problem about a curve

[logistic_guy]

here is the question

A closed curve \(\displaystyle \bold{r}\) is given by \(\displaystyle \bold{r}(t) = (x(t), y(t))\) and \(\displaystyle x\cos t + y\sin t = p\) where \(\displaystyle p(t)\) is the distance from the origin to the tangent line at \(\displaystyle \bold{r}(t)\).

Show that:
(i) \(\displaystyle \bold{r}(t) = \left(p\cos t - \frac{dp}{dt}\sin t, p\sin t + \frac{dp}{dt}\cos t\right)\)
(ii) \(\displaystyle |\bold{r}'(t)| = \left|\frac{dp}{dt} + \frac{d^2p}{dt^2}\right|\)


if i take the derivative of \(\displaystyle p\), i don't get any thing useful

\(\displaystyle -x\sin t + y\cos t = \frac{dp}{dt}\)

 

[mario99]

if i take the derivative of \(\displaystyle p\), i don't get any thing useful

I don't understand the purpose of taking the derivative. But let us assume that you want to take the derivative. It is given in the vector that [imath]x[/imath] and [imath]y[/imath] are functions of [imath]t[/imath], but you treated them like constants when you took the derivative.

Do this and recalculate:

[imath]\displaystyle x(t)\cos t + y(t)\sin t = p(t)[/imath]

 

[logistic_guy]

I don't understand the purpose of taking the derivative. But let us assume that you want to take the derivative. It is given in the vector that [imath]x[/imath] and [imath]y[/imath] are functions of [imath]t[/imath], but you treated them like constants when you took the derivative.

Do this and recalculate:

[imath]\displaystyle x(t)\cos t + y(t)\sin t = p(t)[/imath]

thanks

Product Rule - Formula, Proof, Definition, Examples

Product rule in calculus is a method to find the derivative or differentiation of a function given in the form of a ratio or division of two differentiable functions. Understand the method using the product rule formula and derivations.

www.cuemath.com

i'm not good in complicate derivative. i need to use rule up, correc?

\(\displaystyle x'(t)\cos t + -x(t)\sin t + y'(t)\sin t + y(t)\cos t = p'(t)\)

now i'll convert to my notation

\(\displaystyle \frac{dx(t)}{dt}\cos t + -x(t)\sin t + \frac{dy(t)}{dt}\sin t + y(t)\cos t = \frac{dp(t)}{dt}\)

i'm still stuck. what i do next? do i have to proof the product rule because i'm using it?

 

[BigBeachBanana]

do i have to proof the product rule because i'm using it?

There seems to be a lack of knowledge in your foundation. It is concerning that you're doing vector calculus but asking about algebra and product rules (often introduced in introductory-level calculus). It's difficult to discuss higher-level topics while you're still struggling with basic ones. Are you following a school curriculum or are you studying this on your own?

 

[mario99]

i'm still stuck. what i do next? do i have to proof the product rule because i'm using it?

I think that you are a good student because I have told you before a long story about when to apply the product rule and now I can see that you have learnt from it. This tells me that you are reading what I am writing.

I will be honest with you, I still did not understand the OP. But it seems that you are correct that there is no way to get the parameters of the vector without playing around with derivatives. So far I don't see a way to go to the next step as we have two unknowns and one equation. I will try to work on it more and when I come up with something I will post. You also keep trying and when you get new ideas post them, so I can come up with a way to go to the next step.

And for proving the product rule, I don't think that it is necessary unless you are explicitly told to do so. Also, I have an advice for you which is related to post #4 written by BigBeachBanana. If you don't know the basics of mathematics, not only you will struggle, but we will also waste a precious time to make you understand [imath]1 + 1[/imath] while the main topic is vector calculus!

Note: There are no promises that I will be able to help you in this problem, so don't only depend on me!

 

[logistic_guy]

There seems to be a lack of knowledge in your foundation. It is concerning that you're doing vector calculus but asking about algebra and product rules (often introduced in introductory-level calculus). It's difficult to discuss higher-level topics while you're still struggling with basic ones. Are you following a school curriculum or are you studying this on your own?

course. engineer student 3rd year

i'm try to learn the basics. i read algebra everyday. you mean lack of algebra or calculus? i think calculus is easy. algebra is my problem

I think that you are a good student because I have told you before a long story about when to apply the product rule and now I can see that you have learnt from it. This tells me that you are reading what I am writing.

