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Chapter 13: Problem 6

Find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point. $$f(x, y)=\tan ^{-1} \frac{\sqrt{x}}{y}, \quad(4,-2)$$

Short Answer

Expert verified
Gradient is \((-\frac{1}{16}, -\frac{1}{4})\) at point (4,-2).

Step by step solution

01

Find the Partial Derivatives

To find the gradient, we first need the partial derivatives of the function given. The function is \( f(x, y) = \tan^{-1} \frac{\sqrt{x}}{y} \). Let's find the partial derivative with respect to \(x\): The derivative of \( \tan^{-1} u \) with respect to \( u \) is \( \frac{1}{1+u^2} \). Here, \( u = \frac{\sqrt{x}}{y} \), so the partial derivative with respect to \( x \) is:\[ \frac{\partial f}{\partial x} = \frac{1}{1 + (\frac{\sqrt{x}}{y})^2} \cdot \frac{1}{2\sqrt{x}y}. \]Now, the partial derivative with respect to \( y \): \[ \frac{\partial f}{\partial y} = \frac{1}{1 + (\frac{\sqrt{x}}{y})^2} \cdot \frac{-\sqrt{x}}{y^2}. \]
02

Evaluate the Partial Derivatives at (4, -2)

Next, we evaluate the partial derivatives at the point \((4, -2)\).For \(\frac{\partial f}{\partial x}\):- Substitute \(x = 4\) and \(y = -2\):- \( u = \frac{\sqrt{4}}{-2} = -1 \)\[ \frac{\partial f}{\partial x} = \frac{1}{1 + (-1)^2} \cdot \frac{1}{2\cdot 2\cdot (-2)} = \frac{1}{2} \cdot \frac{1}{-8} = -\frac{1}{16}. \]For \(\frac{\partial f}{\partial y}\):- Using \( u = -1 \):\[ \frac{\partial f}{\partial y} = \frac{1}{1 + (-1)^2} \cdot \frac{-2}{(-2)^2} = \frac{1}{2} \cdot \frac{-2}{4} = -\frac{1}{4}. \]
03

Write the Gradient Vector

The gradient vector \( abla f \) at any point \((x, y)\) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).From our calculations:\[ abla f(4, -2) = \left( -\frac{1}{16}, -\frac{1}{4} \right). \]
04

Sketch the Level Curve and Gradient

The level curve for a function \( f(x, y) = c \) is the set of all points \((x, y)\) where the function equals a constant \( c \). Since the point \((4, -2)\) is on the level curve, we must have:\[ \tan^{-1} \left( \frac{\sqrt{4}}{-2} \right) = c. \]This simplifies to \( c = \tan^{-1}(-1) = -\frac{\pi}{4} \).The gradient vector is \( \left( -\frac{1}{16}, -\frac{1}{4} \right) \), which can be drawn as an arrow originating from \((4, -2)\) and pointing in the direction of steeper ascent on the surface.On a sketch, the level curve would be an implicit equation and the vector would point perpendicularly to this curve at the point \((4, -2)\).

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Key Concepts

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

Partial Derivatives
Partial derivatives help us understand how a function changes as each variable changes while keeping the others constant. Imagine a multivariable function like \( f(x, y) = \tan^{-1} \frac{\sqrt{x}}{y} \), which depends on both \( x \) and \( y \). To find how the function behaves with respect to \( x \) alone, we compute its partial derivative \( \frac{\partial f}{\partial x} \).
  • The formula for the derivative of the arctangent function, \( \tan^{-1}(u) \), with respect to \( u \) is \( \frac{1}{1+u^2} \).
  • In our function, \( u = \frac{\sqrt{x}}{y} \). For the partial derivative with respect to \( x \), we also include the derivative of the inside expression \( \frac{\sqrt{x}}{y} \) with respect to \( x \), which is \( \frac{1}{2\sqrt{x}y} \).
  • The partial derivative with respect to \( y \) involves a similar process, resulting in \( \frac{-\sqrt{x}}{y^2} \).
These derivatives help build the gradient, which is a vector pointing in the direction of the greatest rate of increase of the function.
Level Curves
Level curves are a fascinating concept in calculus that offer insights into the nature of multivariable functions. They are the set of points \((x, y)\) where a function \(f(x, y)\) equals a constant \(c\). You might think of them as the topographical lines on a map representing elevation.
  • For our function, \(f(x, y) = \tan^{-1} \frac{\sqrt{x}}{y}\), a level curve can be defined where the function equals \(-\frac{\pi}{4}\).
  • By solving \(\tan^{-1}\left(\frac{\sqrt{x}}{y}\right) = -\frac{\pi}{4}\), we identify all points \((x, y)\) that lie on this curve. This corresponds to the point where our original point \((4, -2)\) lies.
  • On a graph, level curves help visualize how the function behaves in different regions of the \(xy\)-plane.
The gradient is always perpendicular to the level curves, providing a visual representation of the direction of steepest ascent.
Arctangent Function
The arctangent function, expressed as \( \tan^{-1}(x) \), is the inverse of the tangent function. It gives the angle whose tangent is \(x\). This makes the arctangent crucial in trigonometry and calculus, especially when dealing with angles and slopes.
  • In our function \( f(x, y) = \tan^{-1} \frac{\sqrt{x}}{y} \), the arctangent helps describe the angle formed by the ratio \( \frac{\sqrt{x}}{y} \).
  • This function is bounded between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), providing clear geometric interpretations.
  • Calculating \( \tan^{-1}(-1) \) as \(-\frac{\pi}{4}\) helps us identify the specific level curve through the point \((4, -2)\).
Understanding the arctangent in this context deepens our comprehension of how angles relate to changes in the function, crucial for sketching gradients and level curves.

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Most popular questions from this chapter

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) \(f(x, y, z)=x y+2 y z-3 x z\) at \(P_{0}(1,1,0)\) \(R: |x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z| \leq 0.01\)

Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I\), you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$\begin{aligned} &x=u \cos v, \quad y=u \sin v, \quad z=v, \quad 0 \leq u \leq 2\\\ &0 \leq v \leq 2 \pi \end{aligned}$$

You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer.

Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{aligned}&f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3,-3 / 2 \leq x \leq 3 / 2\\\ &-3 / 2 \leq y \leq 3 / 2\end{aligned}$$

Find the linearizations \(L(x, y, z)\) of the functions at the given points. \(f(x, y, z)=\tan ^{-1}(x y z)\) at a. (1,0,0) b. (1,1,0) c. (1,1,1)
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