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Magazine Chapter 10: Problem 2
Find \(d y / d x\) $$x=1 / t, \quad y=\sqrt{t} e^{-t}$$
Short Answer
Expert verified
\( \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \)
Step by step solution
01
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:\[ \frac{d x}{d t} = -\frac{1}{t^2} \]
02
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:\[ \frac{d y}{d t} = \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t}e^{-t} \]
03
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:\[ \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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, \(t\).
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.
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 \(y = \sqrt{t} e^{-t}\):
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 \( v' = -e^{-t} \).
- \( \frac{d y}{d t} = \left(\frac{1}{2\sqrt{t}}\right) e^{-t} + \sqrt{t}(-e^{-t}) \).
- Simplifying this, we get \( \frac{e^{-t}}{2\sqrt{t}} - \sqrt{t} e^{-t} \), which is key in further calculations leading to \( \frac{d y}{d x} \).
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:
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 \( \frac{d y}{d x} = -t \left( \frac{1 - 2t^3}{2} \right)e^{-t} \).
