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Find an equation of the plane tangent to the following surfaces at the given points (two planes and two equations). $$z=2 \cos (x-y)+2 ;\left(\frac{\pi}{6},-\frac{\pi}{6}, 3\right) \text { and }\left(\frac{\pi}{3}, \frac{\pi}{3}, 4\right)$$

Short Answer

Expert verified
Answer: The equations of the tangent planes at the given points are $$\sqrt{3}(x-\frac{\pi}{6})-\sqrt{3}(y+\frac{\pi}{6}) = 0$$ and $$z = 4$$.

Step by step solution

01

Find the gradient of the surface

To find the gradient, we will take the partial derivative of the surface with respect to x and y. So for the surface $$z=2 \cos (x-y)+2$$, we have: $$ \frac{\partial{z}}{\partial{x}} = -2\sin(x-y)$$ $$ \frac{\partial{z}}{\partial{y}} = 2\sin(x-y)$$
02

Evaluate the gradient at the first point

Now, we will evaluate the gradient at the first point, $$\left(\frac{\pi}{6},-\frac{\pi}{6}, 3\right)$$. This gives us: $$\frac{\partial{z}}{\partial{x}}\Big|_{ \left(\frac{\pi}{6},-\frac{\pi}{6}\right)} = 2\sin\left(\frac{\pi}{6}+\frac{\pi}{6}\right) = 2\sin\left(\frac{\pi}{3}\right) = \sqrt{3}$$ $$\frac{\partial{z}}{\partial{y}}\Big|_{ \left(\frac{\pi}{6},-\frac{\pi}{6}\right)} = -2\sin\left(\frac{\pi}{6}+\frac{\pi}{6}\right) = -2\sin\left(\frac{\pi}{3}\right) = -\sqrt{3}$$
03

Find the equation of the tangent plane at the first point

Using the gradient at the first point and the point itself, we can find the equation of the tangent plane with the formula $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ where $$\textbf{n} = \langle A, B, C \rangle$$ is the gradient of the surface. Inserting our values, we get $$ \sqrt{3}(x-\frac{\pi}{6})-\sqrt{3}(y+\frac{\pi}{6}) +0(z-3) = 0$$.
04

Evaluate the gradient at the second point

Now, we will evaluate the gradient at the second point, $$\left(\frac{\pi}{3},\frac{\pi}{3}, 4\right)$$. This gives us: $$\frac{\partial{z}}{\partial{x}}\Big|_{ \left(\frac{\pi}{3},\frac{\pi}{3}\right)} = -2\sin\left(\frac{\pi}{3}-\frac{\pi}{3}\right) = 0$$ $$\frac{\partial{z}}{\partial{y}}\Big|_{ \left(\frac{\pi}{3},\frac{\pi}{3}\right)} = 2\sin\left(\frac{\pi}{3}-\frac{\pi}{3}\right) = 0$$
05

Find the equation of the tangent plane at the second point

Using the gradient at the second point and the point itself, we can find the equation of the tangent plane with the formula $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ where $$\textbf{n} = \langle A, B, C \rangle$$ is the gradient of the surface. Inserting our values, we get $$ 0(x-\frac{\pi}{3})+0(y-\frac{\pi}{3}) +C(z-4) = 0$$. However, since the partial derivatives with respect to $$x$$ and $$y$$ are zero, we have a degenerate tangent plane. In this case, the tangent plane is the same as the plane $$z = 4$$ parallel to the $$xy$$-plane. Hence, the equations of the tangent planes are: $$ \sqrt{3}(x-\frac{\pi}{6})-\sqrt{3}(y+\frac{\pi}{6}) = 0$$ $$ z = 4$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are at the heart of understanding the behavior of functions in multivariable calculus. Imagine you're on a hilly terrain, and you want to know how steep the hill is if you walk directly north or east. This is what partial derivatives tell us: how a function, representing the height of the terrain, changes as you move in one direction, keeping the other fixed.

For a function like z = f(x, y), the partial derivative with respect to x is represented by \( \frac{\partial{z}}{\partial{x}} \) and tells us how the function changes as x changes while y stays constant. Similarly, \( \frac{\partial{z}}{\partial{y}} \) shows the change with respect to y while x is constant.

In our example, we calculated \( \frac{\partial{z}}{\partial{x}} = -2\sin(x-y) \) and \( \frac{\partial{z}}{\partial{y}} = 2\sin(x-y) \). These derivatives inform us how the surface described by z=2 \cos (x-y)+2 slopes along the x and y axis, respectively.
Gradient of a Surface
In multivariable calculus, the gradient becomes a crucial concept when determining the direction of steepest ascent on a surface. If partial derivatives describe how steep a hill is towards the north or east, the gradient combines these two directions, pointing towards the steepest uphill path from where you're standing.

The gradient of a surface is a vector composed of all its first partial derivatives. For a function z = f(x, y), the gradient vector is denoted as \( abla f = \left\langle \frac{\partial{z}}{\partial{x}}, \frac{\partial{z}}{\partial{y}} \right\rangle \), where \( abla \) is the gradient operator. In the given problem, after finding the partial derivatives, we take their values at given points to get the direction of the gradient of the surface at those points, which eventually gives us the normal vector to the tangent plane.
Multivariable Calculus
Multivariable calculus extends the ideas of single-variable calculus into higher dimensions, which is essential for dealing with functions of several variables. Just as in single-variable calculus, where you study the rate of change and areas under curves, in multivariable calculus, you analyze surfaces, contour maps, and changes across multidimensional terrains.

The concept of taking partial derivatives and finding gradients are part of the foundational tools in this field. They help in understanding and solving problems related to rates of change in multiple directions and optimization problems in spaces that are not merely lines or curves but entire areas or volumes.
Tangent Plane to a Surface
The tangent plane to a surface at a given point is, in essence, the best linear approximation to the surface near that point. Imagine setting a flat, inflexible sheet of paper to just touch a model of a hill at one point without cutting into the hill. This paper represents the tangent plane at that contact point.

The equation of the tangent plane can be found using the point of tangency and the gradient of the surface at that point. The equation is typically written as A(x - x_0) + B(y - y_0) + C(z - z_0) = 0, where (x_0, y_0, z_0) is the point of tangency, and \( \textbf{n} = \langle A, B, C \rangle \) is a vector normal to the tangent plane—which can be found using the gradient. In the provided exercise, we derived such an equation for two specific points on the given surface.

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Most popular questions from this chapter

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