Problem 264 - Calculus 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v} .\) $$ h(x, y)=e^{x} \sin y, P\left(1, \frac{\pi}{2}\right), \mathbf{v}=-\mathbf{i} $$

Step-by-Step Solution

Answer

The directional derivative is \(-e\).

Step 1: Calculate the Partial Derivatives

First, we find the partial derivatives of the function \(h(x, y) = e^x \sin y\) with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is: \[\frac{\partial h}{\partial x} = \frac{d}{dx}(e^x \sin y) = e^x \sin y.\]The partial derivative with respect to \(y\) is: \[\frac{\partial h}{\partial y} = \frac{d}{dy}(e^x \sin y) = e^x \cos y.\]

Step 2: Evaluate the Gradients at the Point

Next, we evaluate these partial derivatives at the point \(P(1, \frac{\pi}{2})\).Substitute \(x = 1\) and \(y = \frac{\pi}{2}\) into the partial derivatives:\[\frac{\partial h}{\partial x}(1, \frac{\pi}{2}) = e^1 \cdot \sin \left(\frac{\pi}{2}\right) = e\]\[\frac{\partial h}{\partial y}(1, \frac{\pi}{2}) = e^1 \cdot \cos \left(\frac{\pi}{2}\right) = 0.\]

Step 3: Formulate the Gradient Vector

The gradient vector of \(h\) at point \(P(1, \frac{\pi}{2})\) is composed of the evaluated partial derivatives:\[abla h(1, \frac{\pi}{2}) = \left( e, 0 \right).\]

Step 4: Normalize the Direction Vector

The given direction vector is \(\mathbf{v} = -\mathbf{i} = (-1, 0)\). Calculate its magnitude:\[\| \mathbf{v} \| = \sqrt{(-1)^2 + 0^2} = 1.\]Since the magnitude is 1, the vector \(\mathbf{v}\) is already a unit vector.

Step 5: Calculate the Directional Derivative

The directional derivative of \(h\) in the direction of \(\mathbf{v}\) at the point \(P\) is given by the dot product of the gradient vector and the direction vector:\[D_{\mathbf{v}} h = abla h \cdot \mathbf{v} = (e, 0) \cdot (-1, 0) = -e \times 1 + 0 \times 0 = -e.\]

Key Concepts

Gradient Vector

A gradient vector is like a compass for multivariable functions. It points in the steepest uphill direction of a function at a given point. This means that when you're at a hill, the gradient vector tells you which way is straight uphill. For the function \( h(x, y) = e^x \sin y \), the gradient vector at point \( P(1, \frac{\pi}{2}) \) is found using the partial derivatives calculated at this point:

• The partial derivative with respect to \( x \) gives the gradient's \( x \)-component: \( e \).
• The partial derivative with respect to \( y \) gives the gradient's \( y \)-component: \( 0 \).

So, the gradient vector at \( P(1, \frac{\pi}{2}) \) is \( (e, 0) \). This vector shows that the function is changing most rapidly in the positive \( x \)-direction, as indicated by the non-zero value \( e \).

Partial Derivatives

Partial derivatives are like the slices of a multivariable function, showing how the function changes with respect to one variable at a time while keeping the others constant. In the exercise, we look at two partial derivatives:

• With respect to \( x \), we have \( \frac{\partial h}{\partial x} = e^x \sin y \). This tells us how \( h \) changes as \( x \) changes, while \( y \) stays fixed.
• With respect to \( y \), \( \frac{\partial h}{\partial y} = e^x \cos y \). Similarly, this shows how \( h \) varies as \( y \) changes, holding \( x \) constant.

By evaluating these derivatives at a specific point \( P(1, \frac{\pi}{2}) \), we determine how the function behaves locally around that point. Thus, partial derivatives play a crucial role in constructing the gradient vector, which captures the overall directional change.

Unit Vector

A unit vector is a vector with a magnitude of 1. It serves as a direction indicator without affecting the scale or length of what it's applied to. When considering the direction for a directional derivative, the unit vector ensures that only the direction contributes, not any scaling.In the exercise, the directional vector \( \mathbf{v} = -\mathbf{i} = (-1, 0) \) is already a unit vector. The magnitude is calculated as:\[\| \mathbf{v} \| = \sqrt{(-1)^2 + 0^2} = 1.\]This means that \( \mathbf{v} \) requires no further adjustment or scaling. Because \( \mathbf{v} \) maintains a unit length, it allows us to compute the directional derivative accurately by multiplying the gradient vector \( abla h \) with \( \mathbf{v} \) through a dot product, focusing solely on direction.