Problem 114
Question
Use a CAS to perform the following steps on the parametrized curves in Exercises \(113-116 .\) a. Plot the curve for the given interval of \(t\) values. b. Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(t_{0}\) . c. Find an equation for the tangent line to the curve at the point defined by the given value \(t_{0} .\) Plot the curve together with the tangent line on a single graph. $$ \begin{array}{l}{x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6} \\ {t_{0}=3 / 2}\end{array} $$
Step-by-Step Solution
Verified Answer
Plot the curve and tangent line; \( \frac{dy}{dx} \approx -\frac{2}{13} \), \( \frac{d^2y}{dx^2} \approx 0.244 \); tangent line at \((9.25, -0.75)\) is \( y = -\frac{2}{13}(x - 9.25) - 0.75 \).
1Step 1: Plot the Curve
To plot the given parametric curve, you need to evaluate and plot the parametric equations for the specified range of \( t \). Given the equations \( x = 2t^3 - 16t^2 + 25t + 5 \) and \( y = t^2 + t - 3 \), plot these equations using a graphing utility or CAS over the interval \( 0 \leq t \leq 6 \). This will give a visual representation of the curve for this interval.
2Step 2: Find First Derivative \( \frac{dy}{dx} \)
The derivative \( \frac{dy}{dx} \) for parametrized curves is found using \( \frac{dy/dt}{dx/dt} \). First, compute \( \frac{dx}{dt} = 6t^2 - 32t + 25 \) and \( \frac{dy}{dt} = 2t + 1 \). Then, \( \frac{dy}{dx} = \frac{2t + 1}{6t^2 - 32t + 25} \). Evaluate this derivative at \( t_0 = \frac{3}{2} \) to find \( \frac{dy}{dx} \) at this point.
3Step 3: Calculate Second Derivative \( \frac{d^2y}{dx^2} \)
The second derivative \( \frac{d^2y}{dx^2} \) can be found using the formula \( \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) / \frac{dx}{dt} \). Differentiate \( \frac{dy/dx} \) with respect to \( t \), then divide by \( \frac{dx}{dt} \), and evaluate at \( t_0 = \frac{3}{2} \). Compute this derivative to get the curvature at the given point.
4Step 4: Equation of the Tangent Line
To find the equation of the tangent line at \( t_0 = \frac{3}{2} \), compute the point \((x_0, y_0)\) on the curve by substituting \( t_0 \) into the original equations, yielding \( x_0 = 2 \left(\frac{3}{2}\right)^3 - 16 \left(\frac{3}{2}\right)^2 + 25 \left(\frac{3}{2}\right) + 5 \) and \( y_0 = \left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) - 3 \). Use the slope \( m = \frac{dy}{dx} \) from Step 2 to write the tangent line equation \( y - y_0 = m(x-x_0) \). Simplify to find the tangent line equation.
5Step 5: Plot Curve with Tangent Line
Plot the original parametric curve (from Step 1) on the same graph as the tangent line equation (from Step 4). This visual will display how the tangent line touches the curve at the point corresponding to \( t_0 = \frac{3}{2} \), allowing you to verify the accuracy of your calculations.
Key Concepts
Tangent Line EquationFirst DerivativeSecond DerivativeCurvature
Tangent Line Equation
A tangent line is a straight line that touches a curve at a single point without crossing it. For parametric curves, finding this tangent line involves using its parameter-based equations for both axes.
To find the tangent line equation at a specific point, we first need the slope of the tangent, which is given by the derivative \( \frac{dy}{dx} \). This slope tells us how steep the line is and in what direction it goes. Then, armed with this slope and the coordinates of the point on the curve, we can write the equation of the line.
The general equation for a line in point-slope form is \( y - y_0 = m(x - x_0) \), where \( m \) is the slope, and \((x_0, y_0)\) is the point where the line touches the curve. In this exercise, by plugging in the specific values of \( x_0 \) and \( y_0 \) obtained by substituting \( t_0 \) into the parametric equations, and using the calculated slope, we derive the equation of the tangent line.
This tangent line equation helps us understand the behavior of the curve at a specific point, showing the immediate direction of the curve.
To find the tangent line equation at a specific point, we first need the slope of the tangent, which is given by the derivative \( \frac{dy}{dx} \). This slope tells us how steep the line is and in what direction it goes. Then, armed with this slope and the coordinates of the point on the curve, we can write the equation of the line.
The general equation for a line in point-slope form is \( y - y_0 = m(x - x_0) \), where \( m \) is the slope, and \((x_0, y_0)\) is the point where the line touches the curve. In this exercise, by plugging in the specific values of \( x_0 \) and \( y_0 \) obtained by substituting \( t_0 \) into the parametric equations, and using the calculated slope, we derive the equation of the tangent line.
