Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The planes tangent to the cylinder \(x^{2}+y^{2}=1\) in \(R^{3}\) all have the form \(a x+b z+c=0\) b. Suppose \(w=x y / z,\) for \(x>0, y>0,\) and \(z>0 .\) A decrease in \(z\) with \(x\) and \(y\) fixed results in an increase in \(w\) c. The gradient \(\nabla F(a, b, c)\) lies in the plane tangent to the surface \(F(x, y, z)=0\) at \((a, b, c)\)

Short answer

#Question# Evaluate the truth of the following statements: a. All tangent planes to the given cylinder \(x^2 + y^2 = 1\) have the form \(ax + bz + c = 0\). b. When \(w = \frac{xy}{z}\) and \(z\) decreases while keeping \(x\) and \(y\) constant, \(w\) increases. c. The gradient vector of \(F\) at (a, b, c) lies in the tangent plane of the surface \(F(x, y, z) = 0\) at (a, b, c). #Answer# a. False b. True c. False

Step-by-step solution

Part a: Equation of Tangent Planes

We know that the equation of the given cylinder is \(x^2 + y^2 = 1\). To determine if the tangent planes have the form \(ax + bz + c = 0\), we need to find an expression for a plane tangent to the cylinder. Since the equation is in terms of \(x\) and \(y\), we can differentiate it with respect to \(x\) and \(y\), respectively. This will give us the direction vectors of the tangent plane. Taking the partial derivative with respect to \(x\): \(\frac{\partial}{\partial x}(x^2 + y^2) = 2x\) Taking the partial derivative with respect to \(y\): \(\frac{\partial}{\partial y}(x^2 + y^2) = 2y\) Now, we can create an equation for the tangent plane at a point \((x_0, y_0, z_0)\) on the surface of the cylinder using the dot product of the gradients and the vector pointing from \((x_0, y_0, z_0)\) to \((x, y, z)\). \((x - x_0, y - y_0, z - z_0) \cdot (2x_0, 2y_0, 0) = 0\) \(2x_0(x - x_0) + 2y_0(y - y_0) = 0\) Now, we can rearrange the equation to form a linear equation. \(2x_0x + 2y_0y = 2x_0^2 + 2y_0^2\) Since \(x_0^2 + y_0^2 = 1\) as the point lies on the surface of the cylinder, we have: \(2x_0x + 2y_0y = 2\) Dividing by 2: \(x_0x + y_0y = 1\) Comparing this equation to the general form of the plane equation \(ax + bz + c = 0\), we can see that they are not the same. The statement is false. The tangent planes of the given cylinder do not have the form \(ax + bz + c = 0\).

Part b: Effect of Decreasing z

We are given \(w = \frac{xy}{z}\), and we need to determine the effect of decreasing \(z\) while keeping \(x\) and \(y\) constant. We can analyze this statement using basic properties of fractions. Since \(x\) and \(y\) are positive and fixed, the expression \(\frac{xy}{z}\) will increase when the denominator \(z\) decreases, and \(x\) and \(y\) remain constant. This is because the value of the fraction becomes larger as the denominator becomes smaller (and greater than 0). So, the statement is true. For a decrease in \(z\) with \(x\) and \(y\) fixed, \(w\) will increase.

Part c: The Gradient and the Tangent Plane

We need to determine if the gradient \(\nabla F(a, b, c)\) lies in the tangent plane of the surface \(F(x, y, z) = 0\) at \((a, b, c)\). Recall the definition of the gradient vector: \(\nabla F(x, y, z) = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)\) Now, consider the tangent plane to the surface at \((a, b, c)\). A normal vector to the tangent plane is given by the gradient vector of \(F\) at \((a, b, c)\), i.e., \(\nabla F(a, b, c)\). Thus, as normal vectors are always perpendicular to the plane, the gradient vector \(\nabla F(a, b, c)\) does not lie in the tangent plane to the surface at \((a, b, c)\). Therefore, the statement is false. The gradient \(\nabla F(a, b, c)\) does not lie in the tangent plane to the surface \(F(x, y, z) = 0\) at \((a, b, c)\).

Key concepts

Normal Vector

A normal vector is fundamentally important in understanding the properties of a surface in three-dimensional space. It's a vector that is perpendicular to the surface at a given point. In the context of the cylinder with equation x^2 + y^2 = 1, the normal vector at any point on the surface provides us with the direction that is orthogonal to the tangent plane at that point.

In our exercise, the tangent plane to a surface can be found by taking the dot product of the normal vector at a given point with the difference between an arbitrary point on the plane and the given point. Using the gradient, which gives the components of the normal vector, we were able to conclude that the planes tangent to the given cylinder are not of the form ax + bz + c = 0, as they don't account for the y-component which should be present due to the symmetry of the cylinder about the x-axis.

Gradient Vector

The gradient vector is the vector consisting of the partial derivatives of a function with respect to all of its variables. This vector points in the direction of the greatest rate of increase of a function. In our step-by-step solution for part c, we utilized this concept to determine the orientation of the gradient vector with respect to the tangent plane of the surface defined by F(x, y, z) = 0.

The gradient vector \( abla F(a, b, c) \) was found to be normal to the tangent plane at the point (a, b, c). This is crucial to grasp in calculus, as the gradient provides useful information about the surface’s behavior at a specific location and why it does not lie within the plane itself, contrary to what might have been assumed.

Partial Derivatives

Partial derivatives represent how a function changes as one of the variables is varied while the others are kept constant. They are essential for finding gradient vectors and for understanding the shape and properties of surfaces in multivariable calculus.

In the cylinder example, specifically, we looked at the partial derivatives with respect to x and y, which gave us the slopes of the tangent plane in the direction of the x and y axes, respectively. These partial derivatives informed us of how the function - and thus the cylinder's surface - changes along the x and y axes and was used to explicitly determine that the tangent planes do not conform to the simplified form ax + bz + c = 0.

Tangent to Cylinder

The concept of a tangent to a cylinder is intricately related to the properties of normal vectors and partial derivatives. A tangent plane at any point on a surface like a cylinder should ideally contain all the tangent lines that pass through that point. For the cylinder given by x^2 + y^2 = 1, the tangent plane at a given point is defined by the set of all points whose displacement vector from the point of tangency is orthogonal to the normal vector calculated from the cylinder's surface.

In our example's step-by-step solution, we determined that the tangent plane at any point on our cylinder cannot be correctly described by the proposed form ax + bz + c = 0 since each plane would miss the y-component. This insight demystifies the tangent planes to a cylinder, allowing students to visualize how these planes touch the cylinder along a line and not just at a single point.