \[\frac{dx}{dt} = \frac{1}{2}t^{-1/2}\]Also, given \(y = \sqrt{t-1} = (t-1)^{1/2}\), thus \[\frac{dy}{dt} = \frac{1}{2}(t-1)^{-1/2}\]

Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule, \[\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2}(t-1)^{-1/2}}{\frac{1}{2}t^{-1/2}}\]which simplifies to\[\frac{dy}{dx} = \frac{(t-1)^{-1/2}}{t^{-1/2}} = \frac{t^{1/2}}{(t-1)^{1/2}} = \frac{\sqrt{t}}{\sqrt{t-1}}.\]

Step 3: Find \(\frac{d^2y}{dx^2}\)

To find the second derivative, differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide it by \(\frac{dx}{dt}\).When computing the derivative of \[\frac{\sqrt{t}}{\sqrt{t-1}},\]this results in\[\frac{d^2y}{dx^2} = \frac{(t-1)^{-1/2} - 1/2t^{1/2}(t-1)^{-3/2}}{1/2t^{-1/2}}\]which simplifies as\[\frac{d^2y}{dx^2}= 2(t-1) - t(t-1)^2.\]

Step 4: Evaluate the derivatives

Evaluate \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) at \(t=2\).So, \(\frac{dy}{dx}\) at \(t=2\) can be found as:\[\frac{dy}{dx} = \frac{\sqrt{2}}{\sqrt{2-1}} = \sqrt{2}.\]Similarly, \(\frac{d^2y}{dx^2}\) at \(t=2\) becomes:\[\frac{d^2y}{dx^2} = 2(2 - 1) - 2(2 - 1)^2 = 2 - 2 = 0.\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions.
It helps find the derivative of a function that is nested within another function.
In simple terms, if you have a function inside another function, like \(u = g(t)\) and \(y = f(u)\), the derivative of \(y\) with respect to \(t\) can be found using the chain rule which states:

• \(\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}\)

In our exercise, we are working with parametric equations \(x = \sqrt{t}\) and \(y = \sqrt{t-1}\).
To find \(\frac{dy}{dx}\), we first differentiate \(x\) and \(y\) with respect to \(t\):

• \(\frac{dx}{dt} = \frac{1}{2}t^{-1/2}\)
• \(\frac{dy}{dt} = \frac{1}{2}(t-1)^{-1/2}\)

Then, using the chain rule, the derivative \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), which simplifies to \(\frac{\sqrt{t}}{\sqrt{t-1}}\).
This shows how the rate of change of \(y\) with respect to \(x\) varies, given the parameter \(t\).

Second Derivative

Understanding the second derivative is important as it reveals information about the function's acceleration or change of the slope.
The second derivative is the derivative of the first derivative and is denoted as \(\frac{d^2y}{dx^2}\).When finding the second derivative in parametric form, it involves taking the derivative of \(\frac{dy}{dx}\) with respect to \(t\) again and dividing it by \(\frac{dx}{dt}\) once more.
In our step-by-step solution, we start by differentiating \(\frac{\sqrt{t}}{\sqrt{t-1}}\) with respect to \(t\).
This leads to a computation involving the derivatives:

• The resulting formula is \(2(t-1) - t(t-1)^2\).

Evaluating this at \(t=2\), gives us \(0\), indicating that the rate of change in the slope is neither increasing nor decreasing at this particular point.

Slope and Concavity

The concepts of slope and concavity help describe the shape of a graph.
While the slope of a curve tells us how steep it is at a certain point, concavity tells us how the curve bends.**Slope**
In the parametric form, the slope is given by the first derivative \(\frac{dy}{dx}\).
For \(t=2\), we found that \(\frac{dy}{dx} = \sqrt{2}\).
This positive value indicates that the tangent line at \(t=2\) is sloping upwards.**Concavity**
Concavity is determined by the sign of the second derivative \(\frac{d^2y}{dx^2}\).
When \(\frac{d^2y}{dx^2} > 0\), the graph is concave up (like a cup); when \(\frac{d^2y}{dx^2} < 0\), it's concave down (like a cap).
At \(t=2\), since \(\frac{d^2y}{dx^2} = 0\), this means the graph is at an inflection point, where its concavity might change.
This sort of detailed analysis of slope and concavity provides a clearer picture of the overall behavior of a function's graph.