ambigus question

[logistic_guy]

here is the question

A circle has its center at the point $\displaystyle (0,1)$ is placed there stationary. Then it was given a push and it moved one full revolution without slipping. An observer traced the point $\displaystyle (0,0)$ on the circle and noticed that it made a cycloid during the full revolution. What is the length of this cycloid?

i say the question is ambigus because it don't give me any formula of curve to calculate. i know the formula of how to calcate the length of a curve and the only inoformation i was given the circle. i'll try to permerterize it

$\displaystyle x^2 + (y - 1)^2 = 1$

can i say $\displaystyle x = \cos t$ and $\displaystyle y = \sin t$?

$\displaystyle \bold{r}(t) = (\cos t, \sin t)$

i'm thinking also of doing this $\displaystyle \bold{r}(t) = (\cos t, \sin t - 1)$

 

[lev888]

here is the question

A circle has its center at the point $\displaystyle (0,1)$ is placed there stationary. Then it was given a push and it moved one full revolution without slipping. An observer traced the point $\displaystyle (0,0)$ on the circle and noticed that it made a cycloid during the full revolution. What is the length of this cycloid?

i say the question is ambigus because it don't give me any formula of curve to calculate. i know the formula of how to calcate the length of a curve and the only inoformation i was given the circle. i'll try to permerterize it

$\displaystyle x^2 + (y - 1)^2 = 1$

can i say $\displaystyle x = \cos t$ and $\displaystyle y = \sin t$?

$\displaystyle \bold{r}(t) = (\cos t, \sin t)$

i'm thinking also of doing this $\displaystyle \bold{r}(t) = (\cos t, \sin t - 1)$

You are supposed to figure out the equation of the curve, which is the path of a point on the circle as the circle moves after the push. The problem is that I don't see anything describing the size of the circle and what the "ground" is.

 

[logistic_guy]

You are supposed to figure out the equation of the curve, which is the path of a point on the circle as the circle moves after the push. The problem is that I don't see anything describing the size of the circle and what the "ground" is.

i know the size of the circle. it is a unit circle shifted one unit up. i just don't know how to parametrize it

 

[Dr.Peterson]

here is the question

A circle has its center at the point $\displaystyle (0,1)$ is placed there stationary. Then it was given a push and it moved one full revolution without slipping. An observer traced the point $\displaystyle (0,0)$ on the circle and noticed that it made a cycloid during the full revolution. What is the length of this cycloid?

i say the question is ambigus because it don't give me any formula of curve to calculate. i know the formula of how to calcate the length of a curve and the only inoformation i was given the circle. i'll try to permerterize it

$\displaystyle x^2 + (y - 1)^2 = 1$

can i say $\displaystyle x = \cos t$ and $\displaystyle y = \sin t$?

$\displaystyle \bold{r}(t) = (\cos t, \sin t)$

i'm thinking also of doing this $\displaystyle \bold{r}(t) = (\cos t, \sin t - 1)$

Have you looked up "cycloid" to see how it is defined in terms of a rolling circle?

Presumably the question is intended to say this:

A unit circle has its center at the point $\displaystyle (0,1)$. It is placed there stationary, and thought of as a wheel resting on the x-axis, which is the ground. Then it was given a push and it moved one full revolution without slipping. An observer traced the point $\displaystyle (0,0)$ on the circle and noticed that it made a cycloid during the full revolution. What is the length of this cycloid?​

Since it rolls without slipping, the position of the moving point will be the changing position of the center of the circle, plus a vector rotating around it.

 

[logistic_guy]

Have you looked up "cycloid" to see how it is defined in terms of a rolling circle?

Presumably the question is intended to say this:

A unit circle has its center at the point $\displaystyle (0,1)$. It is placed there stationary, and thought of as a wheel resting on the x-axis, which is the ground. Then it was given a push and it moved one full revolution without slipping. An observer traced the point $\displaystyle (0,0)$ on the circle and noticed that it made a cycloid during the full revolution. What is the length of this cycloid?​

Since it rolls without slipping, the position of the moving point will be the changing position of the center of the circle, plus a vector rotating around it.

i don't understand everything in looked up link but i see this $\displaystyle x = a(t - \sin t)$ and $\displaystyle y = a(1 - \cos t)$

is this how shifted unit circle parapetrization?

 

[Dr.Peterson]

i don't understand everything in looked up link but i see this $\displaystyle x = a(t - \sin t)$ and $\displaystyle y = a(1 - \cos t)$

is this how shifted unit circle parametrization?

Where is the center of the circle at time t? (This is a shift.)

What angle has the point that started at the origin changed to at time t? So what vector do we add to the center of the circle? (This is a rotation.)

Look at the animation.

