Problem 31 - 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 \(P\). $$F(x, y)=e^{-x^{2} / 2-y^{2} / 2} ; P(-1,1)$$

Step-by-Step Solution

Answer

In summary, at point \(P(-1, 1)\), the unit vectors in the direction of steepest ascent and descent are \(\frac{1}{\sqrt{2}} (1, -1)\) and \(\frac{1}{\sqrt{2}}(-1, 1)\), respectively. Also, a vector pointing in a direction of no change in the function at point \(P\) is \((e^{-1}, e^{-1})\).

Step 1: Find the gradient of the function

To find the gradient of the function, we need to find the partial derivatives with respect to \(x\) and \(y\), denoted as \(\nabla F\). $$\nabla F = \Big(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}\Big)$$ To find the partial derivatives, we get: $$\frac{\partial F}{\partial x} = -x e^{-x^{2} / 2 - y^{2} / 2}$$ $$\frac{\partial F}{\partial y} = -y e^{-x^{2} / 2 - y^{2} / 2}$$ So, the gradient is: $$\nabla F = \Big(-x e^{-x^{2} / 2 - y^{2} / 2}, -y e^{-x^{2} / 2 - y^{2} / 2}\Big)$$

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

Now, we must evaluate the gradient at the given point \(P(-1, 1)\). Plugging in the coordinates of \(P\) into the gradient, we get: $$\nabla F_P = \Big(-(-1) e^{-(-1)^{2} / 2 - 1^{2} / 2}, -(1) e^{-(-1)^{2} / 2 - 1^{2} / 2}\Big)$$ After simplifying the expression, we obtain: $$\nabla F_P = \Big(e^{-1}, -e^{-1}\Big)$$

Step 3: Find the unit vectors in the direction of steepest ascent and descent

In order to find the unit vectors in the direction of steepest ascent and descent, we must normalize the gradient vector found in step 2. The magnitude of \(\nabla F_P\) is: $$\norm{\nabla F_P} = \sqrt{(e^{-1})^2 + (-e^{-1})^2} = e^{-1} \sqrt{2}$$ Now, we find the unit vectors by dividing the gradient vector by its magnitude: $$\hat{v}_{ascent} = \frac{1}{e^{-1}\sqrt{2}} (e^{-1}, -e^{-1}) = \frac{1}{\sqrt{2}} (1, -1)$$ Since the direction of steepest descent is opposite to the direction of steepest ascent, we can find it by multiplying the unit vector of steepest ascent by -1: $$\hat{v}_{descent} = -\hat{v}_{ascent} = \frac{1}{\sqrt{2}}(-1, 1)$$

Step 4: Find a vector pointing in a direction of no change in the function at point \(P\)

In order to find a vector pointing in a direction of no change in the function at the given point, we can compute the cross product between the gradient of the function and a unit vector to obtain a vector orthogonal to the gradient vector. Since we have a 2-dimensional gradient vector, we can use a simple orthogonal vector calculation: $$\vec{v}_{no\_change} = (-\frac{\partial F}{\partial y}, \frac{\partial F}{\partial x})$$ Evaluating this expression at point \(P\), we get: $$\vec{v}_{no\_change} = (e^{-1}, e^{-1})$$ Thus, the vector pointing in a direction of no change in the function at point \(P\) is \((e^{-1}, e^{-1})\).

Key Concepts

Partial Derivatives

In calculus, a partial derivative is a derivative where we hold some variables constant and differentiate others. When working with multivariable functions, partial derivatives help us analyze how the function might change when only one of the input variables changes. Consider the function in the exercise, \[ F(x, y) = e^{-x^2/2 - y^2/2} \]This function depends on two variables, \(x\) and \(y\). To find the partial derivative with respect to \(x\), treat \(y\) as a constant. Similarly, to find the partial derivative with respect to \(y\), treat \(x\) as a constant.

• The partial derivative with respect to \(x\) is given by: \[ \frac{\partial F}{\partial x} = -x e^{-x^2/2 - y^2/2} \]


• The partial derivative with respect to \(y\) is: \[ \frac{\partial F}{\partial y} = -y e^{-x^2/2 - y^2/2} \]

These derivatives form the components of the gradient vector, which points in the direction of the greatest rate of increase of the function and is key to understanding changes in the function's values in various directions.

Directional Derivatives

The concept of directional derivatives extends the idea of partial derivatives. It lets us find how a function changes along any direction at a given point. Unlike partial derivatives, which check change along the axes, directional derivatives offer a more comprehensive picture.

• To compute a directional derivative, you use the gradient of the function, which is a vector consisting of all its partial derivatives.


• The gradient vector for the function is: \[ abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y} \right) \ = \left( -x e^{-x^2/2 - y^2/2}, -y e^{-x^2/2 - y^2/2} \right) \]

Evaluating this gradient at point \(P(-1, 1)\), we derive: \[ abla F_P = \left( e^{-1}, -e^{-1} \right) \]
By taking the dot product of this gradient with any unit vector, we can find how the function \(F(x, y)\) changes in that specific direction. The directional derivative shows the rate of change of the function in the direction of the unit vector. A positive directional derivative indicates an increase, while a negative one indicates a decrease.

Vector Calculus

Vector calculus is the mathematical toolset that deals with vector functions and their differentiation and integration. It's particularly useful in fields where direction matters, like physics and engineering. In the context of this exercise, vector calculus helps us understand gradients and movement on surfaces.

• One central concept is the gradient, which represents a multi-dimensional rate of change. Here, the gradient is expressed via partial derivatives: \[ abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y} \right) \]


• Another key component of vector calculus applied here is directionality. This involves understanding vectors pointing in directions of ascent, descent, and no change:

• **Steepest Ascent and Descent:** The vector of steepest ascent aligns with the gradient direction. It's calculated by normalizing the gradient: \[ \hat{v}_{ascent} = \frac{1}{\sqrt{2}}(1, -1) \]


• The steepest descent is simply the opposite: \[ \hat{v}_{descent} = \frac{1}{\sqrt{2}}(-1, 1) \]


• **No Change:** A direction where the function remains constant. For the problem, it involves finding a vector perpendicular to the gradient: \[ \vec{v}_{no\_change} = (e^{-1}, e^{-1}) \]

By employing vector calculus, we can skillfully navigate through the multi-dimensional landscape, understanding where and how functions change across various paths.