I will be honest with you, I still did not understand the OP. But it seems that you are correct that there is no way to get the parameters of the vector without playing around with derivatives. So far I don't see a way to go to the next step as we have two unknowns and one equation. I will try to work on it more and when I come up with something I will post. You also keep trying and when you get new ideas post them, so I can come up with a way to go to the next step.

And for proving the product rule, I don't think that it is necessary unless you are explicitly told to do so. Also, I have an advice for you which is related to post #4 written by BigBeachBanana. If you don't know the basics of mathematics, not only you will struggle, but we will also waste a precious time to make you understand [imath]1 + 1[/imath] while the main topic is vector calculus!

Note: There are no promises that I will be able to help you in this problem, so don't only depend on me!

thank for the advise. i'm studying basic everyday

i'm think to use \(\displaystyle x = \frac{p - y\sin t}{\cos t}\)

\(\displaystyle \frac{dx(t)}{dt}\cos t + -\left(\frac{p - y\sin t}{\cos t}\right)\sin t + \frac{dy(t)}{dt}\sin t + y(t)\cos t = \frac{dp(t)}{dt}\)

i need simplification for this. can you show me? i learn to proof the product rule if you want to see

 

[mario99]

thank for the advise. i'm studying basic everyday

i'm think to use \(\displaystyle x = \frac{p - y\sin t}{\cos t}\)

\(\displaystyle \frac{dx(t)}{dt}\cos t + -\left(\frac{p - y\sin t}{\cos t}\right)\sin t + \frac{dy(t)}{dt}\sin t + y(t)\cos t = \frac{dp(t)}{dt}\)

i need simplification for this. can you show me? i learn to proof the product rule if you want to see

Well that was a good way to start with, but unfortunately it will not help us to get rid of the derivatives to meet the required form. I am thinking of the chain rule. Why not to try it?

i learn to proof the product rule if you want to see

No need. Instead of that try to apply my idea that I have given you above.

 

[blamocur]

I don't understand the purpose of taking the derivative. But let us assume that you want to take the derivative. It is given in the vector that xxx and yyy are functions of ttt, but you treated them like constants when you took the derivative.

Taking derivative of [imath]p[/imath] can be helpful, especially if following your advice and doing it correctly. One can plug the expression for [imath]p^\prime[/imath] into the right hand side of (i) while taking into account the fact that [imath]x^\prime \cos t + y^\prime\sin t = 0[/imath].

 

[mario99]

Taking derivative of [imath]p[/imath] can be helpful, especially if following your advice and doing it correctly. One can plug the expression for [imath]p^\prime[/imath] into the right hand side of (i) while taking into account the fact that [imath]x^\prime \cos t + y^\prime\sin t = 0[/imath].

Thank you professor blamocur for passing by. This is exactly what I have arrived to. The OP claimed that calculus is easy, so let's hope that he understands the chain rule!

 

[logistic_guy]

yes i know the chain rule. which part of the question i do chain rule?

 

[mario99]

yes i know the chain rule. which part of the question i do chain rule?

I don't want to make your life difficult, so I will tell you exactly what to do.

[imath]\displaystyle \frac{dy}{dt} = \ ?[/imath]

Apply the chain rule to the derivative above. What do you get?

 

[blamocur]

Some closing notes at the end of one week. Distance [imath]p[/imath] to a line can be expressed as [math]p = x n_x + y n_y[/math] where [imath](x,y)[/imath] is any point on the line and [imath](n_x,n_y[/imath]) is the line's normal. This means that [imath](\cos t, \sin t)[/imath] is normal to the line and thus [imath]x'\cos t + y' \sin t = 0[/imath].
[math]p' = x'\cos t - x\sin t + y' \sin t + y\cos t[/math]Now look at the first coordinate of the right hand side of (i):
[math]p\cos t - p'\sin t = x\cos^2 t + y\sin t \cos t - \sin t (x'\cos t - x\sin t + y' \sin t + y\cos t) =[/math][math]= x\cos^2 t + y\sin t \cos t - (x'\sin t\cos t - x\sin^2 t + y' \sin^2 t + y\sin t\cos t) =[/math][math]= x - \sin t (x'\cos t + y' \sin t) = x[/math]A similar procedure shows that the second coordinate is equal to [imath]y[/imath] thus proving part (i).