Problem 36 - Thomas Calculus

Question

A cone The line segment joining the origin to the point \((h, r)\) is revolved about the \(x\) -axis to generate a cone of height \(h\) and base radius \(r .\) Find the cone's surface area with the parametric equations \(x=h t, y=r t, 0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi r(\) slant height\().\)

Step-by-Step Solution

Answer

The surface area of the cone is \( \pi r \sqrt{h^2 + r^2} \). This matches the geometric formula.

Step 1: Identify Parametric Equations

The given parametric equations are: \[ x = ht, \quad y = rt \quad \text{where} \quad 0 \leq t \leq 1 \] These parametrize the line segment from the origin \((0,0)\) to the point \((h,r)\).

Step 2: Express Derivatives in Terms of t

Differentiate both parametric equations with respect to \(t\): \[ \frac{dx}{dt} = h \] \[ \frac{dy}{dt} = r \] These derivatives represent the rate of change of \(x\) and \(y\) with respect to \(t\).

Step 3: Calculate Differential Arc Length

The differential arc length \(ds\) for a parametric curve is given by: \[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] Substitute \(\frac{dx}{dt} = h\) and \(\frac{dy}{dt} = r\): \[ ds = \sqrt{h^2 + r^2} \, dt \]

Step 4: Calculate Surface Area

The surface area of the cone when the curve is revolved about the \(x\)-axis is given by: \[ A = \int_{t=0}^{t=1} 2 \pi y \, ds \] Substituting \(y = rt\) and \(ds = \sqrt{h^2 + r^2} \, dt\): \[ A = \int_{t=0}^{t=1} 2 \pi rt \sqrt{h^2 + r^2} \, dt \] \[ A = 2 \pi r \sqrt{h^2 + r^2} \int_{t=0}^{t=1} t \, dt \]

Step 5: Complete the Integration

Calculate the integral \(\int t \, dt\) from \(t=0\) to \(t=1\): \[ \int_{t=0}^{t=1} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{1} = \frac{1}{2} \] So, the surface area \(A\) becomes: \[ A = 2 \pi r \sqrt{h^2 + r^2} \times \frac{1}{2} = \pi r \sqrt{h^2 + r^2} \]

Step 6: Verify with Geometry Formula

The formula for the surface area of a cone is \( \pi r \times \text{slant height} \). The slant height of the cone is \( \sqrt{h^2 + r^2} \). Thus, the geometric surface area formula \( \pi r \sqrt{h^2 + r^2} \) matches our calculation.

Key Concepts

Parametric Equations

Parametric equations are a crucial mathematical concept used to describe curves and surfaces in a more flexible way than traditional equations. Instead of expressing one variable in terms of another, parametric equations express each of the coordinates as independent functions of a parameter, often denoted as \( t \). This method is especially useful in representing curves like the one used in the cone problem, allowing for a description of objects and paths that are more complex.

Consider the parametric equations \( x = ht \) and \( y = rt \), where \( 0 \leq t \leq 1 \). Here, \( t \) acts as the parameter that controls the movement from the origin \( (0,0) \) to the point \( (h,r) \).

• Each point on the line segment from the origin to \((h, r)\) is described as \( t \) varies between 0 and 1.
• These equations effectively translate the linear path into a manageable form for further calculations, such as finding the surface area of the cone, by focusing on how \( x \) and \( y \) change over the interval.

Differential Arc Length

The differential arc length is a way to compute the length of a segment of a curve. It is particularly useful in problems involving curves defined by parametric equations. The formula for the differential arc length \( ds \) involves both the derivative of the \( x \) and of the \( y \) components with respect to the parameter \( t \).

Given the derivatives \( \frac{dx}{dt} = h \) and \( \frac{dy}{dt} = r \) from our parametric equations, we find:

\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \sqrt{h^2 + r^2} \, dt\]

• This formula captures how the curve stretches and bends between every infinitesimal increment of \( t \).
• By squaring and adding the derivatives, we ensure that \( ds \) accounts for changes in both directions (x and y) as we trace the curve.
• The constant \( \sqrt{h^2 + r^2} \) reflects that for this linear segment, the full length is uniform for any subdivided section, which simplifies its integration over \( t \).

Surface Area of Revolution

The surface area of revolution involves creating a surface by rotating a curve around an axis, often used to visualize and calculate the surface area of objects with rotational symmetry, such as cones. In this problem, we revolve the line segment defined by \( x = ht \), \( y = rt \) around the \( x \)-axis.
To calculate the surface area, we use the formula:
\[A = \int 2 \pi y \, ds\]Substitute \( y = rt \) and the simplified differential arc length \( ds = \sqrt{h^2 + r^2} \, dt \):

\[A = \int_{t=0}^{t=1} 2 \pi rt \sqrt{h^2 + r^2} \, dt = 2 \pi r \sqrt{h^2 + r^2} \int_{t=0}^{t=1} t \, dt\]

• The calculation reveals how stretching the infinitesimal arc length \( ds \) around the \( x \)-axis contributes to the surface area.
• The integration of \( t \) from 0 to 1 calculates the cumulative effect of rotating each infinitesimal element of the line, culminating in half the full area \( \left( \frac{1}{2} \right) \).
• By substituting values directly into the integral and performing the calculations, we correctly verify the result matches the geometric formula for a cone's lateral surface area \( \pi r \sqrt{h^2 + r^2} \).