Chapter 9: Problem 6
Find the gradient of the given function at the indicated point. $$ f(x, y)=\sqrt{x^{3} y-y^{4}} ;(3,2) $$
Short Answer
Expert verified
The gradient at (3,2) is \( \left( \frac{27}{\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \).
Step by step solution
01
Find the partial derivative with respect to x
To find the gradient, we first need to determine the partial derivative of the function with respect to x. Consider the function: \[ f(x, y) = \sqrt{x^3 y - y^4} \]The partial derivative with respect to x is:\[ \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^3 y - y^4}} \cdot \left(3x^2 y\right) \]
02
Find the partial derivative with respect to y
Next, we find the partial derivative of the function with respect to y. Using the same function:\[ f(x, y) = \sqrt{x^3 y - y^4} \]The partial derivative with respect to y is:\[ \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^3 y - y^4}} \cdot \left(x^3 - 4y^3\right) \]
03
Evaluate the partial derivatives at the given point
Now that we have both partial derivatives, we evaluate them at the point \((3, 2)\). For the partial derivative with respect to x:\[ \frac{\partial f}{\partial x}(3,2) = \frac{1}{2\sqrt{3^3 (2) - 2^4}} \cdot (3 \cdot 3^2 \cdot 2) = \frac{1}{2\sqrt{54 - 16}} \cdot 54 \]This simplifies to \[ \frac{1}{2\sqrt{38}} \cdot 54 \]For the partial derivative with respect to y:\[ \frac{\partial f}{\partial y}(3,2) = \frac{1}{2\sqrt{3^3 (2) - 2^4}} \cdot (3^3 - 4\cdot 2^3) = \frac{1}{2\sqrt{38}} \cdot (27 - 32) \]This simplifies to \[ \frac{1}{2\sqrt{38}} \cdot (-5) \]
04
Comprehend the gradient vector
The gradient vector \(abla f\) is comprised of these partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}(3,2), \frac{\partial f}{\partial y}(3,2) \right) \]Putting both values together:\[ abla f = \left( \frac{54}{2\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \]This can also be expressed as:\[ abla f = \frac{1}{2\sqrt{38}} (54, -5) \]
05
Simplify the gradient
To express the gradient in a more concise manner, we rewrite it using a common coefficient:\[ abla f = \left( \frac{27}{\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a key concept in multivariable calculus. They provide us with the rate of change of a function concerning one variable while keeping other variables constant. In the context of the original exercise, we have a function \( f(x, y) = \sqrt{x^3 y - y^4} \). To find the gradient at a specific point, we need to calculate its partial derivatives.
- The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), gives us the rate at which \( f \) changes as \( x \) changes, keeping \( y \) constant.
- Similarly, the partial derivative with respect to \( y \), denoted as \( \frac{\partial f}{\partial y} \), indicates how \( f \) changes as \( y \) varies while \( x \) remains constant.
Multivariable Calculus
Multivariable calculus expands the concepts of calculus into functions of several variables, like the one we encountered in this exercise. Unlike single-variable calculus dealing with curves, multivariable calculus helps analyze surfaces and higher dimensional analogs.In our case, the function \( f(x, y) = \sqrt{x^3 y - y^4} \) is a representation of a surface in three-dimensional space. Multivariable calculus allows us to explore many real-world phenomena, such as:
- Temperature variations across a geographical location.
- Pressure changes in different regions of the atmosphere.
- Market trends depending on various economic parameters.
Gradient Vector
The gradient vector, often symbolized as \( abla f \), is an essential tool when working with multivariable functions. It combines all the partial derivatives into a single entity, thereby giving a direction in which the function \( f \) increases most rapidly.In the original problem, after calculating the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), the gradient vector at point \((3, 2)\) is constructed as:\[abla f = \left( \frac{54}{2\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right)\]The gradient vector tells us:
- The direction in which increases in the input yield the steepest increase in \( f \).
- The magnitude of \( abla f \) signifies the steepness of this increase.