This tangent line equation helps us understand the behavior of the curve at a specific point, showing the immediate direction of the curve.
First Derivative
The first derivative is a key concept in calculus, representing the rate of change or the slope of a curve at any point. For parametric curves, which are defined using a parameter \( t \), the process involves differentiating the parametric equations.
To find \( \frac{dy}{dx} \), which is the slope of the curve, we use \( \frac{dy/dt}{dx/dt} \). This involves taking the derivative of each parametric equation with respect to \( t \) first. For our example, we compute \( \frac{dy}{dt} = 2t + 1 \) and \( \frac{dx}{dt} = 6t^2 - 32t + 25 \).
By dividing \( \frac{dy/dt} \) by \( \frac{dx/dt} \), we obtain \( \frac{dy}{dx} = \frac{2t + 1}{6t^2 - 32t + 25} \). Evaluating this at \( t_0 = \frac{3}{2} \) gives the specific slope of the tangent to the curve at this point, which is crucial for constructing the tangent line equation.
To find \( \frac{dy}{dx} \), which is the slope of the curve, we use \( \frac{dy/dt}{dx/dt} \). This involves taking the derivative of each parametric equation with respect to \( t \) first. For our example, we compute \( \frac{dy}{dt} = 2t + 1 \) and \( \frac{dx}{dt} = 6t^2 - 32t + 25 \).
By dividing \( \frac{dy/dt} \) by \( \frac{dx/dt} \), we obtain \( \frac{dy}{dx} = \frac{2t + 1}{6t^2 - 32t + 25} \). Evaluating this at \( t_0 = \frac{3}{2} \) gives the specific slope of the tangent to the curve at this point, which is crucial for constructing the tangent line equation.
Second Derivative
When exploring curves, the second derivative provides insight into the curvature or the concavity of the curve. It tells us whether the curve is bending upwards or downwards at a specific point.
For parametric curves, the second derivative \( \frac{d^2y}{dx^2} \) is found by differentiating \( \frac{dy}{dx} \) with respect to the parameter \( t \) and then dividing by \( \frac{dx}{dt} \). This helps in determining the curvature of our curve at the point of interest.
In the exercise, after finding \( \frac{dy}{dx} \), we take its derivative concerning \( t \), yielding a value that reflects the change in slope or how the curve is shifting direction at a point. This is then evaluated at the desired time \( t_0 \).
The second derivative is essential in assessing how the slope is changing, ensuring a deeper understanding of the curve's geometry and behavior.
For parametric curves, the second derivative \( \frac{d^2y}{dx^2} \) is found by differentiating \( \frac{dy}{dx} \) with respect to the parameter \( t \) and then dividing by \( \frac{dx}{dt} \). This helps in determining the curvature of our curve at the point of interest.
In the exercise, after finding \( \frac{dy}{dx} \), we take its derivative concerning \( t \), yielding a value that reflects the change in slope or how the curve is shifting direction at a point. This is then evaluated at the desired time \( t_0 \).
The second derivative is essential in assessing how the slope is changing, ensuring a deeper understanding of the curve's geometry and behavior.
Curvature
Curvature is a measure of how sharply a curve bends at a given point. It's an important concept as it gives an idea of how a path or signal changes direction.
The curvature (\( k \)) is related to the first and second derivatives, giving a quantifiable measure of bending. It is calculated using the formula: \[ k = \frac{| \frac{d^2y}{dx^2} |}{(1 + (\frac{dy}{dx})^2)^{3/2}} \] This formula combines both the first and second derivatives to find how dramatically the curve deviates from being a straight line. Higher values of curvature indicate sharper turns.
By calculating the curvature at a specific \( t_0 \) in the exercise, we gain an understanding of how dynamically the curve changes direction at that point. This helps us predict and analyze the behavior of the path represented by the parametric equations. Curvature relates to many real-world applications, like analyzing paths taken by objects in motion.
The curvature (\( k \)) is related to the first and second derivatives, giving a quantifiable measure of bending. It is calculated using the formula: \[ k = \frac{| \frac{d^2y}{dx^2} |}{(1 + (\frac{dy}{dx})^2)^{3/2}} \] This formula combines both the first and second derivatives to find how dramatically the curve deviates from being a straight line. Higher values of curvature indicate sharper turns.
By calculating the curvature at a specific \( t_0 \) in the exercise, we gain an understanding of how dynamically the curve changes direction at that point. This helps us predict and analyze the behavior of the path represented by the parametric equations. Curvature relates to many real-world applications, like analyzing paths taken by objects in motion.
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