Problem 9 - Calculus with Concepts in Calculus

Question

Find the gradient of the function at the given point. $$ f(x, y)=\frac{x+3 y}{5 x+2 y} ;\left(-1, \frac{3}{2}\right) $$

Step-by-Step Solution

Answer

Gradient at \((-1, \frac{3}{2})\) is \((-4.875, -3.25)\).

Step 1: Understand the gradient concept

The gradient of a function \( f(x, y) \) is a vector of its partial derivatives, denoted as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). It represents the direction of the steepest ascent of the function.

Step 2: Calculate the partial derivative with respect to x

The function is \( f(x, y)=\frac{x+3y}{5x+2y} \). To find \( \frac{\partial f}{\partial x} \), use the quotient rule: if \( f(x) = \frac{u(x)}{v(x)} \), then \( f' = \frac{u'v - uv'}{v^2} \). Here, \( u(x, y) = x + 3y \) and \( v(x, y) = 5x + 2y \). Compute the derivative with respect to \( x \):- \( u_x = 1 \)- \( v_x = 5 \)Apply the quotient rule:\[ \frac{\partial f}{\partial x} = \frac{(1)(5x+2y) - (x+3y)(5)}{(5x+2y)^2} = \frac{5x + 2y - 5x - 15y}{(5x+2y)^2} = \frac{2y - 15y}{(5x+2y)^2} = \frac{-13y}{(5x+2y)^2} \]

Step 3: Calculate the partial derivative with respect to y

For the same function \( f(x, y) \), now compute \( \frac{\partial f}{\partial y} \) using the quotient rule:- \( u_y = 3 \)- \( v_y = 2 \)Therefore,\[ \frac{\partial f}{\partial y} = \frac{(3)(5x+2y) - (x+3y)(2)}{(5x+2y)^2} = \frac{15x + 6y - 2x - 6y}{(5x+2y)^2} = \frac{13x}{(5x+2y)^2} \]

Step 4: Evaluate the gradient at the point

Now substitute the point \( (-1, \frac{3}{2}) \) into the gradient vector \( abla f(x, y) = \left( \frac{-13y}{(5x+2y)^2}, \frac{13x}{(5x+2y)^2} \right) \):- Substitute into \( \frac{-13y}{(5x+2y)^2} \): \[ \frac{-13(\frac{3}{2})}{(5(-1) + 2(\frac{3}{2}))^2} = \frac{-19.5}{(-5+3)^2} = \frac{-19.5}{4} = -4.875 \]- Substitute into \( \frac{13x}{(5x+2y)^2} \): \[ \frac{13(-1)}{(5(-1) + 2(\frac{3}{2}))^2} = \frac{-13}{(-5+3)^2} = \frac{-13}{4} = -3.25 \]

Step 5: Write the final gradient vector

The gradient vector at the point \((-1, \frac{3}{2})\) is:\[ abla f(-1, \frac{3}{2}) = (-4.875, -3.25) \]

Key Concepts

Partial Derivatives

Partial derivatives are fundamental to understanding functions with multiple variables, like those you encounter in multivariable calculus. In this context, a partial derivative of a function is the derivative with respect to one variable while keeping the other variables constant. For example, if you have a function \(f(x, y)\), the partial derivative with respect to \(x\), written as \(\frac{\partial f}{\partial x}\), examines how the function changes as \(x\) changes, holding \(y\) constant. Similarly, the partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), examines the function's response to changes in \(y\) alone.

• This approach helps isolate the effect each variable has on the function.
• Partial derivatives are used to find the gradient of a function, which is essential in optimization problems.

By understanding partial derivatives, you can dissect complex functions and understand the influence of each variable independently.

Quotient Rule in Calculus

The quotient rule is a vital tool when dealing with derivatives of functions that are ratios of two differentiable functions. In calculus, if a function \(f(x)\) is expressed as a ratio \(\frac{u(x)}{v(x)}\), where both \(u(x)\) and \(v(x)\) are differentiable, the quotient rule helps compute its derivative. The formula is: \[ f'(x) = \frac{u'v - uv'}{v^2} \]This means you differentiate \(u(x)\) and \(v(x)\) separately, and then apply the formula above. When applying it to multivariable functions like \(f(x, y) = \frac{x+3y}{5x+2y}\), it requires identifying both \(u\) and \(v\), and their partial derivatives with respect to each variable.

• For instance, if \(u(x, y) = x + 3y\) and \(v(x, y) = 5x + 2y\), the quotient rule requires differentiation regarding both \(x\) and \(y\).
• Applying these derivatives enables you to calculate \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

The quotient rule is indispensable for functions expressed as quotients, facilitating precise calculation of their derivatives.

Evaluating at a Point

Evaluating a function or its derivatives at a specific point gives us concrete values that reveal the function's behavior at that exact location in its domain. In the given exercise, once partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) are found, they need to be evaluated at the point \((-1, \frac{3}{2})\). This involves plugging the point's coordinates into the expressions for the partial derivatives.

• For instance, substituting \((-1, \frac{3}{2})\) into \(\frac{-13y}{(5x+2y)^2}\), and \(\frac{13x}{(5x+2y)^2}\), results in exact numerical values.
• This step is critical to find the gradient at a specific point, which can indicate the direction and rate of steepest ascent or descent from that point.

Evaluating derivatives at a point translates theoretical expressions into actionable information, aiding in deeper understanding and practical application of the function's behaviors.