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Magazine Chapter 1: Problem 73
For the following exercises, find all points on the curve that have the given slope. $$ x=t+\frac{1}{t}, \quad y=t-\frac{1}{t}, \text { slope }=1 $$
Short Answer
Expert verified
No real solution exists for the given slope of 1.
Step by step solution
01
Understand Parametric Equations
We have two parametric equations: \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \). These equations express \( x \) and \( y \) in terms of a parameter \( t \). Our goal is to find the points on the curve where the slope of the tangent is 1.
02
Find Derivatives of Parametric Equations
Calculate the derivatives of \( x \) and \( y \) with respect to \( t \). The derivatives are: \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \) and \( \frac{dy}{dt} = 1 + \frac{1}{t^2} \).
03
Express the Slope of the Curve
The slope of the curve is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \).
04
Set the Slope Equal to 1
We need the slope \( \frac{dy}{dx} \) to be equal to 1. Therefore, set the equation \( \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = 1 \) and solve for \( t \).
05
Solve the Equation for \( t \)
To solve \( \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = 1 \), cross multiply to get \( 1 + \frac{1}{t^2} = 1 - \frac{1}{t^2} \). Simplify to obtain \( \frac{2}{t^2} = 0 \), which does not hold for any real \( t \). This means the assumption was incorrect, indicating there was a miscalculation.
06
Correct the Approach
Upon correcting, setting the correct balance \( 1 + \frac{1}{t^2} = 1 - \frac{1}{t^2} \) should actually lead to \( \frac{2}{t^2} = 0 \), originally seen as incorrect, but a reevaluation confirms no real \( t \) exists for equal slope to 1.
07
Consider Other Adjustments, Reevaluate if Errors Occur
Since true balance or correction in steps lead to inequalities \((\frac{2}{t^2} = 0)\) for physical solutions of real numbers, evaluate solutions reiterate confirmature corrective calculus initiated with original steps ensuring precise alternative derivations if needed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope of Parametric Curve
The slope of a parametric curve expresses how steep the curve is at a particular point and is crucial in understanding the direction of the curve. Unlike standard Cartesian equations, the slope \(m\) is not given directly by \(\frac{dy}{dx}\); instead, when working with parametric equations like \(x = f(t)\) and \(y = g(t)\), the slope is found using the derivatives with respect to the parameter \(t\).
To express the slope, it is calculated using the ratio of the derivatives:
To express the slope, it is calculated using the ratio of the derivatives:
- First, we find \( \frac{dx}{dt} = f'(t) \)
- Next, we calculate \( \frac{dy}{dt} = g'(t) \)
- Finally, the slope \(m\) is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
Derivatives of Parametric Equations
Derivatives of parametric equations enable us to analyze the behavior of curves in much the same way as traditional calculus does for functions of a single variable. Calculating these derivatives helps in finding the rate at which each coordinate changes.
For our exercise, the derivatives of the given parametric equations are calculated as follows:
For our exercise, the derivatives of the given parametric equations are calculated as follows:
- The derivative of \(x=t+\frac{1}{t}\) gives \(\frac{dx}{dt} = 1 - \frac{1}{t^2}\).
- Similarly, for \(y=t-\frac{1}{t}\), the derivative is \(\frac{dy}{dt} = 1 + \frac{1}{t^2}\).
Solving Parametric Equations
Solving parametric equations involves finding values of the parameter \(t\) for which certain conditions on the curve are satisfied. In our case, we want the slope \(\frac{dy}{dx}\) to equal 1 to identify specific points on the curve.
This is achieved by equating the parametric derivative expression to the desired slope value:
This is achieved by equating the parametric derivative expression to the desired slope value:
- Set the derived slope \(\frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = 1\).
- Cross multiply to obtain \(1 + \frac{1}{t^2} = 1 - \frac{1}{t^2}\).
- This simplifies to \(\frac{2}{t^2} = 0\), indicating no valid real \(t\) satisfies the condition.
Calculus Problem Solving
Solving calculus problems with parametric curves often requires a blend of creativity and precision. Understanding how derivatives interact with parameterized functions is just part of the equation.
Here are key takeaways for tackling such problems:
Here are key takeaways for tackling such problems:
- Always verify your calculations at each step. Ensuring no errors in derivative calculations and solving method can prevent wrong conclusions.
- Parametric equations might have no solutions in some cases, as demonstrated here. This does not mean your method is incorrect but rather that the scenario requires further analysis.
- Sometimes assumptions need revisiting鈥攅rrors in setting conditions or solving can lead to misleading results.
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Parametric curves in calculus introduce unique challenges, and mastering them deepens your overall understanding of mathematical curves and slopes.
