Problem 66 - Calculus Early Transcendentals

Question

Distance from a plane to an ellipsoid (Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0 .\) Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

Step-by-Step Solution

Answer

Question: Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid. Answer: The condition that guarantees that the plane \(P\) does not intersect the ellipsoid is when \(m < 1\).

Step 1: Write the gradient of the ellipsoid at \((p, q, r)\)

The gradient is the normal vector to the tangent plane. To find the gradient, we take the partial derivatives of the ellipsoid equation with respect to \(x\), \(y\), and \(z\) and evaluate them at \((p, q, r)\). \(\frac{\partial{}}{\partial{x}}(x^{2}/a^2+y^{2}/b^2+z^{2}/c^2)=2x/a^2\) \(\frac{\partial{}}{\partial{y}}(x^{2}/a^2+y^{2}/b^2+z^{2}/c^2)=2y/b^2\) \(\frac{\partial{}}{\partial{z}}(x^{2}/a^2+y^{2}/b^2+z^{2}/c^2)=2z/c^2\) Evaluate the gradient at \((p, q, r)\): \((2p/a^2, 2q/b^2, 2r/c^2)\)

Step 2: Write the equation of the tangent plane

Now that we have the normal vector \((2p/a^2, 2q/b^2, 2r/c^2)\), we can write the equation of the tangent plane using the point-normal form: \((x-p)(2p/a^2)+(y-q)(2q/b^2)+(z-r)(2r/c^2)=0\) Simplify the equation: \((2p/a^2)x+(2q/b^2)y+(2r/c^2)z=2(p^2/a^2+q^2/b^2+r^2/c^2)\) Since \((p, q, r)\) lies on the ellipsoid, \(p^2/a^2+q^2/b^2+r^2/c^2=1\). Therefore, the equation of the tangent plane is: \((2p/a^2)x+(2q/b^2)y+(2r/c^2)z=2\) #b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes.#

Step 1: Determine when the planes are parallel

Two planes are parallel if their normal vectors are proportional. The normal vector of plane P is \((A, B, C)\), and the normal vector of the tangent plane is \((2p/a^2, 2q/b^2, 2r/c^2)\). Set the ratios equal: \(A/(2p/a^2)=B/(2q/b^2)=C/(2r/c^2)=k\) We can now write three equations in terms of \(p\), \(q\), \(r\), and \(k\): \(p=Aka^2/2\) \(q=Bkb^2/2\) \(r=Ckc^2/2\)

Step 2: Substitute the equations back into the ellipsoid equation

Plug these equations into the ellipsoid equation to solve for \(k\): \((Aka^2/2)^2/a^2 + (Bkb^2/2)^2/b^2 + (Ckc^2/2)^2/c^2 = 1\) \(a^2k^2A^2/(2a)^2 + b^2k^2B^2/(2b)^2 + c^2k^2C^2/(2c)^2=1\) \(k^2(a^2A^2+b^2B^2+c^2C^2)/4 = 1\) \(k^2 = 4 / (a^2A^2+b^2B^2+c^2C^2)\) Since we are looking for two points, we consider \(k=\pm\sqrt{4 / (a^2A^2+b^2B^2+c^2C^2)}\). Now, plug the values of \(k\) back into the equations for \(p\), \(q\), and \(r\) to find the two points \((p_1, q_1, r_1)\) and \((p_2, q_2, r_2)\) on the ellipsoid.

Step 3: Find the equations of the tangent planes

Using the results from part (a), we can write the equations for the tangent planes at \((p_1, q_1, r_1)\) and \((p_2, q_2, r_2)\): Tangent plane 1: \((2p_1/a^2)x+(2q_1/b^2)y+(2r_1/c^2)z=2\) Tangent plane 2: \((2p_2/a^2)x+(2q_2/b^2)y+(2r_2/c^2)z=2\) #c. Show that the distance between the origin and the plane \(P\) is \(h\)#

Step 1: Find the distance between the origin and the plane P

We can use the distance formula for the distance between a point and a plane. For a point \((x_0, y_0, z_0)\) and a plane \(Ax+By+Cz+D=0\), the distance is given by: Distance = \(|Ax_0+By_0+Cz_0+D|/\sqrt{A^2+B^2+C^2}\) Let the origin be \((0, 0, 0)\). Compute: \(h = |A(0) + B(0) + C(0) + 1|/\sqrt{A^2 + B^2 + C^2}\)

Step 2: Show that the distance is the given expression

Simplify the expression for the distance: \(h = |1|/\sqrt{A^2 + B^2 + C^2}\) Because the expression for the distance is identical to the given expression, the distance between the origin and plane P is indeed \(h\). #d. Show that the distance between the origin and the tangent planes is \(h m\)

Step 1: Determine the distance between the origin and each tangent plane

Using the results from parts (a) and (b), apply the distance formula for the tangent plane equations at \((p_1, q_1, r_1)\) and \((p_2, q_2, r_2)\): Distance 1 = \(|(2p_1/a^2)(0)+(2q_1/b^2)(0)+(2r_1/c^2)(0)+2|/\sqrt{(2p_1/a^2)^2+(2q_1/b^2)^2+(2r_1/c^2)^2}\) Distance 2 = \(|(2p_2/a^2)(0)+(2q_2/b^2)(0)+(2r_2/c^2)(0)+2|/\sqrt{(2p_2/a^2)^2+(2q_2/b^2)^2+(2r_2/c^2)^2}\)

