Why does the speed remain the same?

[lioric]

You learned the dot product (scalar product) of two vectors, and you should know that the dot product of a vector with itself is the square of the magnitude of the vector. So $\vec b \cdot \vec b = |\vec b|^2$, or you can write it in the form $a=\sqrt {{\vec b} ^2}$, $b$ meaning the magnitude of $\vec b$.

You have the time dependent velocity vector $\vec v(t)$ . The speed is its magnitude, $v(t)= \sqrt {{\vec v(t)} ^2}$
v(t) is just a function of $\vec v$, and $\vec v$ is a function of t, Determine dv/dt, using the chain rule. What do you get?
It might be confusing that $\vec v$ is vector. How would you differentiate a function F(t)=f(g^2(t)) with respect to time?

Df/dt right?
It would be g^2 since t has no power and the differentiation is with respect to t I don't need to do anything with g

 

[ehild]

Do you know the chain rule? What is the derivative of $f(t) = \sqrt{sin(t)}$ with respect to t, for example?

 

[lioric]

Do you know the chain rule? What is the derivative of $f(t) = \sqrt{sin(t)}$ with respect to t, for example?

Sorry my bad

Is this ok?

 

[Mark44]

Sorry my bad

View attachment 97676
Is this ok?

No. Somehow you ended up with the correct derivative, but the work you show is wrong.
In the second line of your attached image, you have
$$\frac{df(t)}{dt} = (\sin t)^{1/2} \cdot \sin t$$
This is wrong. On the right side, you haven't taken the derivative yet -- $(\sin t)^{1/2}$ is the same as $\sqrt{\sin t}$. Also, where did the extra factor of $\sin(t)$ come from?
The third line is correct, and so is the last line.

 

[ehild]

View attachment 97676
Is this ok?

The end is OK. Now you have the function $f(t)= {\vec v(t)} ^2$. What is df/dt? Do not worry about that the inner function is a vector, just apply the chain rule.

 

[lioric]

The end is OK. Now you have the function $f(t)= {\vec v(t)} ^2$. What is df/dt? Do not worry about that the inner function is a vector, just apply the chain rule.

Do/dt=2v(t)

 

[ehild]

Do/dt=2v(t)

No, you have to differentiate the inner function, too. If v(t)= sin(t), for example, what is the derivative of f(t)=(sin(t))² ?

 

[lioric]

No, you have to differentiate the inner function, too. If v(t)= sin(t), for example, what is the derivative of f(t)=(sin(t))² ?

2(sin (t))cos t

 

[ehild]

2(sin (t))cos t

Good. cos(t) is the derivative of the inner function. So how do you write the derivative of f(t)=(V(t))² in general?

 

[lioric]

Good. cos(t) is the derivative of the inner function. So how do you write the derivative of f(t)=(V(t))² in general?

2(V(t))V

 

[Orodruin]

2(V(t))V

No, you did not take the derivative for the inner derivative.

I am sorry to say so, but your attempts here indicate to me that you are not mathematically prepared to take the physics class you are in. A word of advice would therefore be to go back to your single and multi variable calculus before trying to go further. Continuing without the prerequisite math is only going to come back to bite you in the end.

 

[lioric]

No, you did not take the derivative for the inner derivative.

I am sorry to say so, but your attempts here indicate to me that you are not mathematically prepared to take the physics class you are in. A word of advice would therefore be to go back to your single and multi variable calculus before trying to go further. Continuing without the prerequisite math is only going to come back to bite you in the end.

I know you guys have tried very hard to help me. I'm sorry for being such a pain
And I know there is no escape from maths. I just have hard time understanding the question.
Thank you for all you have done I'll try to get by some other way.

 

[Orodruin]

I just have hard time understanding the question.

This here is exactly why I recommend you to revisit your calculus. If you do not even understand the question, how can you hope to understand the answer? There is no shame in going back to repeat the basics, you will need the basics to build on that knowledge.