Is This Answer Correct? | Exploring Accuracy in Question Responses

[ertagon2]

Is the answer to this question correct?
View attachment 7783

 

[MarkFL]

Well, let's see...let:

\(\displaystyle u=1-\cos(3x)\implies du=3\sin(3x)\,dx\)

So, the integral becomes:

\(\displaystyle I=\frac{1}{3}\int_0^1 u\,du=\frac{1}{6}(1^2-0^2)=\frac{1}{6}\quad\checkmark\)

 

[tkhunny]

Is the answer to this question correct?

Or, $(1-\cos(3x))\cdot\sin(3x) = \sin(3x) - \cos(3x)\cdot\sin(3x) = \sin(3x) - \dfrac{1}{2}\sin(6x)$

Keep all your options open. Don't get stuck in just one way of thinking.

 

[ertagon2]

Well, let's see...let:

\(\displaystyle u=1-\cos(3x)\implies du=3\sin(3x)\,dx\)

So, the integral becomes:

\(\displaystyle I=\frac{1}{3}\int_0^1 u\,du=\frac{1}{6}(1^2-0^2)=\frac{1}{6}\quad\checkmark\)

Side question how is this [M]du=3sin(3x)dx[/M]allowed ? I thought that [M]du/dx is[/M] more of a notation than a fraction.

 

[tkhunny]

Side question how is this [M]du=3sin(3x)dx[/M]allowed ? I thought that [M]du/dx is[/M] more of a notation than a fraction.

You can think that if you like. However, it makes little or no difference whether we see the fraction version or the differential version - so long as the context is clear. The goal is to be clear and comprehensible. Notation is meant to express clearly what is wanted and to assist the communicator, not to become excessively cumbersome. Keep in mind, though that it may not be a good idea to invent new notation. There is already way too much. :-)

 

[tkhunny]

Perhaps we can get some professional educators on here to give us a clue how tricky this is.

Q for Professional Educator: When a beginning calculus student first encounters the differential concept, is this usually an easy transition?

Learning the Derivative
If y = 2x^2, then dy/dx = 4x. I see, so dy/dx means the derivative of a function y, with respect to x.

Learning Linearization or Linear Differentials
If y = 2x^2, then dy = 4x*dx. Wait, can you do that?

 

[Ackbach]

Perhaps we can get some professional educators on here to give us a clue how tricky this is.

Q for Professional Educator: When a beginning calculus student first encounters the differential concept, is this usually an easy transition?

Learning the Derivative
If y = 2x^2, then dy/dx = 4x. I see, so dy/dx means the derivative of a function y, with respect to x.

Learning Linearization or Linear Differentials
If y = 2x^2, then dy = 4x*dx. Wait, can you do that?

I was a professional educator. In my experience, the differential is not terribly difficult to use, primarily because it's straight-forward to remember. The identity $dy=\dfrac{dy}{dx} \, dx$ is easy because it LOOKS like you're canceling, even though you're not. To really UNDERSTAND requires more work, and that's not straight-forward. You have to understand limits, which is easily the most difficult concept in all of calculus. That's my opinion.