Question

Find the slopes of the surface in the \(x\) - and \(y\) -directions at the given point. $$ \begin{array}{l} z=\cos (2 x-y) \\ \left(\frac{\pi}{4}, \frac{\pi}{3}, \frac{\sqrt{3}}{2}\right) \end{array} $$

Short answer

The slopes of the surface at the given point in the \( x \) and \( y \) directions are given by \( \frac{∂z}{∂x} \) and \( \frac{∂z}{∂y} \) respectively, each evaluated at the point \( (\frac{\pi}{4}, \frac{\pi}{3}, \frac{\sqrt{3}}{2}) \). Evaluate these expressions to find the final slopes.

Step-by-step solution

Find the partial derivative with respect to x

Apply the chain rule to differentiate \( \cos(2x - y) \) with respect to \(x\). The derivative of the cosine of a function is the negative of the sine of that function times the derivative of the inner function. So, \( \frac{∂z}{∂x} = - \sin(2x - y) * 2 \).

Evaluate the partial derivative at the given point

Now, substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{3}\) into \( \frac{∂z}{∂x} = - \sin(2x - y) * 2 \). Then, \( \frac{∂z}{∂x} = - \sin(2*\frac{\pi}{4} - \frac{\pi}{3}) * 2 \). Evaluate the above expression to obtain \( \frac{∂z}{∂x} \) at the given point.

Find the partial derivative with respect to y

Again, differentiate \( \cos(2x - y) \) with respect to \(y\). The chain rule leads to \( \frac{∂z}{∂y} = \sin(2x - y) \). There is no additional factor because the derivative of \( -y \) with respect to \( y \) is \( -1 \).

Evaluate the partial derivative at the given point

Finally, substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{3}\) into \( \frac{∂z}{∂y} = \sin(2x - y) \). Then, \( \frac{∂z}{∂y} = \sin(2*\frac{\pi}{4} - \frac{\pi}{3}) \). Evaluate the above expression to obtain \( \frac{∂z}{∂y} \) at the given point.

Key concepts

Chain Rule

Understanding the chain rule is vital when working with functions composed of other functions, particularly in calculus. In our exercise, the chain rule is applied to calculate the slope of the surface described by the function z = \( \text{cos}(2x - y) \) .

When we differentiate the function with respect to x, the outer function is \( \text{cos}(u) \), where u = 2x - y is the inner function. The chain rule states that the derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function with respect to x. In this case, the derivative of \( \text{cos}(u) \) with respect to u is -\( \text{sin}(u) \), and the derivative of u with respect to x is 2.

This process makes it possible to handle more complex derivatives by breaking them down into simpler components that can be more easily solved.

Partial Derivative Calculation

Partial derivative calculation is a core element in multivariable calculus, used to determine the rate of change of a function in relation to one variable while keeping other variables constant. In our exercise, we look at the function z depending on both x and y.

The steps described provide a methodical approach to finding \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), the partial derivatives with respect to x and y respectively. The calculations involve applying derivation rules to the expression \( \text{cos}(2x - y) \) for each variable in turn. These values indicate how the function's value changes in the direction of each axis (x or y), with application at a specific point on the surface.

Slope of a Surface

The slope of a surface at a given point represents the steepness or incline of the surface in a particular direction. In the context of the provided exercise, we are interested in the slopes in the x- and y-directions, or the \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) at the point \( \left(\frac{\pi}{4}, \frac{\pi}{3}, \frac{\sqrt{3}}{2}\right) \).

Using the calculated partial derivatives, the slopes give us an insight into how steep the surface is if we moved along the x-axis or the y-axis from that point. Specifically, these values delineate the tangential slope of the surface along the direction of the respective axes. The value of the slope can be positive, negative, or zero, indicating an uphill, downhill, or level surface direction respectively at the point of interest.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. The problem in question is an illustration of how multivariable calculus is used to explore and analyze the properties of surfaces and functions in a multi-dimensional space.

In multivariable calculus, the idea of a derivative expands to partial derivatives, which are used to determine slopes and rates of change along different dimensions. Furthermore, concepts like the chain rule become more complex but equally more powerful, as they allow us to differentiate functions that are composed of other functions that have more than one variable.

The exercise showcases these principles by computing the partial derivatives with respect to x and y, offering a glimpse into the practical aspects of multivariable calculus, such as describing the geometry of surfaces and its application in fields like physics, engineering, economics, and beyond.