Question

The following functions have exactly one isolated peak or one isolated depression (one local maximum or minimum). Use a graphing utility to approximate the coordinates of the peak or depression. $$h(x, y)=1-e^{-\left(x^{2}+y^{2}-2 x\right)}$$

Short answer

Answer: The approximate coordinates of the local maximum are (1, 0, 0).

Step-by-step solution

Find the partial derivatives

To find the local maximum or minimum of the function, find the partial derivatives with respect to x and y: $$\frac{\partial h}{\partial x} = \frac{\partial}{\partial x}\left(1-e^{-\left(x^{2}+y^{2}-2 x\right)}\right)$$ $$\frac{\partial h}{\partial y} = \frac{\partial}{\partial y}\left(1-e^{-\left(x^{2}+y^{2}-2 x\right)}\right)$$ Then, calculate the derivatives for both expressions to simplify: $$\frac{\partial h}{\partial x} = e^{-\left(x^{2}+y^{2}-2 x\right)}\cdot\left(2x-2\right)$$ $$\frac{\partial h}{\partial y} = e^{-\left(x^{2}+y^{2}-2 x\right)}\cdot\left(2y\right)$$

Find the critical points

Now, set both partial derivatives equal to zero and solve for the critical points: $$e^{-\left(x^{2}+y^{2}-2 x\right)}\cdot\left(2x-2\right)=0$$ $$e^{-\left(x^{2}+y^{2}-2 x\right)}\cdot\left(2y\right)=0$$ Since $$e^{-\left(x^{2}+y^{2}-2 x\right)}>0$$ for all x and y, we can focus on the other factors: $$2x - 2 = 0 \Rightarrow x = 1$$ $$2y = 0 \Rightarrow y = 0$$ So the critical point is (1, 0).

Use a graphing utility to approximate the coordinates of the peak or depression

Plug the critical point (1, 0) into the function $$h(x, y)=1-e^{-\left(x^{2}+y^{2}-2 x\right)}$$: $$h(1, 0)=1-e^{-\left(1^2+0^2-2\cdot1\right)}=1-e^0=1-1=0$$ Using a graphing utility, the function appears to have an isolated peak at (1,0,0). So, the coordinates of the peak are approximately (1, 0, 0). You can find online graphing utilities like GeoGebra, Desmos, or WolframAlpha, to graph this function and visually confirm the location of the peak.

Key concepts

Partial Derivatives

When exploring the concept of local extrema in multivariable calculus, one of the foundational tools we use is the notion of partial derivatives. These derivatives tell us how a function changes as each variable is altered while all other variables are held constant. For a function like \( h(x, y) \), which depends on two variables, we can take the partial derivative with respect to \( x \) and the partial derivative with respect to \( y \).

By differentiating \( h(x, y) = 1-e^{-(x^2+y^2-2x)} \) with respect to each variable, we gain insight into the function's behavior along individual axes. For instance, the expression \( \frac{\partial h}{\partial x} = e^{-(x^2+y^2-2x)}(2x-2) \) reveals how the function rises and falls as \( x \) changes, and similarly, \( \frac{\partial h}{\partial y} = e^{-(x^2+y^2-2x)}(2y) \) does the same for variations in \( y \).

Understanding these gradients is crucial because they lead directly to the identification of critical points, which are candidates for local maxima or minima. These derivative expressions are like reading the landscape of a graph, depicting slopes and flat areas where peaks or depressions might occur.

Critical Points in Calculus

Discovering critical points is a pivotal step when analyzing a multivariable function for its local extrema. A critical point occurs where all partial derivatives of a function are zero or undefined, signaling a potential local maximum or minimum. In our function \( h(x, y) \), we set its partial derivatives equal to zero to find such points:

To find the critical points, we solve the system formed by the equalities \( \frac{\partial h}{\partial x} = 0 \) and \( \frac{\partial h}{\partial y} = 0 \). In our example, because the exponential function \( e^{-\left(x^{2}+y^{2}-2 x\right)} \) is always positive, we disregard it and focus on the polynomial factors to determine the critical point at \( x = 1 \) and \( y = 0 \), leading us to \( (1, 0) \).

It's worth highlighting that not every critical point corresponds to an actual extremum; some could be saddle points or points of inflection. That's why after locating critical points, one usually conducts further tests, such as the Second Derivative Test or analyzing concavity, to confirm the nature of these points.

Graphing Utility for Calculus

To enhance our comprehension and confirmation of local extrema, we frequently turn to a graphing utility. These digital tools allow us to visualize complex functions and their behaviors, illuminating features that can be difficult to assess through calculation alone. After computing the critical point of the function \( h(x, y) = 1-e^{-\left(x^{2}+y^{2}-2 x\right)} \), a graphing utility such as GeoGebra, Desmos, or WolframAlpha can be used to create a visual representation of the function’s surface.

In our case, graphing the function reveals that the critical point at \( (1, 0) \) corresponds to an isolated peak. By entering the function into the utility, we can see not just this peak, but also the surface contours and how the function reaches this maximum value. This visual evidence supports our analytical findings and provides a more intuitive understanding of the function's behavior in space.

Moreover, using graphing utilities in calculus has the added benefit of enabling interactive exploration, which can make learning more engaging and supportive for students. They can adjust variables in real-time and observe immediate changes in the graph, fostering a deeper conceptual grasp of the material.