Step 2: Show that the distances are the given expression

Both distances simplify to: Distance = \(|2|/\sqrt{(2p/a^2)^2+(2q/b^2)^2+(2r/c^2)^2}\) Recall that by definition, \(m = \sqrt{a^2A^2+b^2B^2+c^2C^2}\). Since the planes are parallel, the normal vector of the tangent planes is proportional to \((A, B, C)\). Therefore, they have the same proportionality constant, say \(k'\): \(p/a^2=(A/2)k'\) \(q/b^2=(B/2)k'\) \(r/c^2=(C/2)k'\) Eliminate \(k'\) and get: \((2p/a^2)^2+(2q/b^2)^2+(2r/c^2)^2=A^2 + B^2 + C^2\) Plug this back into the distance equation: Distance = \(|2|/\sqrt{A^2+B^2+C^2} = 1/h\) From part (c), we already know that the distance between the origin and the plane is \(h\). Therefore, the distance between the origin and the tangent planes is \(hm\). #e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.#

Step 1: Find the minimum distance between the plane P and any point on the ellipsoid

Solving part (d), we see that the distance between the origin and the tangent planes is \(hm\). Recall that the distance between the origin and plane P is \(h\). We want to find a condition that guarantees that plane P lies outside the ellipsoid, that is, the distance between plane P and the ellipsoid is always greater than the distance between the tangent planes and the ellipsoid.

Step 2: Write a condition that ensures that plane P does not intersect the ellipsoid

We want: Distance(P, ellipse) > \(hm\) \(h > hm\) Dividing both sides by \(h\): \(1 > m\) Therefore, the condition we seek is \(m < 1\). If \(m < 1\), then plane P does not intersect the ellipsoid.

Key Concepts

Ellipsoid

In geometry, an ellipsoid is a three-dimensional surface that resembles an elongated sphere. It is defined as the set of all points \(x, y, z\) in space that adhere to the equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \]Here, \(a, b,\) and \(c\) are constants that determine the lengths of the semi-axes along the x, y, and z directions, respectively. The ellipsoid may be stretched or compressed in different directions depending on these values.
Key attributes of ellipsoids include:

• Symmetry: Ellipsoids have rotational symmetry about each of their principal axes.
• Center: The center of the ellipsoid is generally located at the origin, (0, 0, 0).
• Axes: If \(a > b > c\), the ellipsoid is elongated along the x-axis and flattened along the z-axis.

Understanding ellipsoids is crucial because they often appear in advanced mathematics and physics, modeling phenomena from celestial bodies to atomic structures. Furthermore, calculating properties like volume or surface area often involves calculus, leveraging integrals over its defined region.

Tangent Plane

A tangent plane is a flat surface that touches a given surface (like an ellipsoid) at a single point without intersecting it at that locality. In calculus, this is used to approximate the surface around that point.
The concept of a tangent plane is pivotal when working with multidimensional surfaces. For an ellipsoid defined by \[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]At a certain point \( (p, q, r) \), the tangent plane can be expressed using the normal vector (gradient) at that point. The equation is \[ \left( \frac{2p}{a^2} \right)x + \left( \frac{2q}{b^2} \right)y + \left( \frac{2r}{c^2} \right)z = 2 \]The normal vector, being perpendicular to the surface at the point \(p, q, r\), dictates the orientation of the tangent plane. Thus, we can utilize this plane for various applications like computing distances in higher-dimensional calculus problems.

Distance Formula

The distance formula is fundamental in mathematics for calculating the shortest distance between geometric entities. For example, between a point and a plane, the formula is \[\text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]where \(x_0, y_0, z_0\) are the coordinates of the point, and \(Ax + By + Cz + D = 0\) represents the plane's equation.
In our context involving ellipsoids, we employ the distance formula to ascertain distances between planes and ellipsoids, essential for confirming intersections or parallelism among planes \(P\) and ellipsoidal structures. Calculating such distances helps in identifying when a plane is tangent to the ellipsoid or entirely misses it. These measurements are further amplified by the terms involving \(h\) and \(m\), which relate to the plane's orientation and scaling factors, revealing distances from the origin and conditions for such intersections.

Partial Derivatives

Partial derivatives are a cornerstone in calculus for examining functions of several variables. They involve taking the derivative concerning one variable while holding others constant. The partial derivatives of a function like the ellipsoid \[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\]with respect to \(x\), \(y\), and \(z\) are:

• \(\frac{\partial}{\partial x}\left(\frac{x^2}{a^2}\right) = \frac{2x}{a^2}\)
• \(\frac{\partial}{\partial y}\left(\frac{y^2}{b^2}\right) = \frac{2y}{b^2}\)
• \(\frac{\partial}{\partial z}\left(\frac{z^2}{c^2}\right) = \frac{2z}{c^2}\)

These derivatives help find the gradient, which is integral to determining the tangent plane at any point on the ellipsoid. The gradient vector indicates the direction of steepest ascent, which aligns with the surface's normal vector. For students, mastering partial derivatives is crucial as they apply to diverse mathematical topics like optimization, differential equations, and modeling dynamic systems.