Problem 20 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$z=\sin x y+2 ;(1,0,2) \text { and }(0,5,2)$$

Step-by-Step Solution

Answer

The equations of the tangent planes are: 1. For the tangent plane at the point \((1,0,2)\): \(y = 0\) 2. For the tangent plane at the point \((0,5,2)\): \(x = 0\)

Step 1: Compute the gradient of the given function

Given a function \(z = f(x,y)\), the gradient vector can be found by taking the partial derivative with respect to both x and y. Our function is $$z = \sin(x y) + 2.$$ To compute its gradient, we need the partial derivatives: $$\frac{\partial z}{\partial x} = \cos(x y) \cdot y$$ $$\frac{\partial z}{\partial y} = \cos(x y) \cdot x$$ Thus, the gradient of the function is given by the vector: $$\nabla z = \left\langle\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\right\rangle = \langle y\cos(x y), x\cos(x y)\rangle$$.

Step 2: Evaluate the gradient at each point

Now, we need to evaluate the gradient at each of the given points. For \(P_1 = (1,0,2)\): $$\nabla z(1,0) = \langle 0\cos(1\cdot0), 1\cos(1\cdot0)\rangle = \langle 0,1 \rangle$$ For \(P_2 = (0,5,2)\): $$\nabla z(0,5) = \langle 5\cos(0\cdot5), 0\cos(0\cdot5)\rangle = \langle 5,0 \rangle$$

Step 3: Apply the point-normal form of the plane equation

The point-normal form of the plane equation is given by: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ where \((A, B, C)\) is the normal vector and \((x_0, y_0, z_0)\) is a point on the plane. For the tangent plane at \(P_1 = (1,0,2)\), the normal vector is the gradient evaluated at \(P_1\), \(\nabla z(1,0) = \langle 0,1 \rangle\). The point-normal form of the tangent plane at \(P_1\) becomes: $$0(x-1) + 1(y-0) + 0(z-2) = 0$$ which simplifies to: $$y = 0$$ For the tangent plane at \(P_2 = (0,5,2)\), the normal vector is the gradient evaluated at \(P_2\), \(\nabla z(0,5) = \langle 5,0 \rangle\). The point-normal form of the tangent plane at \(P_2\) becomes: $$5(x-0) + 0(y-5) + 0(z-2) = 0$$ which simplifies to: $$5x = 0$$ or $$x = 0$$

Step 4: Final Answer

The equations of the tangent planes are: 1. For the tangent plane at the point \((1,0,2)\): $$y = 0$$ 2. For the tangent plane at the point \((0,5,2)\): $$x = 0$$

Key Concepts

Partial Derivative

The concept of a partial derivative is like a window into the way functions interact in a multi-dimensional space. Imagine you have a surface hovering in a three-dimensional space, defined by a function in terms of two variables, such as \(z = \sin(xy) + 2\). Here, \(x\) and \(y\) are your main actors. Now, the partial derivative tells you how this function changes if you tweak one of these actors while holding the other steady.

If you take the partial derivative with respect to \(x\), symbolized as \(\frac{\partial z}{\partial x}\), you're focusing on how changes in \(x\) impact \(z\), assuming \(y\) stands still. As a result, the tangled expressions that emerge reflect the drama between \(z\) and \(x\), leaving \(y\) in a supporting role. In our specific example, this results in \(\frac{\partial z}{\partial x} = y \cdot \cos(xy)\).

Conversely, if you shift your gaze to \(\frac{\partial z}{\partial y}\), you're now following how variations in \(y\) sway \(z\), while \(x\) remains unmoved. This gives us \(\frac{\partial z}{\partial y} = x \cdot \cos(xy)\).

By taking these partial derivatives, you are essentially unraveling the complex narrative of how each variable influences the behavior of the surface, providing the building blocks necessary to understand gradients and, eventually, the formation of tangent planes.

Gradient Vector

The gradient vector works as a directional arrow towards steepest ascent of a function. Composed of all the first partial derivatives, this vector succinctly encapsulates both the direction and rate of change of a surface at any given point. In mathematical form, if you have \(z = f(x, y)\), the gradient vector, denoted as \(abla z\), is expressed as \(\langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \rangle\).

This vector isn't just a bundle of numbers: it provides a comprehensive view of a surface's terrain at a snapshot moment. Each component within the gradient reflects the slope along its respective axis. The role of the gradient is crucial when you're hunting for steep paths or trying to balance precariously across a surface.

In the context of our problem, the gradient vector at any point \((x, y)\) on the surface \(z = \sin(xy) + 2\) is \(abla z = \langle y \cos(xy), x \cos(xy) \rangle\). Evaluating this at given points like \((1, 0)\) yields \(\langle 0, 1 \rangle\), while at \((0, 5)\) it gives \(\langle 5, 0 \rangle\). These results represent the normals to the tangent planes at these respective points, serving as guides in the construction of the tangent plane equations.

Point-Normal Form of Plane Equation

The point-normal form is a valuable tool in describing planes. This elegant formula connects the dots between a normal vector that points perpendicularly to the plane and a specific point that the plane passes through. The equation takes the form \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\), where \((A, B, C)\) is the normal vector, and \((x_0, y_0, z_0)\) is a given point on the plane.

The beauty of this formula lies in its simplicity. With the right components plugged in, it describes an infinite flat surface in three-dimensional space, all centered around the chosen point and oriented by the normal vector. The trick to using this form is understanding that the normal vector \(\langle A, B, C \rangle\) gives your plane its unique tilt and direction.

For example, at the point \((1, 0, 2)\) on the surface \(z = \sin(xy) + 2\), we have a gradient vector, our normal, of \(\langle 0, 1 \rangle\). Substituting into the equation, we find that only the \(y\)-component survives, neatly collapsing the formula to \(y = 0\) at this point.

Similarly, at \((0, 5, 2)\), the normal vector is \(\langle 5, 0 \rangle\), resulting in the straightforward plane equation \(5x = 0\), or equivalently, \(x = 0\). Thus, each equation reflects a specific plane tangent at the given points-with normal vectors guiding their formation like invisible architects.