Tag: pre calculus

Implicit Differentiation in 6 Steps

Sometimes in calculus, equations can’t be solved easily since x and y don’t explicitly define y as a function x.

In such cases, we imply that a function y= f(x)to satisfy the given equation. In short words this is implicit differentiation, giving you the ability to find the derivative of y with respect to x without solving the given equation for y.

Let’s look at an example problem solved that implements this concept.

Solved: Implicit Differentiation Problem

This problem has a few stages which we have to consider. First thing we have to do is perform implicit differentiation to determine the following using the given equation:

After that we’ll determine the value of at the given point mentioned above which will give us the slope of the tangent line.

From there, we can determine the equation of the tangent line and then determine the equation of the normal line. So let’s break it down each part of the problem step by step:

1)Let’s start by differentiating both sides of the equation with respect to x. If we have a function of y, we have to apply the chain rule and we’ll have an extra factor of the following:

Differentiating both sides with respect to x:

Now we need to apply the product rule:

This is the derivative of the first function times the derivative of the second function.

Now differentiating the second function:

3) Now let’s solve this equation for dy/dx. Since the dy/dx terms are both negative, we can move them to the right of the equation.

4) We can now factor out dy/dx on the right:

Taking cos(y) as common factor in both the sides,

Eliminating cosine terms in both the sides,

Now we will write the equation for y¹

5) Let’s find the slope of the tangent line at the point (0,2π) To do this, we need to evaluate y’at the point. In this case we need to substitute 0 for x and 2π for y and simplify:

Now we know the slope of the tangent line is m=0. This also indicates that the tangent line is y=2π.

*We can also simplify to get the tangent line y=2π and m=0 using point-slope form:

6) If the tangent line is a horizontal line and we know that the regular line is a vertical line. The vertical line passing through (0,22π) must be x=0.

Therefore our final answer is:

Tangent line: y=2π

Normal line: x=0

We can also look at it graphically:

Conclusion

This was one example of how you can apply implicit differentiation in calculus. If you’re struggling with it- keep practicing. Or better yet, take a look at a high quality survival guide for your textbook along with on demand help with solutions.

May 2, 2022
Applications of Integrals in Mathematics
Applications of integrals can be tricky- it can be applied to math, science, and engineering, economics etc. If you don’t have the right teacher or textbook survival guide, getting lost before exam can be easy.

In this guide you’re going to learn how integration is applied in various scenarios, the different types of integrals, and some examples.

Applications of Integrals: What does it mean?

Integrals can be applied in many different areas like math, science, and engineering, statistics,and economics.

In math, we use it to find:
- The center of mass (Centroid) of an area with curved sides
- The area between two curves
- The area under a curve
- The average value of a curve
- Integral transforms like Laplace and Fourier transforms.
When speaking about integrals, there are two different types: definite integrals (also called Riemann Integral) and indefinite integrals.

In simple terms, a definite integral has definite limits ( defined upper and lower limits/bounds) which can be represented by:

In contrast to definite integrals, an indefinite integral is an integral which has no limits. This can be represented by:

                                               (C is the constant of integration)

Solved: Example Problem of Applications of Integrals

The given function is a type of parabola, the graph is as shown below:

First, take a look at our notes below:

We will integrate the given function to get . We know that using which we will compute the value of the constant of integration. Once we know the value of constant of integration, the next step is to find f(4).

1) Knowing all that, let’s integrate .To do that, we need to apply the power rule of integration which would be:

2) Using the power rule,

                                                  ……………(1)

Thus we got the result of integration.

3) Now we will find the value of the constant of integration,

We know that , hence applying this value to equation(1) we get











Therefore we found the value of the constant of integration which is

(4) Now we will apply this value of integration constant to

Conclusion

This is just one example of many different scenarios. We have to remember in math alone there are numerous applications and in physics even more! Applications of integrals is a topic that takes practice with the right survival guides and experts to help you with solutions– you’ll be on the right track.
April 20, 2022