Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

A bee sat at the point \((1,2,1)\) on the ellipsoid \(x^{2}+y^{2}+2 z^{2}=6\) (distances in feet). At \(t=0\), it took off along the normal line at a speed of 4 feet per second. Where and when did it hit the plane \(2 x+3 y+z=49\) ?

Step-by-Step Solution

Answer

The bee hits the plane at \((4.9, 9.8, 8.8)\) after 1.95 seconds.

Step 1: Find the normal vector

The normal to the surface of the ellipsoid at the point where the bee is sitting can be found by taking the gradient of the function defining the ellipsoid. The equation of the ellipsoid is given by \(f(x, y, z) = x^2 + y^2 + 2z^2 - 6\). Thus, the gradient (normal vector) is \(abla f(x, y, z) = (2x, 2y, 4z)\). Evaluating this at the point \((1, 2, 1)\), the normal vector is \((2 \cdot 1, 2 \cdot 2, 4 \cdot 1) = (2, 4, 4)\).

Step 2: Determine the parametric equation of the path

The bee travels along the line in the direction of the normal vector \((2, 4, 4)\), starting from the point \((1, 2, 1)\). The parametric equations for the path are therefore \(x = 1 + 2t\), \(y = 2 + 4t\), and \(z = 1 + 4t\), where \(t\) represents time in seconds and the speed is 4 feet per second.

Step 3: Find the intersection point with the plane

The plane is given by the equation \(2x + 3y + z = 49\). Substitute the parametric expressions for \(x\), \(y\), and \(z\) into the plane equation: \(2(1 + 2t) + 3(2 + 4t) + (1 + 4t) = 49\). Simplify to get \(2 + 4t + 6 + 12t + 1 + 4t = 49\). Combining terms results in \(10 + 20t = 49\).

Step 4: Solve for time \(t\)

Subtract 10 from both sides of the equation to obtain \(20t = 39\). Dividing both sides by 20 gives \(t = \frac{39}{20} = 1.95\) seconds. This is the time when the bee hits the plane.

Step 5: Calculate the position on the plane

Substitute \(t = 1.95\) back into the parametric equations: \(x = 1 + 2(1.95) = 4.9\), \(y = 2 + 4(1.95) = 9.8\), \(z = 1 + 4(1.95) = 8.8\). Thus, the point of intersection is \((4.9, 9.8, 8.8)\).

Key Concepts

Ellipsoid

An ellipsoid is a 3D geometric shape resembling a stretched or compressed sphere. It is defined mathematically by an equation of the form:

• Standard form: \frac{x²}{a²} + \frac{y²}{b²} + \frac{z²}{c²} = 1
• General form: Ax² + By² + Cz² = D

In this exercise, we have the equation of an ellipsoid: \(x^2 + y^2 + 2z^2 = 6\). Notice how the coefficients of \(x^2\), \(y^2\), and \(z^2\) might differ; this is what causes the `"ellipsoidal"` shape, rather than a perfect sphere. The term \(2z^2\) indicates that distances along the z-axis are scaled differently, impacting the overall shape.

Understanding ellipsoids helps in visualizing the problem of a bee sitting on its surface. By recognizing this mathematical structure, you can comprehend how normals and intersections behave with such surfaces.

Normal Vector

To understand how a normal vector works, let's start by defining it. A normal vector is a perpendicular vector to a surface at a given point. For an ellipsoid, obtaining the normal vector requires differentiation, often through finding the gradient of its equation.

Here, we have the ellipsoid equation \(f(x, y, z) = x^2 + y^2 + 2z^2 - 6\).

• The gradient, \(abla f(x, y, z)\), gives the normal vector: \((2x, 2y, 4z)\).

At point \((1, 2, 1)\), substitute to find the specific normal vector: \((2\cdot1, 2\cdot2, 4\cdot1) = (2, 4, 4)\). This vector is crucial as it indicates the direction in which the bee starts its flight path.

Understanding normal vectors is vital, especially when analyzing objects like ellipsoids in physics and engineering, as they provide insights into directions of tangential forces and potential paths.

Parametric Equation

Parametric equations offer a way to describe the path of a moving object using parameters, typically time \(t\). They are especially useful in describing curves and surfaces in physics and mathematics.

Considering a bee's path, we know its starting point \((1, 2, 1)\) and direction given by the normal vector \((2, 4, 4)\). With a speed of 4 feet per second, the parametric equations are formed by:

• \(x = 1 + 2t\)
• \(y = 2 + 4t\)
• \(z = 1 + 4t\)

Each variable is expressed as a function of \(t\), the parameter. This clear breakdown provides a mathematical representation of the bee's trajectory, pivotal when calculating where it will intersect other geometric entities, like planes.

Parametric representations facilitate complex motion analysis, extending beyond simple linear paths to accommodate curves and more intricate movement patterns.

Intersection with Plane

An intersection with a plane often involves checking where a path, as described by parametric equations, meets the plane's equation. It's an essential concept in more advanced geometry and calculus.

Here, the plane equation is \(2x + 3y + z = 49\). To find the intersection, substitute the parametric equations into the plane's equation. This means:

• Substitute \(x = 1 + 2t\), \(y = 2 + 4t\), \(z = 1 + 4t\) into the plane's equation.
• The equation becomes \(2(1 + 2t) + 3(2 + 4t) + (1 + 4t) = 49\).

Simplifying this expression leads to an equation in terms of \(t\), \(10 + 20t = 49\). Solving gives \(t = \frac{39}{20} = 1.95\) seconds. You can then find the intersection point by substituting back into the parametric equations: \((4.9, 9.8, 8.8)\).

This process evaluates spatial relationships critical in computer graphics, physics simulations, and precise engineering design.