Problem 21 - Advanced Engineering Mathematics

Question

In Problems, consider the plane through the points \(P\) and \(Q\) that is perpendicular to the \(x y\) -plane. Find the slope of the tangent at the indicated point to the curve of intersection of this plane and the graph of the given function in the direction of \(Q\). $$ f(x, y)=(x-y)^{2} ; P(4,2), Q(0,1) ;(4,2,4) $$

Step-by-Step Solution

Answer

The slope is \(-\frac{12}{\sqrt{17}}\).

Step 1: Understand the Problem

We need to find the curve of intersection between a given plane and the graph of a function. The plane is through points \(P(4,2)\) and \(Q(0,1)\) and is perpendicular to the \(xy\)-plane. We need to find the slope of the tangent at the point \((4,2,4)\) on this curve in the direction of point \(Q\).

Step 2: Identify the Plane Equation

Since the plane is perpendicular to the \(xy\)-plane, its normal vector must be parallel to the \(z\)-axis, meaning the plane equation is of the form \(z = c\). The point \((4,2,4)\) on the graph of the function suggests \(z = 4\), so the plane equation is \(z = 4\).

Step 3: Determine the Curve of Intersection

The graph of the function \(f(x, y) = (x-y)^2\) intersects the plane \(z = 4\) where \((x-y)^2 = 4\). Solving \((x-y)^2 = 4\) gives the equations \(x-y = 2\) and \(x-y = -2\). These represent two intersections (curves) on this plane.

Step 4: Determine the Direction Vector

The direction from \(P(4,2)\) to \(Q(0,1)\) is given by the vector \((0-4, 1-2) = (-4, -1)\).

Step 5: Find the Slope

Since the direction vector is \((-4, -1)\), we need to consider the derivative of \(f(x,y)\) in this direction. Use the gradient \( abla f(x, y) = (2(x-y), -2(x-y)) = (2d, -2d)\) where \(d = x-y\). For the curve \(x-y = 2\), \(d = 2\), the gradient is \((4, -4)\). The slope in the direction \((-4, -1)\) is:\[\frac{abla f \cdot \mathbf{v}}{\|\mathbf{v}\|} = \frac{4(-4) + (-4)(-1)}{\sqrt{(-4)^2 + (-1)^2}} = \frac{-16 + 4}{\sqrt{17}} = \frac{-12}{\sqrt{17}}\]

Step 6: Conclude with the Result

The slope of the tangent at the point \((4,2,4)\) in the direction of point \(Q\) is \(-\frac{12}{\sqrt{17}}\).

Key Concepts

Plane Intersection

When solving problems involving plane intersections, it is important to understand how a plane interacts with the graph of a function. In this exercise, the plane is perpendicular to the xy-plane, meaning it stands straight up like a wall. This specific orientation ensures that its normal vector aligns with the z-axis. Since we have a point \(4,2,4\) on this plane, it becomes clear that the plane is defined by the equation \(z = 4\). Hence, the plane intersects the graph of the function \(f(x, y) = (x-y)^2\) where \(z = 4\). This results in solving the equation \( (x-y)^2 = 4\) to find the curve(s) at their meeting point.

• The vertical orientation simplifies the plane's equation to one variable, \"z = 4\".
• The interaction forms curves described by simpler linear equations.

Slope Calculation

Understanding the slope in mathematical terms helps determine how steep a curve is at a given point. Calculating the slope of a tangent to a curve can reveal the direction and steepness of the curve at a specific point. In this context, to find the slope at the intersection, one must compute how the curve behaves in a given direction. Here, we find the slope of the tangent at point \(4,2,4\) using the gradient of the function and direction derived from points \(P\) and \(Q\). For computing this, we employ the dot product of the gradient vector and the direction vector, then normalize it:

• Evaluate the gradient of the function to understand change rates.
• Use directional vectors to pin down specific slope needs.
• Apply vector operations like the dot product for precise slope values.

Gradient Vector

The gradient vector is a crucial concept for determining how a function changes at any given point. Essentially, the gradient shows the direction of steepest ascent on the graph of the function. Here, the gradient vector of the function \(f(x,y) = (x-y)^2\) is calculated as \(abla f(x, y) = (2(x-y), -2(x-y))\). Substituting values at \(x-y = 2\), it simplifies to \( (4, -4) \). This vector tells us the fastest increase direction from that point.

• The gradient vector helps visualize change on the function's surface.
• Directional derivatives are strongly dependent on gradients.
• Effective calculation allows identifying steep or gentle slopes.

Vector Direction

The concept of vector direction is essential for guiding how we measure changes along a curve. When determining the slope in a specific direction, the direction vector plays a vital role. In this problem, the vector from point \(P(4,2)\) to \(Q(0,1)\) gives us \((-4, -1)\) as the direction vector. This vector provides the path along which we want to measure the slope of the tangent.

• Direction vectors help guide slope calculations in precise directions.
• Understanding geometry allows for better vector application.
• Aligns with gradient use to provide meaningful tangent insights.