Problem 2 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$x=1 / t, \quad y=\sqrt{t} e^{-t}$$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \)

Step 1: Differentiate x with respect to t

The equation provided is \(x = \frac{1}{t}\). We need to find \( \frac{d x}{d t} \). Differentiating \( x \) with respect to \( t \), we get:\[ \frac{d x}{d t} = -\frac{1}{t^2} \]

Step 2: Differentiate y with respect to t

The equation for \( y \) is \( y = \sqrt{t} e^{-t} \). We'll use the product rule \((uv)' = u'v + uv'\) to differentiate. Let \( u = \sqrt{t} \) and \( v = e^{-t} \).\( u' = \frac{1}{2\sqrt{t}} \) and \( v' = -e^{-t} \), so:\[ \frac{d y}{d t} = \left(\frac{1}{2\sqrt{t}}\right) e^{-t} + \sqrt{t}(-e^{-t}) \] Simplifying:\[ \frac{d y}{d t} = \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \]

Step 3: Find dy/dx using the chain rule

Now that we have \( \frac{d x}{d t} = -\frac{1}{t^2} \) and \( \frac{d y}{d t} = \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \), apply the chain rule:\[ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \]Substitute the derivatives found:\[ \frac{d y}{d x} = \frac{\frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t}}{-\frac{1}{t^2}} \] Simplify the expression:\[ \frac{d y}{d x} = -t^2 \left( \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \right) \] Factor and simplify further:\[ \frac{d y}{d x} = -t^2 \left( \frac{1}{2\sqrt{t}} - \sqrt{t} \right)e^{-t} \]which becomes:\[ \frac{d y}{d x} = -t^2 \left( \frac{1}{2t} - t \right)e^{-t} \] After simplifying, this becomes:\[ \frac{d y}{d x} = -t^2 \left( \frac{1}{2t} - \frac{t^2}{t} \right)e^{-t} = -t^2 \left( \frac{1 - 2t^3}{2t} \right)e^{-t} \] Finally:\[ \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \]

Key Concepts

Differentiation

Differentiation is a core concept in calculus that focuses on understanding how a function changes as its input changes. It is essentially the process of finding the derivative of a function, which represents the rate of change or the slope of the function at any given point.
In the context of this exercise, we have two equations: \(x = \frac{1}{t}\) and \(y = \sqrt{t} e^{-t}\). To find how \(y\) changes with respect to \(x\), we must first differentiate both \(x\) and \(y\) with respect to another variable, \(t\).

• This involves using basic differentiation rules, such as the power and exponential rules, to calculate \( \frac{d x}{d t}\) and \( \frac{d y}{d t} \).
• The result for \( \frac{d x}{d t} \) from the given equation \( x = \frac{1}{t} \) is \( -\frac{1}{t^2} \).
• For \( y \), we need to use the product rule for its differentiation as it is a product of two functions.

The process of differentiation here sets the stage for applying the chain rule to find the desired \( \frac{d y}{d x} \).

Product Rule

The product rule is a technique in differentiation used when you have to differentiate products of two or more functions. In mathematical terms, if you have a function \(y = uv\), where both \(u\) and \(v\) are functions of \(t\), the derivative is given by \((uv)' = u'v + uv'\). This rule allows differentiation of the product by treating each function separately.
Applying the product rule to our original function \(y = \sqrt{t} e^{-t}\):

• Let \( u = \sqrt{t} \), which gives \( u' = \frac{1}{2\sqrt{t}} \).
• Let \( v = e^{-t} \), which gives \( v' = -e^{-t} \).

Combining these, we calculate the derivative \( \frac{d y}{d t} \) using the product rule:

• \( \frac{d y}{d t} = \left(\frac{1}{2\sqrt{t}}\right) e^{-t} + \sqrt{t}(-e^{-t}) \).
• Simplifying this, we get \( \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t} e^{-t} \), which is key in further calculations leading to \( \frac{d y}{d x} \).

This illustrates how the product rule breaks down the complexity of differentiating compounded functions.

dy/dx Calculation

The calculation of \( \frac{d y}{d x} \) in calculus often involves multiple steps, especially when dealing with composite functions. Here, the chain rule is employed to find \( \frac{d y}{d x} \). The chain rule states that \( \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \) whenever \(x\) and \(y\) are both functions of \(t\).
From our exercise, after finding \( \frac{d x}{d t} = -\frac{1}{t^2} \) and \( \frac{d y}{d t} = \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \), we apply the chain rule:

• \( \frac{d y}{d x} = \frac{\left( \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \right)}{-\frac{1}{t^2}} \).
• Simplifying leads to \( -t^2 \left( \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \right) \).
• Further simplification results in \( \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \).

This result encapsulates the changes in \(y\) with respect to \(x\) by considering each variable relative to \(t\). Understanding this multi-step calculation process is crucial for mastering calculus and its applications.