Problem 31
Question
In Problems 31-34, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=t^{2}, y=t^{3}, t=2 $$
Step-by-Step Solution
Verified Answer
The equation of the tangent line is \(y = 3x - 4\).
1Step 1: Determine the Parametric Equations
The given parametric equations are \(x = t^2\) and \(y = t^3\). We need to find the tangent line at \(t = 2\).
2Step 2: Find the Point on the Curve
Substitute \(t = 2\) into the parametric equations to find the point on the curve. \(x = 2^2 = 4\) and \(y = 2^3 = 8\). Hence, the point on the curve is \((4, 8)\).
3Step 3: Calculate Derivatives dx/dt and dy/dt
Differentiate \(x = t^2\) to get \(\frac{dx}{dt} = 2t\) and \(y = t^3\) to get \(\frac{dy}{dt} = 3t^2\).
4Step 4: Compute Slope of Tangent Line
The slope of the tangent line is given by \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3t}{2}\). Substitute \(t = 2\) to find the slope: \(\frac{3 \times 2}{2} = 3\).
5Step 5: Write Equation of Tangent Line
The equation of the tangent line is given by \(y - y_0 = m(x - x_0)\) where \((x_0, y_0)\) is the point on the curve and \(m\) is the slope. Substituting \((4, 8)\) and slope \(3\) yields \(y - 8 = 3(x - 4)\). Expand to get \(y = 3x - 4\).
6Step 6: Sketching the Graph (Visual Guide)
Make a rough sketch of the curve in the plane. Mark the point (4, 8) and draw the tangent line with slope 3 through this point, which illustrates the steepness and direction of the tangent.
Key Concepts
Parametric EquationsSlope of Tangent LineDerivatives CalculationEquation of Tangent Line
Parametric Equations
Parametric equations are a way to express a set of related quantities as explicit functions of a parameter.
In this exercise, we're given the equations \(x = t^2\) and \(y = t^3\). These equations allow each variable, \(x\) and \(y\), to depend on a third variable, \(t\).
This method provides an effective way to describe curves that might be difficult to define with a standard function like \(y = f(x)\). For example, the route an object travels through a wind path can be described more naturally in parametric form.
To find the position of the object at different moments in time, simply substitute specific values of \(t\) into the equations. Here, when \(t = 2\), substituting in these equations gives us the point \((4, 8)\) on the plane.
In this exercise, we're given the equations \(x = t^2\) and \(y = t^3\). These equations allow each variable, \(x\) and \(y\), to depend on a third variable, \(t\).
This method provides an effective way to describe curves that might be difficult to define with a standard function like \(y = f(x)\). For example, the route an object travels through a wind path can be described more naturally in parametric form.
To find the position of the object at different moments in time, simply substitute specific values of \(t\) into the equations. Here, when \(t = 2\), substituting in these equations gives us the point \((4, 8)\) on the plane.
Slope of Tangent Line
Determining the slope of a tangent line to a curve at a specific point is crucial for understanding how the curve behaves at that point.
For parametric curves, this involves finding the rate of change of \(y\) with respect to \(x\), which is defined by the ratio \(\frac{dy}{dx}\).
In this exercise, the slope is obtained using the derivatives of the given parametric equations.
Evaluating this at \(t = 2\) results in a slope of 3, which gives us the steepness of the line.
For parametric curves, this involves finding the rate of change of \(y\) with respect to \(x\), which is defined by the ratio \(\frac{dy}{dx}\).
In this exercise, the slope is obtained using the derivatives of the given parametric equations.
- Differentiate \(x = t^2\) to get \(\frac{dx}{dt} = 2t\).
- Differentiate \(y = t^3\) to get \(\frac{dy}{dt} = 3t^2\).
Evaluating this at \(t = 2\) results in a slope of 3, which gives us the steepness of the line.
Derivatives Calculation
Understanding how to calculate derivatives is essential for finding the slope of tangent lines for parametric equations.
Let's take a closer look at how derivatives were computed in this example.
For \(x = t^2\), apply the basic power rule for differentiation. The derivative \(\frac{dx}{dt} = 2t\) arises because the power rule simplifies differentiation to multiplying by the power and reducing it by one.
For \(y = t^3\), using the same power rule results in \(\frac{dy}{dt} = 3t^2\).
These steps reveal how variables change in relation to the parameter \(t\), enabling us to measure how one quantity changes relative to another through the use of parametric equations. Calculating these derivatives correctly is pivotal as they directly impact the accurate calculation of the slope of the tangent line.
Let's take a closer look at how derivatives were computed in this example.
For \(x = t^2\), apply the basic power rule for differentiation. The derivative \(\frac{dx}{dt} = 2t\) arises because the power rule simplifies differentiation to multiplying by the power and reducing it by one.
For \(y = t^3\), using the same power rule results in \(\frac{dy}{dt} = 3t^2\).
These steps reveal how variables change in relation to the parameter \(t\), enabling us to measure how one quantity changes relative to another through the use of parametric equations. Calculating these derivatives correctly is pivotal as they directly impact the accurate calculation of the slope of the tangent line.
Equation of Tangent Line
Finding the equation of a tangent line involves applying the point-slope form of a line, which is a straightforward way to illustrate the tangent at a particular point on the curve.
In this exercise, with the known slope of 3 and the point \((4, 8)\) from the curve, the equation is constructed as follows:
The point-slope formula for a line is: \(y - y_0 = m(x - x_0)\), where \((x_0, y_0)\) is the known point and \(m\) is the slope.
Substituting the values from our problem gives \(y - 8 = 3(x - 4)\).
Simplifying and solving for \(y\) yields \(y = 3x - 4\).
This equation is essential for accurately representing the direction and rate of change at the point where the tangent meets the curve, providing valuable insight into the curve's behavior at that specific location.
In this exercise, with the known slope of 3 and the point \((4, 8)\) from the curve, the equation is constructed as follows:
The point-slope formula for a line is: \(y - y_0 = m(x - x_0)\), where \((x_0, y_0)\) is the known point and \(m\) is the slope.
Substituting the values from our problem gives \(y - 8 = 3(x - 4)\).
Simplifying and solving for \(y\) yields \(y = 3x - 4\).
This equation is essential for accurately representing the direction and rate of change at the point where the tangent meets the curve, providing valuable insight into the curve's behavior at that specific location.
Other exercises in this chapter
Problem 30
Any line segment through the focus of a parabola, with end points on the parabola, is a focal chord. Prove that the tangent lines to a parabola at the end point
View solution Problem 31
In Problems 23-36, name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. \(r=\frac{6}{2+\sin \theta}\)
View solution Problem 31
The sum of the distances of \(P\) from \((0, \pm 9)\) is 26 .
View solution Problem 31
Find the foci of the ellipse $$ 16(x-1)^{2}+25(y+2)^{2}=400 $$
View solution