Problem 32 - Calculus Early Transcendentals

Question

Direction of steepest ascent and descent Consider the following functions and points \(P\). a. Find the unit vectors that give the direction of steepest ascent and steepest descent at \(P\) b. Find a vector that points in a direction of no change in the function at \(P\). $$f(x, y)=2 \sin (2 x-3 y) ; P(0, \pi)$$

Step-by-Step Solution

Answer

Answer: a. Steepest Ascent: \(\left(\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}\right)\) Steepest Descent: \(\left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)\) b. Direction of no change: \((3, 2)\)

Step 1: Find the partial derivatives

First, we need to find the partial derivatives of the function \(f(x, y)\). The partial derivative with respect to \(x\): $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (2\sin(2x -3y)) = 2\cos(2x - 3y) * 2$$ The partial derivative with respect to \(y\): $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (2\sin(2x -3y)) = 2\cos(2x - 3y) * (-3)$$

Step 2: Evaluate the gradient at \(P\)

Now we need to evaluate the gradient of \(f(x, y)\) at the point \(P(0, \pi)\): $$\nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$ At the point \((0, \pi)\): $$\nabla f(0, \pi) = (4\cos(2(0) - 3(\pi)), -6\cos(2(0) - 3(\pi))) = (4\cos(-6\pi), -6\cos(-6\pi))$$ Since \(\cos(-6\pi) = \cos(6\pi) = 1\), we have: $$\nabla f(0, \pi) = (4, -6)$$

Step 3: Determine the unit vectors for steepest ascent and descent

Since the direction of steepest ascent is given by the gradient, its unit vector is simply the normalized gradient vector: $$\text{Ascent Unit Vector} = \frac{\nabla f(0, \pi)}{\left\|\nabla f(0, \pi) \right\|} = \frac{(4, -6)}{\sqrt{4^2 + (-6)^2}} = \frac{(4, -6)}{\sqrt{52}} = \left(\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}\right)$$ The direction of steepest descent is the opposite of the gradient, so its unit vector is simply the negative of the normalized gradient vector: $$\text{Descent Unit Vector} = - \frac{\nabla f(0, \pi)}{\left\|\nabla f(0, \pi) \right\|} = \left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)$$

Step 4: Determine a vector in the direction of no change

To find a direction of no change, we need to find a vector that is orthogonal to the gradient vector. We can do this using the dot product. Let \(v = (a, b)\) be the vector orthogonal to the gradient. Then, $$\nabla f(0, \pi) \cdot v = (4, -6) \cdot (a, b) = 4a - 6b = 0$$ By solving for a variable, say \(a\), we can find vectors that satisfy this condition: $$a = \frac{3}{2}b$$ For instance, we can choose \(b = 2\), which gives the vector in the direction of no change as: $$\text{No Change Direction} = \left(\frac{3}{2}(2), 2\right) = (3, 2)$$ So, our final answers are: a. Steepest Ascent: \(\left(\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}\right)\) Steepest Descent: \(\left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)\) b. Direction of no change: \((3, 2)\)

Key Concepts

Partial Derivatives

Partial derivatives are foundational when working with functions of several variables. They help you determine how the function changes as you vary one variable while keeping the others constant. Think of it as slicing through a multi-dimensional landscape to see the change along a specific direction.

For a function like \(f(x, y) = 2\sin(2x - 3y)\), partial derivatives with respect to \(x\) and \(y\) give:

• \(\frac{\partial f}{\partial x} = 4\cos(2x - 3y)\)
• \(\frac{\partial f}{\partial y} = -6\cos(2x - 3y)\)

These results are achieved by treating one variable as constant while differentiating with respect to the other. Doing so, you uncover how changes in \(x\) or \(y\) alone impact the function. Each partial derivative reflects the function slope along its specific axis.

Steepest Ascent

The gradient reflects the direction of steepest ascent of a function. This means if you follow the gradient direction, you'll climb a surface, reaching the highest increase per step. The vector gradient is made from the partial derivatives:

• For \(f(x, y) = 2\sin(2x - 3y)\), at point \((0, \pi)\), the gradient \(abla f\) is \((4, -6)\).

To find the direction of steepest ascent in standard form, calculate the unit vector by dividing the gradient by its magnitude:

• Steepest ascent unit vector: \(\left(\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}\right)\).

This unit vector points to the path of quickest increase.

Steepest Descent

The concept of steepest descent is the opposite of steepest ascent. It concerns itself with the direction that drops you down the fastest from a given point on a surface. To find this in terms of the gradient, simply negate the ascent unit vector.

• Given the gradient \((4, -6)\), the ascent unit vector \(\left(\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}\right)\) becomes \(\left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)\) for descent.

This tells you where to go if you want to head downhill most rapidly, much like a marble rolling down the steeper path.

Vector Orthogonality

In multivariable calculus, a vector orthogonal to the gradient indicates no change in function value as you move in that direction. This orthogonal relationship is found using the dot product:

• For example, with the gradient \((4, -6)\), a vector \(v = (a, b)\) is orthogonal if \(4a - 6b = 0\).

Solving this equation gives you a family of vectors, each pointing along a line of constant function value. For instance, choosing \(a = \frac{3}{2}b\), and setting \(b = 2\), yields the orthogonal direction \((3, 2)\). This vector implies zero gradient, meaning a horizontal line on the surface.