Chapter 10: Problem 32
, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=3 t, y=8 t^{3} ; t=-\frac{1}{2} $$
Short Answer
Expert verified
The tangent line at \( t = -\frac{1}{2} \) is \( y = 2x + 2 \).
Step by step solution
01
Find Point on the Curve
To find the point on the curve corresponding to the given value of \( t = -\frac{1}{2} \), substitute \( t \) into the parametric equations. \( x = 3 (-\frac{1}{2}) = -\frac{3}{2} \) and \( y = 8 (-\frac{1}{2})^3 = -\frac{8}{8} = -1 \). Thus, the point is \((-\frac{3}{2}, -1)\).
02
Differentiate Parametric Equations
To find the slope of the tangent line, differentiate both parametric equations with respect to \( t \). The derivative \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 24t^2 \).
03
Calculate Slope of Tangent Line
The slope \( m \) of the tangent line is given by the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24t^2}{3} = 8t^2 \). Substitute \( t = -\frac{1}{2} \) to find \( m: 8(-\frac{1}{2})^2 = 8 \cdot \frac{1}{4} = 2 \).
04
Write Equation of the Tangent Line
Use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \) with \( m = 2 \), \((x_1, y_1) = (-\frac{3}{2}, -1)\). The equation becomes \( y + 1 = 2(x + \frac{3}{2}) \). Simplify to get \( y = 2x + 2 \).
05
Sketch the Curve and Tangent Line
Plot the parametric curve using the equations \( x = 3t \) and \( y = 8t^3 \). The tangent line at \( t = -\frac{1}{2} \) should intersect the curve at \((-\frac{3}{2}, -1)\) and have a slope of 2, so it should rise 2 units for every 1 unit it moves to the right.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
In calculus, parametric equations are a way to define a curve using parameters rather than explicit functions of one variable. Here, variables like \( x \) and \( y \) are expressed in terms of a third variable, typically \( t \), called the parameter. This method is especially useful when a curve cannot be easily represented by an equation \( y = f(x) \) or \( x = g(y) \).
For example, consider the parametric equations \( x = 3t \) and \( y = 8t^3 \). Here, both \( x \) and \( y \) change as \( t \) changes, thus describing a curve as \( t \) varies over all possible values. The parameter \( t \) can be thought of as a sort of "time" that determines the position of a point tracing the path of the curve.
One key benefit of parametric equations is their flexibility. They are helpful for describing complex shapes and trajectories, and often simplify calculations in physics and engineering.
For example, consider the parametric equations \( x = 3t \) and \( y = 8t^3 \). Here, both \( x \) and \( y \) change as \( t \) changes, thus describing a curve as \( t \) varies over all possible values. The parameter \( t \) can be thought of as a sort of "time" that determines the position of a point tracing the path of the curve.
One key benefit of parametric equations is their flexibility. They are helpful for describing complex shapes and trajectories, and often simplify calculations in physics and engineering.
Tangent Line
A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point without crossing it. It represents the direction the curve is heading at exactly that point.
To find the equation of a tangent line in parametric form, we need to first determine a specific point on the curve, then compute its slope at that point, and finally apply the point-slope form of a line equation.
For instance, in this exercise, we found the point on the curve at \( t = -\frac{1}{2} \) using the parametric equations, yielding \(-\frac{3}{2}, -1)\). Knowing the specific point allows us to fit the tangent line precisely to the curve at that location.
To find the equation of a tangent line in parametric form, we need to first determine a specific point on the curve, then compute its slope at that point, and finally apply the point-slope form of a line equation.
For instance, in this exercise, we found the point on the curve at \( t = -\frac{1}{2} \) using the parametric equations, yielding \(-\frac{3}{2}, -1)\). Knowing the specific point allows us to fit the tangent line precisely to the curve at that location.
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which one quantity changes with respect to another. This rate of change is represented by a derivative.
In terms of parametric equations, differentiation allows us to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) which are essential for further calculations. These derivatives tell us how \( x \) and \( y \) individually change as the parameter \( t \) changes.
For this problem, we found \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 24t^2 \). These expressions are derived through the basic rules of differentiation applied to the parametric equations. With these, we can proceed to find the slope of the tangent line (as tackled in the next section).
In terms of parametric equations, differentiation allows us to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) which are essential for further calculations. These derivatives tell us how \( x \) and \( y \) individually change as the parameter \( t \) changes.
For this problem, we found \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 24t^2 \). These expressions are derived through the basic rules of differentiation applied to the parametric equations. With these, we can proceed to find the slope of the tangent line (as tackled in the next section).
Slope of a Line
The slope of a line represents its steepness and can evaluate how it inclines or declines. In calculus, when dealing with curves and their tangent lines, the slope at a specific point is found using derivatives.
In the context of parametric curves, the slope of the tangent line is provided by the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This formula effectively gives us the rate of change of \( y \) with respect to \( x \), which is crucial for defining the tangent.In our specific example, substituting \( t = -\frac{1}{2} \) into \( \frac{dy}{dx} = 8t^2 \), we determined the slope of the tangent line to be \( 2 \). This tells us that for every unit the line moves horizontally, it rises 2 units vertically, confirming the line's inclination.
In the context of parametric curves, the slope of the tangent line is provided by the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This formula effectively gives us the rate of change of \( y \) with respect to \( x \), which is crucial for defining the tangent.In our specific example, substituting \( t = -\frac{1}{2} \) into \( \frac{dy}{dx} = 8t^2 \), we determined the slope of the tangent line to be \( 2 \). This tells us that for every unit the line moves horizontally, it rises 2 units vertically, confirming the line's inclination.
