Problem 86
Question
Find all points on the curve \(x=t+4, y=t^{3}-3 t\) at which there are vertical and horizontal tangents.
Step-by-Step Solution
Verified Answer
Vertical tangents: None; Horizontal tangents: Points (5, -2) and (3, 2).
1Step 1: Parameterize the Curve
The problem provides the parametric equations of the curve: \(x = t + 4\) and \(y = t^3 - 3t\). These equations describe the curve in terms of the parameter \(t\). To find vertical and horizontal tangents, we need to investigate the derivatives of these equations.
2Step 2: Compute Derivatives
Find the derivatives of the coordinates with respect to \(t\).\[\frac{dx}{dt} = \frac{d(t+4)}{dt} = 1\]\[\frac{dy}{dt} = \frac{d(t^3 - 3t)}{dt} = 3t^2 - 3\]
3Step 3: Find Vertical Tangents
A vertical tangent occurs when \(\frac{dx}{dt} = 0\), but since \(\frac{dx}{dt} = 1\), which is never zero, there are no vertical tangents on this curve.
4Step 4: Find Horizontal Tangents
A horizontal tangent occurs when \(\frac{dy}{dt} = 0\).Set \(3t^2 - 3 = 0\) and solve for \(t\):\[3t^2 - 3 = 0\]\[t^2 = 1\]\[t = 1 \text{ or } t = -1\]
5Step 5: Calculate Points for Horizontal Tangents
Substitute the values of \(t\) found in step 4 back into the parametric equations to find the points on the curve:- For \(t = 1\): \(x = 1 + 4 = 5\), \(y = 1^3 - 3 \cdot 1 = -2\), so the point is \((5, -2)\).- For \(t = -1\): \(x = -1 + 4 = 3\), \(y = (-1)^3 - 3(-1) = 2\), so the point is \((3, 2)\).
Key Concepts
Horizontal TangentsVertical TangentsDerivativesCurve Parameterization
Horizontal Tangents
Horizontal tangents on a curve are spots where the tangent line is flat or horizontal. To locate these points on a parametric curve, we look for where the derivative of the y-coordinate with respect to the parameter \( t \) is zero.
In simpler terms, this means we want \( \frac{dy}{dt} = 0 \). For the given curve equations, the derivative for y was computed as \( \frac{dy}{dt} = 3t^2 - 3 \). By setting this equal to zero, we find the values of \( t \) that make the slope of the tangent zero. Solving the equation \( 3t^2 - 3 = 0 \) gives us \( t = 1 \) and \( t = -1 \).
Now we know these are the points where the curve has horizontal tangents, at which the line touches the curve without running upwards or downwards.
In simpler terms, this means we want \( \frac{dy}{dt} = 0 \). For the given curve equations, the derivative for y was computed as \( \frac{dy}{dt} = 3t^2 - 3 \). By setting this equal to zero, we find the values of \( t \) that make the slope of the tangent zero. Solving the equation \( 3t^2 - 3 = 0 \) gives us \( t = 1 \) and \( t = -1 \).
Now we know these are the points where the curve has horizontal tangents, at which the line touches the curve without running upwards or downwards.
Vertical Tangents
Vertical tangents occur where the tangent line is perfectly vertical, indicating that the y-coordinate changes while the x-coordinate remains constant. For a vertical tangent to exist, the derivative of the x-coordinate with respect to \( t \), or \( \frac{dx}{dt} \), must be zero.
In our curve, \( x = t + 4 \), the derivative of x is \( \frac{dx}{dt} = 1 \). Since this value is always 1 and never zero, there are no vertical tangents for this curve because the x-coordinate never stops changing as \( t \) changes.
Thus, in any case where \( \frac{dx}{dt} \) is not zero, you can confidently state that no vertical tangents exist.
In our curve, \( x = t + 4 \), the derivative of x is \( \frac{dx}{dt} = 1 \). Since this value is always 1 and never zero, there are no vertical tangents for this curve because the x-coordinate never stops changing as \( t \) changes.
Thus, in any case where \( \frac{dx}{dt} \) is not zero, you can confidently state that no vertical tangents exist.
Derivatives
Derivatives are key mathematical tools used to understand the rate of change or slope of functions. In parametric equations, derivatives help us understand how the coordinates \(x\) and \(y\) change with respect to a parameter \(t\).
For the given parametric equations of the curve, we computed:
For the given parametric equations of the curve, we computed:
- \( \frac{dx}{dt} = 1 \)
- \( \frac{dy}{dt} = 3t^2 - 3 \)
Curve Parameterization
Parameterization of a curve involves describing the curve through equations that express the coordinates as functions of a single variable \( t \). This form is especially useful in calculus to dive deeper into the curves' properties.
The given parametric equations are \( x = t + 4 \) and \( y = t^3 - 3t \). These illuminate that both \( x \) and \( y \) depend on the same parameter \( t \). By examining how each coordinate changes as \( t \) changes, we can find derivatives, points of tangency, and other features.
In this exercise, the choice of parameterization directly enabled us to systematically address the problem, calculate derivatives, and evaluate conditions for special tangents. Understanding parameterization thereby aids in solving complex problems involving curves more easily and vividly.
The given parametric equations are \( x = t + 4 \) and \( y = t^3 - 3t \). These illuminate that both \( x \) and \( y \) depend on the same parameter \( t \). By examining how each coordinate changes as \( t \) changes, we can find derivatives, points of tangency, and other features.
In this exercise, the choice of parameterization directly enabled us to systematically address the problem, calculate derivatives, and evaluate conditions for special tangents. Understanding parameterization thereby aids in solving complex problems involving curves more easily and vividly.
Other exercises in this chapter
Problem 83
For the curve \(x=4 t, y=3 t-2,\) find the slope and concavity of the curve at \(t=3\).
View solution Problem 84
For the parametric curve whose equation is \(x=4 \cos \theta, y=4 \sin \theta, \quad\) find the slope and concavity of the curve at \(\theta=\frac{\pi}{4}\).
View solution Problem 87
Find all points on the curve \(x=\sec \theta, y=\tan \theta\) at which horizontal and vertical tangents exist.
View solution Problem 88
Find \(d^{2} y / d x^{2}\). $$ x=t^{4}-1, \quad y=t-t^{2} $$
View solution