# Weird Dot Product Homework Q: Unusual Answer?

• wrong_class
In summary, the conversation discusses finding the line integral of a vector function using two different methods. The first method involves taking the dot product of the vector function and its derivative, while the second method involves taking the magnitude of both the vector function and its derivative. The two methods yield different results, with the second method being correct. There is also a question about finding the line integral when the function returns a scalar or a vector, with the suggestion of using the gradient and dot product.
wrong_class

## Homework Statement

i have the vector m(x(t), y(t)) = r = (s^t cos(t), e^t sin(t)) and want to find the line integral of it

## Homework Equations

1. $$\int m \centerdot r' dt$$
2. $$\int |m| |r'| dt$$

## The Attempt at a Solution

the answer is sqrt(2)/2 (e^pi - 1). when i do the problem the first way, i do not get the sqrt(2)
When i do the problem the first way, the answer is wrong, but when i do it the second way, it is correct.

is the first way even correct? i am told it is, but wolfram alpha is agreeing with me

Last edited:
what are m and r, what are the answers you are getting?

The two integrals are different, why should you get the same result ?

I did the first, assuming m is the vector
$$m = e^t cos\textsl{t}\ \vec{i}+e^t sin\textsl{t}\ \vec{j}$$

$$\int_{0}^{\pi \over 2} m\cdot m' dt = {e^\pi \over 2}- {1 \over 2 }$$

One the second one it is likely a sqrt to pop up.

so the integrals are different. So why did someone who learned this already tell me these two are the same?

thats the answer I am getting.

thanks!

can someone clarify this for me?:

given r(t) is a vector, how do you find the line integral if f(x(t),(y(t)) returns a scalar? A vector? do you always get 2 scalars (magnitude of f and magnitude of r' ) and multiply them? do you take the gradient if f is scalar and then dot deL_f and r' ?

## 1. What is the dot product?

The dot product is a mathematical operation that takes two vectors and returns a scalar value. It is calculated by multiplying the corresponding components of the vectors and then adding the products together.

## 2. How is the dot product different from the cross product?

The dot product returns a scalar value while the cross product returns a vector. Additionally, the dot product is commutative while the cross product is not. The dot product is also used to find the angle between two vectors, while the cross product is used to find a vector that is perpendicular to both input vectors.

## 3. What makes this dot product homework "weird"?

This homework is considered "weird" because it presents an unusual or unexpected question or answer related to the dot product. It may require a different approach or unconventional thinking to solve.

## 4. How can I approach solving this homework?

The best way to approach solving this homework is to first understand the concept of the dot product and how it is calculated. Then, carefully read the question and think critically about the information given. It may also be helpful to break down the problem into smaller parts and use diagrams to visualize the vectors.

## 5. Can I apply the dot product in real-life situations?

Yes, the dot product has many real-life applications, such as calculating work and power in physics, finding the angle between two objects, and even in computer graphics for lighting and shading calculations.

• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
441
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
630
• Calculus and Beyond Homework Help
Replies
9
Views
890
• Calculus and Beyond Homework Help
Replies
2
Views
3K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
506
• Calculus and Beyond Homework Help
Replies
1
Views
1K