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Magazine Chapter 7: Problem 73
For the following exercises, find all points on the curve that have the given slope. \(x=t+\frac{1}{t}, \quad y=t-\frac{1}{t},\) slope \(=1\)
Short Answer
Expert verified
There are no points on the curve with a slope of 1.
Step by step solution
01
Parametric Equations for Derivatives
Given the parametric equations \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \), we need to find their derivatives with respect to \( t \). We have \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \) and \( \frac{dy}{dt} = 1 + \frac{1}{t^2} \).
02
Expression for Slope of Curve
The slope of the tangent to the curve is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute the derivatives from step 1 to obtain the expression: \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \).
03
Set Slope to 1
We are given that the slope \( \frac{dy}{dx} = 1 \). Substitute into the slope expression: \( \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = 1 \). Solving this equation will give us the values of \( t \) where the slope is 1.
04
Solve for t
Multiply both sides by \( 1 - \frac{1}{t^2} \) to clear the fraction: \( 1 + \frac{1}{t^2} = 1 - \frac{1}{t^2} \). Simplifying gives \( \frac{2}{t^2} = 0 \), which implies no real values for \( t \) since \( \frac{2}{t^2} \) cannot be zero.
05
Evaluate for Solutions
The previous steps indicate that there is a mistake: \( \frac{2}{t^2} = 0 \) has no real solutions. We must recompute or verify earlier calculations, but given no alternatives appeared from the algebra, this assertively points to no points on the curve having a slope of 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
When dealing with parametric equations, derivatives play a crucial role in understanding the behavior of curves. Derivatives help us determine essential properties like velocity, acceleration, and most importantly in this case, the slope of a tangent line. In our exercise, we start with the parametric equations for x and y:
- \( x = t + \frac{1}{t} \)
- \( y = t - \frac{1}{t} \)
- The derivative of \( x \), \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \)
- The derivative of \( y \), \( \frac{dy}{dt} = 1 + \frac{1}{t^2} \)
Curve Slope
The slope of a curve at a given point is a vital measure in calculus, representing the rate of change. For parametric equations, the slope of the curve is defined using the relationship between the derivatives of x and y concerning t. Specifically, it is given by the expression \( \frac{dy}{dx} \), which equates to \( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).
In the context of our exercise, we substitute the derivatives found previously:
In the context of our exercise, we substitute the derivatives found previously:
- \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \)
Tangent Line
In calculus, a tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the instantaneous direction of the curve at that particular point. To find where a tangent line has a specific slope鈥攍ike 1 in our exercise鈥攚e equate \( \frac{dy}{dx} \) to the desired slope.
In our example:
In our example:
- We need to solve \( \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = 1 \)
Calculus Problem-Solving
Solving calculus problems requires a systematic approach to applying mathematical concepts and properties. When given parametric equations, it's vital to:
- Differentiate the equations to find rates of change.
- Use derivative expressions to determine curve properties like slopes and tangents.
- Set up equations to solve for specific conditions, such as given slopes.
- Analyze resulting equations for possible solutions, bearing in mind the domain and viability of solutions.
