Problems based on partial derivatives

Problems based on Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus. They allow us to find the rate of change of a function with respect to one variable while holding all other variables constant. In this article, we will explore problems based on partial derivatives and learn how to solve them.

Table of Contents

  1. Introduction to Partial Derivatives
  2. Notation for Partial Derivatives
  3. Solving Problems based on Partial Derivatives
  4. Examples of Problems based on Partial Derivatives
  5. Conclusion

1. Introduction to Partial Derivatives

In calculus, a partial derivative measures the rate at which a function changes with respect to one of its variables while keeping all other variables constant. It allows us to analyze how a function behaves in different directions.

For a function of two variables, say (f(x, y)), we can find the partial derivative of (f) with respect to (x) by treating (y) as a constant and differentiating (f) with respect to (x). Similarly, we can find the partial derivative of (f) with respect to (y) by treating (x) as a constant and differentiating (f) with respect to (y).

2. Notation for Partial Derivatives

The partial derivative of a function (f) with respect to (x) is denoted by (\frac{{\partial f}}{{\partial x}}) or (f_x). Similarly, the partial derivative of (f) with respect to (y) is denoted by (\frac{{\partial f}}{{\partial y}}) or (f_y).

3. Solving Problems based on Partial Derivatives

To solve problems based on partial derivatives, we follow these steps:

  1. Identify the function for which we need to find the partial derivatives.
  2. Determine the variables with respect to which we need to differentiate.
  3. Apply the rules of differentiation to find the partial derivatives.
  4. Simplify the expressions obtained after differentiation if possible.
  5. Interpret the results in the context of the problem.

4. Examples of Problems based on Partial Derivatives

Let's consider a few examples to understand how to solve problems based on partial derivatives.

Example 1:

Find the partial derivatives of the function (f(x, y) = 3x^2y + \sin(xy)) with respect to (x) and (y).

Solution:

To find the partial derivative of (f) with respect to (x), we treat (y) as a constant and differentiate (f) with respect to (x).

(\frac{{\partial f}}{{\partial x}} = \frac{{\partial}}{{\partial x}}(3x^2y + \sin(xy)))

Using the power rule of differentiation, we differentiate each term separately:

(\frac{{\partial}}{{\partial x}}(3x^2y) = 6xy)

(\frac{{\partial}}{{\partial x}}(\sin(xy)) = y\cos(xy))

Therefore, (\frac{{\partial f}}{{\partial x}} = 6xy + y\cos(xy))

To find the partial derivative of (f) with respect to (y), we treat (x) as a constant and differentiate (f) with respect to (y).

(\frac{{\partial f}}{{\partial y}} = \frac{{\partial}}{{\partial y}}(3x^2y + \sin(xy)))

Using the power rule of differentiation, we differentiate each term separately:

(\frac{{\partial}}{{\partial y}}(3x^2y) = 3x^2)

(\frac{{\partial}}{{\partial y}}(\sin(xy)) = x\cos(xy))

Therefore, (\frac{{\partial f}}{{\partial y}} = 3x^2 + x\cos(xy))

Example 2:

A rectangular box has a volume given by (V(x, y, z) = xyz) and a surface area given by (S(x, y, z) = 2(xy + yz + xz)). Find the rate at which the volume is changing with respect to the surface area when (x = 2), (y = 3), and (z = 4).

Solution:

We need to find the rate at which the volume is changing with respect to the surface area, which can be expressed as (\frac{{dV}}{{dS}}).

To find this rate, we need to find the partial derivatives of (V) and (S) with respect to (x), (y), and (z).

(\frac{{\partial V}}{{\partial x}} = yz)

(\frac{{\partial V}}{{\partial y}} = xz)

(\frac{{\partial V}}{{\partial z}} = xy)

(\frac{{\partial S}}{{\partial x}} = 2(y + z))

(\frac{{\partial S}}{{\partial y}} = 2(x + z))

(\frac{{\partial S}}{{\partial z}} = 2(x + y))

Substituting the given values (x = 2), (y = 3), and (z = 4) into the partial derivatives, we get:

(\frac{{\partial V}}{{\partial x}} = 3 \times 4 = 12)

(\frac{{\partial V}}{{\partial y}} = 2 \times 4 = 8)

(\frac{{\partial V}}{{\partial z}} = 2 \times 3 = 6)

(\frac{{\partial S}}{{\partial x}} = 2(3 + 4) = 14)

(\frac{{\partial S}}{{\partial y}} = 2(2 + 4) = 12)

(\frac{{\partial S}}{{\partial z}} = 2(2 + 3) = 10)

Finally, we can calculate the rate at which the volume is changing with respect to the surface area:

(\frac{{dV}}{{dS}} = \frac{{\frac{{\partial V}}{{\partial x}}}}{{\frac{{\partial S}}{{\partial x}}}} = \frac{{12}}{{14}})

Therefore, the rate at which the volume is changing with respect to the surface area is (\frac{{12}}{{14}}).

5. Conclusion

Problems based on partial derivatives are an important aspect of multivariable calculus. They allow us to analyze how a function changes with respect to one variable while holding all other variables constant. By following the steps outlined in this article and practicing with examples, you can become proficient in solving problems based on partial derivatives.