Question

Suppose \(f(x, y)=4 x^{2}+3 y^{2}\) a. Calculate the partial derivatives of \(f\) b. Suppose \(f(x, y)=16 .\) Use the implicit function theorem to calculate \(d y / d x\) c. What is the value of \(d y / d x\) if \(x=1, y=2 ?\) d. Graph your results and use it to interpret the results in parts (b) and (c) of this problem.

Short answer

#Answer# (a) The partial derivatives of f with respect to x and y are \(\frac{\partial f}{\partial x} = 8x\) and \(\frac{\partial f}{\partial y} = 6y\). (b) The derivative of y with respect to x when \(f(x, y) = 16\) is \(\frac{dy}{dx} = -\frac{4x}{3y}\). (c) The value of \(\frac{dy}{dx}\) when \(x = 1\) and \(y = 2\) is \(-\frac{2}{3}\). (d) The graph displays an ellipse described by \(4x^2 + 3y^2 = 16\) and a tangent line at point (1, 2) with a slope of -2/3. It shows that as x increases, y decreases, and the tangent line represents the rate at which the function f(x, y) = 16 changes in the x-direction at point (1, 2).

Step-by-step solution

(a) Compute the partial derivatives of f with respect to x and y

To calculate the partial derivatives, we'll find the derivative of the function concerning x and y, holding other variables constant: \(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (4x^2 + 3y^2) = 8x\) \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (4x^2 + 3y^2) = 6y\)

(b) Use the implicit function theorem to compute dy/dx

We are given that \(f(x, y) = 16\), so \(4x^2 + 3y^2 = 16\). We can use the implicit function theorem to calculate \(\frac{dy}{dx}\) by finding the ratio between partial derivatives: \(\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = -\frac{8x}{6y} = -\frac{4x}{3y}\)

(c) Calculate the value of dy/dx when x=1 and y=2

To find the requested value, plug in the values \(x = 1\) and \(y = 2\) into our equation for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{4(1)}{3(2)} = -\frac{4}{6} = -\frac{2}{3}\)

(d) Graph your results and use it to interpret the results in parts (b) and (c)

To graph the ellipse described by \(4x^2 + 3y^2 = 16\) and the tangent line at the point \((1, 2)\), we can draw the ellipse and the tangent line on a coordinate plane, then make the following observations: 1. The tangent line at point \((1, 2)\) has a slope of \(-\frac{2}{3}\), which was calculated in part (c). 2. The tangent line represents the rate at which the function \(f(x, y) = 16\) is changing in the x-direction at the point \((1, 2)\). 3. The negative slope indicates that as x increases, y decreases. The graph would show the ellipse originating from the function \(f(x, y) = 16\) and the tangent line at point \((1, 2)\) with a slope of \(-\frac{2}{3}\). This visual representation can help to interpret the derivative calculated in part (c) and further understand the relationship between x and y.

Key concepts

Partial Derivatives

Partial derivatives play a crucial role in understanding how a function behaves with respect to each of its variables. Let's take the function f(x, y) = 4x^2 + 3y^2 from the exercise. The partial derivative with respect to x is found by treating y as a constant and differentiating f with respect to x. Conversely, the partial derivative with respect to y is calculated by treating x as constant and differentiating with respect to y.

In our case, the partial derivatives are \(\frac{\partial f}{\partial x} = 8x\) and \(\frac{\partial f}{\partial y} = 6y\). Understanding these derivatives is essential as they indicate the rate of change of the function along the axes of the respective variables. For instance, the derivative \(\frac{\partial f}{\partial x}\) tells us how f changes in the x-direction for a small change in x, holding y constant.

dy/dx Calculation

When dealing with functions involving two variables, such as f(x, y), we often need to understand how these variables are interconnected. The implicit function theorem is a powerful tool that comes in handy, particularly in situations where y is defined implicitly by an equation in terms of x. It provides us with a method to find \(\frac{dy}{dx}\) without the need to solve y explicitly.

The theorem states that under certain conditions, if we have an equation f(x, y) = c, where c is a constant, we can view y as an implicit function of x and determine \(\frac{dy}{dx}\) as -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}. For the given exercise, \(\frac{dy}{dx} = -\frac{8x}{6y}\) simplifies to -\frac{4x}{3y}. This gives us a formula to compute the rate at which y changes with respect to x at any point on the curve f(x, y) = 16.

Graphical Interpretation

Visualizing mathematical concepts can enhance understanding. The function f(x, y) = 4x^2 + 3y^2 forms an ellipse when set equal to a constant, such as 16. Graphing this equation provides a visual representation of the problem.

By plotting the level curve 4x^2 + 3y^2 = 16, we see the ellipse on the coordinate plane. The point (1, 2) is on this ellipse, and the calculated \(\frac{dy}{dx} = -\frac{2}{3}\) at this point signifies the slope of the tangent to the ellipse at that point. Graphically, the negative slope indicates that as we move to the right along the x-axis (x increasing), we move down along the y-axis (y decreasing). This reinforces the derivative's conceptual meaning as the rate of change of y relative to x, and visually shows the direction and steepness of that change at the point of tangency.