 

[logistic_guy]

Where is the center of the circle at time t? (This is a shift.)

i'm looking at animation and i see the center of the circle moving to the right with a fixed height

What angle has the point that started at the origin changed to at time t? So what vector do we add to the center of the circle? (This is a rotation.)

the angle start at $\displaystyle -\frac{\pi}{2}$ and for one red curve it end at $\displaystyle -\frac{\pi}{2} - 2\pi = -\frac{5\pi}{2}$

i think $\displaystyle x = a(t - \sin t)$ and $\displaystyle y = a(1 - \cos t)$ describe the trace on the red line and i can combine them as a vector describing the red curve

$\displaystyle \bold{r}(t) = (a(t - \sin t), a(1 - \cos t))$

i think i'm able to calculate the length of the curve but need the value of $\displaystyle a$? or i ignore it because it is a constant?

 

[khansaheb]

i'm looking at animation and i see the center of the circle moving to the right with a fixed height


the angle start at $\displaystyle -\frac{\pi}{2}$ and for one red curve it end at $\displaystyle -\frac{\pi}{2} - 2\pi = -\frac{5\pi}{2}$

i think $\displaystyle x = a(t - \sin t)$ and $\displaystyle y = a(1 - \cos t)$ describe the trace on the red line and i can combine them as a vector describing the red curve

$\displaystyle \bold{r}(t) = (a(t - \sin t), a(1 - \cos t))$

i think i'm able to calculate the length of the curve but need the value of $\displaystyle a$? or i ignore it because it is a constant?

You do NOT ignore 'a'. Here the length will be a function of 'a'.

 

[Dr.Peterson]

i think i'm able to calculate the length of the curve but need the value of $\displaystyle a$? or i ignore it because it is a constant?

The answer is the word you omitted in your initial statement of the problem: It's a unit circle.

On the page you're looking at, how do they define a?? Read carefully.

 

[logistic_guy]

You do NOT ignore 'a'. Here the length will be a function of 'a'.

thank

The answer is the word you omitted in your initial statement of the problem: It's a unit circle.

On the page you're looking at, how do they define a?? Read carefully.

it say the $\displaystyle a$ is related to the hump. i don't get it. if it is related to the unit circle, it mean only one thing, the radius. is that correct?

 

[Dr.Peterson]

thank


it say the $\displaystyle a$ is related to the hump. i don't get it. if it is related to the unit circle, it mean only one thing, the radius. is that correct?

Here is what I asked you to locate here:



Again, READ. Reading a formula without seeing the definition of the variables in it means MISreading the formula.

 

[logistic_guy]

Here is what I asked you to locate here:

View attachment 38290

Again, READ. Reading a formula without seeing the definition of the variables in it means MISreading the formula.

thank

so $\displaystyle a$ is the radius of the unit circle. i think now i can calculate the length of the curve

$\displaystyle \bold{r}(t) = (a(t - \sin t), a(1 - \cos t))$ it is given $\displaystyle a = 1$

$\displaystyle \bold{r}(t) = (t - \sin t, 1 - \cos t)$

$\displaystyle \bold{r}'(t) = (1 - \cos t, \sin t)$

$\displaystyle |\bold{r}'(t)| = \sqrt{(1 - \cos t)^2 + \sin^2 t} = \sqrt{1 - 2\cos t + \cos^2 t + \sin^2 t} = \sqrt{1 - 2\cos t + 1} = \sqrt{2 - 2\cos t}$

length of cycloid $\displaystyle = \int |\bold{r}'(t)| = \int_{-\frac{\pi}{2}}^{-\frac{5\pi}{2}} \sqrt{2 - 2\cos t}$

is it possible to solve this further or i leave the solution like this?

 

[mario99]

length of cycloid $\displaystyle = \int |\bold{r}'(t)| = \int_{-\frac{\pi}{2}}^{-\frac{5\pi}{2}} \sqrt{2 - 2\cos t}$

is it possible to solve this further or i leave the solution like this?

$\displaystyle \int \sqrt{2 - 2\cos t} \ dt = 2\int \sin\left(\frac{t}{2}\right) \ dt$

 

[Steven G]

$\displaystyle x^2 + (y - 1)^2 = 1$

can i say $\displaystyle x = \cos t$ and $\displaystyle y = \sin t$?

$\displaystyle \bold{r}(t) = (\cos t, \sin t)$

i'm thinking also of doing this $\displaystyle \bold{r}(t) = (\cos t, \sin t - 1)$

One of us is confused, that is for sure. It seems to me that the radius of this circle is constant and always equals 1.
To say that $\displaystyle \bold{r}(t) = (\cos t, \sin t)$ bothers me.

 

[Dr.Peterson]

One of us is confused, that is for sure. It seems to me that the radius of this circle is constant and always equals 1.
To say that $\displaystyle \bold{r}(t) = (\cos t, \sin t)$ bothers me.

