Question

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 \(P\). $$f(x, y)=x^{4}-x^{2} y+y^{2}+6 ; P(-1,1)$$

Short answer

a. The unit vectors giving the direction of steepest ascent and steepest descent at point \(P(-1, 1)\) are: - Steepest ascent: \(\hat{u}_{ascend} = \left\langle \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right\rangle\) - Steepest descent: \(\hat{u}_{descend} = \left\langle \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right\rangle\) b. A vector pointing in a direction of no change in the function at point \(P\) is: \(\hat{u}_{no\_change} = \langle 3, 2 \rangle\).

Step-by-step solution

Calculate the gradient of the function

Calculate the partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\). $$\frac{\partial f}{\partial x} = 4x^3 - 2xy$$ $$\frac{\partial f}{\partial y} = -x^2 + 2y$$

Evaluate the gradient at point P

Evaluate the partial derivatives at the point \(P(-1, 1)\): $$\left. \frac{\partial f}{\partial x}\right|_{(-1, 1)} = (4)(-1)^3 - 2(-1)(1) = -2$$ $$\left. \frac{\partial f}{\partial y}\right|_{(-1, 1)} = (-1)^2 + 2(1) = 3$$ So, the gradient at point \(P\) is: \(\nabla f = \langle -2, 3 \rangle\).

Find the unit vectors for steepest ascent and descent

Calculate the magnitude of the gradient: $$||\nabla f|| = \sqrt{(-2)^2 + (3)^2} = \sqrt{13}$$ Normalize the gradient to find the unit vector for the direction of steepest ascent, which will be in the same direction as the gradient: $$\hat{u}_{ascend} = \frac{\nabla f}{||\nabla f||} = \left\langle \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right\rangle$$ For steepest descent, we take the negative of the gradient's unit vector: $$\hat{u}_{descend} = -\hat{u}_{ascend} = \left\langle \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right\rangle$$

Find a vector for the direction of no change

To find a vector pointing in a direction of no change in the function at \(P\), we need to find a vector orthogonal to the gradient, which means their dot product will be zero: $$\hat{u}_{no\_change} \cdot \nabla f = 0$$ Choose a vector with components \(\langle a, b \rangle\), and we have: $$\langle a, b \rangle \cdot \langle -2, 3 \rangle = -2a + 3b = 0$$ There could be multiple vectors that satisfy this condition. Keeping \(a = 3\), we find \(b = 2\). Therefore, a vector for the direction of no change, which is orthogonal to the gradient is: \(\hat{u}_{no\_change} = \langle 3, 2 \rangle\). #Summary# - The unit vector for the direction of steepest ascent at point \(P\) is: \(\hat{u}_{ascend} = \left\langle \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right\rangle\). - The unit vector for the direction of steepest descent at point \(P\) is: \(\hat{u}_{descend} = \left\langle \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right\rangle\). - A vector pointing in a direction of no change in the function at point \(P\) is: \(\hat{u}_{no\_change} = \langle 3, 2 \rangle\).

Key concepts

Steepest Ascent

The steepest ascent in the context of multivariable calculus refers to the direction where a function increases most rapidly at a given point. This direction aligns with the gradient vector. The gradient vector is composed of partial derivatives of the function with respect to each variable. For our given function, \[f(x, y) = x^{4} - x^{2}y + y^{2} + 6,\] we determined that the gradient at point \(P(-1, 1)\) is \(abla f = \langle -2, 3 \rangle\).
To find the unit vector for the steepest ascent, normalize the gradient by dividing it by its magnitude:

• Gradient magnitude: \(||abla f|| = \sqrt{13}\).
• Unit vector for ascent: \(\hat{u}_{ascend} = \left\langle \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right\rangle\).

The unit vector essentially gives the direction of steepest ascent without scaling the length of the vector, maintaining its direction verified by its unit magnitude.

Steepest Descent

The steepest descent is the direction where the function decreases most rapidly, moving in exactly the opposite direction of the gradient vector. For problems involving optimization, identifying this direction can be crucial, particularly in iterative methods like gradient descent.
In our problem, once we have determined the unit vector for steepest ascent as \(\hat{u}_{ascend} = \left\langle \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right\rangle\), the direction of steepest descent is simple to find. It is the negative of this unit vector:

• Unit vector for descent: \(\hat{u}_{descend} = \left\langle \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right\rangle\).

This inverse nature reflects how steepest descent heads toward decreasing values of the function, just as steep literal descents lead downward. By adhering strictly to this direction, the fastest decrease from the point \(P(-1, 1)\) occurs.

Directional Derivative

The directional derivative represents how a function changes as one moves in a specified direction away from a point, serving as a vital concept in understanding functions with multiple variables. It can be explored through vectors perpendicular or orthogonal to the gradient vector, leading to zero change or no change in the function's value.
In mathematical terms, if \(\hat{u}\) is a unit vector, the directional derivative of a function \(f\) at certain point \(P\) in the direction of \(\hat{u}\) is given by the dot product \(abla f \cdot \hat{u}\). If the dot product is zero, the direction denotes no change:

• For our function, the gradient is \(abla f = \langle -2, 3 \rangle\).
• Finding orthogonal vector: \( \hat{u}_{no\_change} = \langle 3, 2 \rangle\).

Here, \(\langle 3, 2 \rangle\) forms a vector orthogonal to the gradient, ensuring motion in this direction causes no local change in \(f(x, y)\) at the given point \(-1, 1\). Such directions are crucial in optimization and analysis tasks, simplifying multidimensional evaluations.