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Magazine Chapter 11: Problem 13
Find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$x=\frac{1}{t+1}, \quad y=\frac{t}{t-1}, \quad t=2$$
Short Answer
Expert verified
Equation of tangent line: \( y = 9x - 1 \); \( \frac{d^2 y}{dx^2} = 90 \).
Step by step solution
01
Determine the Point on the Curve
To find the point on the curve corresponding to the given value of \( t \), substitute \( t = 2 \) into the expressions for \( x \) and \( y \). Calculate \( x \) and \( y \) as follows:\[ x = \frac{1}{t+1} = \frac{1}{2+1} = \frac{1}{3} \]\[ y = \frac{t}{t-1} = \frac{2}{2-1} = 2 \]Thus, the point is \( \left( \frac{1}{3}, 2 \right) \).
02
Find Derivatives dx/dt and dy/dt
First, find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Differentiating \( x \) with respect to \( t \), we get:\[ \frac{dx}{dt} = \frac{d}{dt}\left(\frac{1}{t+1}\right) = -\frac{1}{(t+1)^2} \]For \( y \):\[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{t}{t-1}\right) = \frac{(t-1)(1) - t(1)}{(t-1)^2} = \frac{-1}{(t-1)^2} \]
03
Calculate the Slope dy/dx
The slope of the tangent line, \( \frac{dy}{dx} \), is given by:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{1}{(t-1)^2}}{-\frac{1}{(t+1)^2}} = \frac{(t+1)^2}{(t-1)^2} \]Evaluate it at \( t = 2 \):\[ \frac{dy}{dx} \bigg|_{t=2} = \frac{(2+1)^2}{(2-1)^2} = \frac{9}{1} = 9 \]
04
Write the Equation of the Tangent Line
Use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \). Here \( (x_1, y_1) = \left( \frac{1}{3}, 2 \right) \) and \( m = 9 \):\[ y - 2 = 9\left(x - \frac{1}{3}\right) \]Simplify the equation:\[ y - 2 = 9x - 3 \]\[ y = 9x - 1 \]This is the equation of the tangent line.
05
Find Second Derivatives d虏y/dt虏 and d虏x/dt虏
Calculate the second derivatives with respect to \( t \):\[ \frac{d^2x}{dt^2} = \frac{d}{dt} \left(-\frac{1}{(t+1)^2}\right) = \frac{2}{(t+1)^3} \]\[ \frac{d^2y}{dt^2} = \frac{d}{dt} \left(-\frac{1}{(t-1)^2}\right) = \frac{2}{(t-1)^3} \]
06
Calculate d虏y/dx虏 at t=2
To find \( \frac{d^2y}{dx^2} \), use:\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]First calculate \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \):\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{(t+1)^2}{(t-1)^2}\right) \]\[ = \frac{2(t+1)(t-1)^2 - 2(t-1)(t+1)^2}{(t-1)^4} \]\[ = \frac{2[(t+1)(t-1)]}{(t-1)^4} \]At \( t = 2 \):\[ = \frac{2[(3)(1) - 2(4)]}{1} = \frac{2(3 - 8)}{1} = \frac{-10}{1} = -10 \]Finally:\[ \frac{d^2y}{dx^2} \bigg|_{t=2} = \frac{-10}{\frac{-1}{9}} = 90 \]
07
Conclusion
The equation of the tangent line at \((\frac{1}{3}, 2)\) is \( y = 9x - 1 \). The value of \( \frac{d^2 y}{dx^2} \) at this point is \( 90 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way of expressing mathematical functions where both the x and y coordinates are defined in terms of another variable, often 't'. This variable is known as the parameter, and it allows us to track a curve as 't progresses'. In our specific problem, the equations are given as:
\[ x = \frac{1}{t+1} \] and \[ y = \frac{t}{t-1} \].
By varying the parameter \( t \), we can find different points on the curve. When \( t \) is the independent variable, the corresponding values of \( x \) and \( y \) can be determined easily, as we did here with \( t = 2 \).