Note the bold; that's (a first attempt at) the position vector representing the curve, not the scalar radius (which we later understand to be 1).

The eventual correct version is the sum of the position vector for the center of the circle, $(t, 1)$ and the vector representing the radius at the appropriate angle, $(-\sin(t), -\cos(t))$: $\displaystyle \bold{r}(t) = (t - \sin t, 1 - \cos t)$.

At that point, yes, both of you were confused.

 

[logistic_guy]

$\displaystyle \int \sqrt{2 - 2\cos t} \ dt = 2\int \sin\left(\frac{t}{2}\right) \ dt$

thank

length of cycloid = $\displaystyle \int_{-\frac{\pi}{2}}^{-\frac{5\pi}{2}} \sqrt{2 - 2\cos t} \ dt = 2\int_{-\frac{\pi}{2}}^{-\frac{5\pi}{2}} \sin\left(\frac{t}{2}\right) \ dt = -4\cos\frac{t}{2} = -4\cos-\frac{5\pi}{4} - \cos-\frac{\pi}{4} = -4\frac{-\sqrt{2}}{2}-\frac{\sqrt{2}}{2} = 4\sqrt{2}$

 

[BigBeachBanana]

The result of the integration does not look right. It should be a negative number based on the limits of the integrand.

 

[BigBeachBanana]

$\displaystyle \int \sqrt{2 - 2\cos t} \ dt = 2\int \sin\left(\frac{t}{2}\right) \ dt$

It should be
$$\displaystyle \int \sqrt{2 - 2\cos t} \ dt = 2\int \left|\sin\left(\frac{t}{2}\right)\right| \ dt$$Considering the definite integral, let $u=\dfrac{t}{2} \implies du = \frac{1}{2}dt, \quad t_1 = -\dfrac{5\pi}{2\cdot 2}, \quad t_2 = -\dfrac{\pi}{2\cdot2}$:
$$4 \int_{-\frac{5\pi}{4}}^{-\frac{\pi}{4}} \left|\sin u\right| \ du$$Apply a phase shift of $\dfrac{\pi}{4}$ to avoid dealing with absolute value. Since $-\pi < u < 0, |\sin u|=-\sin u:$
$$4 \int_{-\pi}^{0} \left|\sin u\right| \ du = -4 \int_{-\pi}^{0} \sin u\ du = 8$$

 

[mario99]

It should be
$$\displaystyle \int \sqrt{2 - 2\cos t} \ dt = 2\int \left|\sin\left(\frac{t}{2}\right)\right| \ dt$$Considering the definite integral, let $u=\dfrac{t}{2} \implies du = \frac{1}{2}dt, \quad t_1 = -\dfrac{5\pi}{2\cdot 2}, \quad t_2 = -\dfrac{\pi}{2\cdot2}$:
$$4 \int_{-\frac{5\pi}{4}}^{-\frac{\pi}{4}} \left|\sin u\right| \ du$$Apply a phase shift of $\dfrac{\pi}{4}$ to avoid dealing with absolute value. Since $-\pi < u < 0, |\sin u|=-\sin u:$
$$4 \int_{-\pi}^{0} \left|\sin u\right| \ du = -4 \int_{-\pi}^{0} \sin u\ du = 8$$

According to the link that Professor Dave sent, the answer is $8a$, which in our case $8(1) = 8$. This agrees with my integral:

$\displaystyle 2\int_{0}^{2\pi} \sin\left(\frac{t}{2}\right) \ dt = 8$

The OP just chose a complicated initial angle, $\displaystyle \theta = -\frac{\pi}{2}$ with which he should be very careful where the curve will be positive and where it will be negative (since he needs to force the negative area to be positive). Well he got it positive but lost some of it due to that the curve [ of $\displaystyle \sin\left(\frac{t}{2}\right)$ ] passed through positive and negative portions.

OP, why do you bother yourself to start with that angle when you are interested in a positive length in a full revolution? Just do the $(0,2\pi)$ interval. I don't blame you to choose that complicated initial angle as the point of interest was at the origin $\left(\displaystyle \theta = -\frac{\pi}{2}\right)$.

 

[BigBeachBanana]

According to the link that Professor Dave sent, the answer is $8a$, which in our case $8(1) = 8$. This agrees with my integral:

$\displaystyle 2\int_{0}^{2\pi} \sin\left(\frac{t}{2}\right) \ dt = 8$

This is only true for certain intervals. $$\sqrt{2-2\cos t} = 2\sin\!\left(\frac{t}{2}\right)$$It'd be wrong to omit the absolute value without good reasons. Your suggestion of picking a different starting angle is essentially applying a phase shift.