This approach is particularly useful in situations where direct relationships between x and y can become cumbersome or even impossible to derive. Think about a train moving on twisted tracks 鈥� using a parameter like time makes it much easier to describe the position of the train.
\[ x = \frac{1}{t+1} \] and \[ y = \frac{t}{t-1} \].
By varying the parameter \( t \), we can find different points on the curve. When \( t \) is the independent variable, the corresponding values of \( x \) and \( y \) can be determined easily, as we did here with \( t = 2 \).
This approach is particularly useful in situations where direct relationships between x and y can become cumbersome or even impossible to derive. Think about a train moving on twisted tracks 鈥� using a parameter like time makes it much easier to describe the position of the train.
Derivatives
Derivatives are crucial as they help us understand how a function changes. When we have parametric equations, we find the derivatives with respect to the parameter \( t \). This involves calculating \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), which are the rates of change of \( x \) and \( y \) with respect to \( t \).
Once we have these, the slope of the curve at any point can be found using \( \frac{dy}{dx} \), the ratio of these two derivatives. Here, we calculated
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \],
which simplifies to show how quickly \( y \) is changing with respect to \( x \).
This concept is like understanding how quickly you're climbing a hill as you walk; the "steepness" at any point gives you a sense of effort, which in mathematical terms, translates into the slope of the function.
Once we have these, the slope of the curve at any point can be found using \( \frac{dy}{dx} \), the ratio of these two derivatives. Here, we calculated
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \],
which simplifies to show how quickly \( y \) is changing with respect to \( x \).
This concept is like understanding how quickly you're climbing a hill as you walk; the "steepness" at any point gives you a sense of effort, which in mathematical terms, translates into the slope of the function.
Second Derivative
The second derivative \( \frac{d^2y}{dx^2} \) offers insights into the curvature of the path described by a parametric equation. It tells us whether the curve is concave up or down at a certain point, much like knowing if you're about to go up another hill or down into a valley.
To find this, we first find \( \frac{d^2y}{dt^2} \) and \( \frac{d^2x}{dt^2} \), then use
\[ \frac{d^2y}{dx^2} = \frac{ \frac{d}{dt}(\frac{dy}{dx}) }{\frac{dx}{dt}} \].
This second derivative aids in understanding the acceleration of change in the curve. In our problem, we calculated \( \frac{d^2y}{dx^2} \) at \( t=2 \). This value can tell us about how sharply the curve bends at that point, similar to your speedometer's acceleration meter catching swift changes in speed.
To find this, we first find \( \frac{d^2y}{dt^2} \) and \( \frac{d^2x}{dt^2} \), then use
\[ \frac{d^2y}{dx^2} = \frac{ \frac{d}{dt}(\frac{dy}{dx}) }{\frac{dx}{dt}} \].
This second derivative aids in understanding the acceleration of change in the curve. In our problem, we calculated \( \frac{d^2y}{dx^2} \) at \( t=2 \). This value can tell us about how sharply the curve bends at that point, similar to your speedometer's acceleration meter catching swift changes in speed.
Point-Slope Form
The point-slope form is a handy way to write the equation of a line when you know a point on the line and its slope. It's expressed as:
\[ y - y_1 = m(x - x_1) \],
where \( m \) represents the slope, and \( (x_1, y_1) \) is the given point through which the line passes.
In our exercise, we used this form to find the equation of the tangent line, utilizing the point \( \left( \frac{1}{3}, 2 \right) \) and slope \( 9 \).
This method is analogous to knowing you're at a certain location and heading in a particular direction; having this information, you can fairly confidently describe your path from here on, just as you can for a line on a graph.
\[ y - y_1 = m(x - x_1) \],
where \( m \) represents the slope, and \( (x_1, y_1) \) is the given point through which the line passes.
In our exercise, we used this form to find the equation of the tangent line, utilizing the point \( \left( \frac{1}{3}, 2 \right) \) and slope \( 9 \).
This method is analogous to knowing you're at a certain location and heading in a particular direction; having this information, you can fairly confidently describe your path from here on, just as you can for a line on a graph.
