calculus problem that look more geometry than calculus

[logistic_guy]

here is the question

The formula \(\displaystyle A = (\cos t, \sin t)\) represents an object moving counterclockwise at a speed of \(\displaystyle 1\) unit per second on the unit circle. At point \(\displaystyle A\) Jo says \(\displaystyle (-\sin t, \cos t)\) is a vector that describes the object's velocity. Let \(\displaystyle P = (0,\sin t)\) be the projection of \(\displaystyle A\) on the y-axis. As time passes, notice that \(\displaystyle P\) moves with varying speed up and down this axis. When \(\displaystyle P\) passes \(\displaystyle (0,0.6)\), its speed is \(\displaystyle 0.8\) unit per second. Make calculations that support this statement.

my attemp
one look at this problem think it's calulus then geometry then calculus. i know it's confusing

i'm not use on this notation. i worked in curves before and i understand \(\displaystyle \bold{r} = (\cos t, \sin t)\) better

i can find the velocity easily \(\displaystyle \bold{r}' = (-\sin t, \cos t)\) this confirm what Jo say is correct

i proof the first part of the question and i'm confused how to start the secobd part

 

[fresh_42]

It is a mixture of geometry and calculus, but the questions and terms it uses are clearly calculus. We have a path [imath] \gamma(t)\, : \,[0,2\pi]\longrightarrow \mathbb{S}^1 \subseteq \mathbb{R}^2 [/imath] around the unit circle [imath] \mathbb{S}^1 [/imath] in the Euclidean plane [imath] \mathbb{R}^2, [/imath] with normally [imath] (x,y) [/imath]-coordinates. Around the circle, we have [imath] x(t)=\cos(t) [/imath] and [imath] y(t)=\sin(t). [/imath] Since the radius is conveniently one, I wouldn't bother much about it and avoid notations with [imath] r. [/imath]

Before calculating anything, we have to answer what is meant by the speed of a point. Physics tells us that velocity is [imath] \dot \gamma (t)=\dfrac{d}{dt} \gamma(t) [/imath] - a vector - and speed its absolute value [imath] \left\|\dot \gamma (t)\right\|=\left\|\dfrac{d}{dt} \gamma(t)\right\| [/imath] - a scalar, a number.

These formulas allow you to check the statements made in the problem text and which you should perform to understand it better.

The formulas are true for any other path [imath] t\longmapsto \gamma(t) ,[/imath] too, in particular for [imath] \gamma_0(t)=(0,\sin(t)). [/imath]

 

[khansaheb]

here is the question

The formula \(\displaystyle A = (\cos t, \sin t)\) represents an object moving counterclockwise at a speed of \(\displaystyle 1\) unit per second on the unit circle. At point \(\displaystyle A\) Jo says \(\displaystyle (-\sin t, \cos t)\) is a vector that describes the object's velocity. Let \(\displaystyle P = (0,\sin t)\) be the projection of \(\displaystyle A\) on the y-axis. As time passes, notice that \(\displaystyle P\) moves with varying speed up and down this axis. When \(\displaystyle P\) passes \(\displaystyle (0,0.6)\), its speed is \(\displaystyle 0.8\) unit per second. Make calculations that support this statement.

my attemp
one look at this problem think it's calulus then geometry then calculus. i know it's confusing

i'm not use on this notation. i worked in curves before and i understand \(\displaystyle \bold{r} = (\cos t, \sin t)\) better

i can find the velocity easily \(\displaystyle \bold{r}' = (-\sin t, \cos t)\) this confirm what Jo say is correct

i proof the first part of the question and i'm confused how to start the secobd part

Please post the EXACT wording of the problem.

 

[logistic_guy]

It is a mixture of geometry and calculus, but the questions and terms it uses are clearly calculus. We have a path [imath] \gamma(t)\, : \,[0,2\pi]\longrightarrow \mathbb{S}^1 \subseteq \mathbb{R}^2 [/imath] around the unit circle [imath] \mathbb{S}^1 [/imath] in the Euclidean plane [imath] \mathbb{R}^2, [/imath] with normally [imath] (x,y) [/imath]-coordinates. Around the circle, we have [imath] x(t)=\cos(t) [/imath] and [imath] y(t)=\sin(t). [/imath] Since the radius is conveniently one, I wouldn't bother much about it and avoid notations with [imath] r. [/imath]

Before calculating anything, we have to answer what is meant by the speed of a point. Physics tells us that velocity is [imath] \dot \gamma (t)=\dfrac{d}{dt} \gamma(t) [/imath] - a vector - and speed its absolute value [imath] \left\|\dot \gamma (t)\right\|=\left\|\dfrac{d}{dt} \gamma(t)\right\| [/imath] - a scalar, a number.

These formulas allow you to check the statements made in the problem text and which you should perform to understand it better.

The formulas are true for any other path [imath] t\longmapsto \gamma(t) ,[/imath] too, in particular for [imath] \gamma_0(t)=(0,\sin(t)). [/imath]

thank

i think if i write the curve as \(\displaystyle A\) or \(\displaystyle \bold{r}\) or \(\displaystyle \bold{r}(t)\) or \(\displaystyle \boldsymbol{\gamma}(t)\) i'm not change the properties of the curve right?

if i study curves they use interchangebly \(\displaystyle \bold{r}(t)\) and \(\displaystyle \boldsymbol{\gamma}(t)\). i think \(\displaystyle \boldsymbol{\gamma}(t)\) is the best notation for curves

you're using the notation of sphere \(\displaystyle \mathbb{S}\) for our curve. why? why it's a subset of \(\displaystyle \mathbb{R}^2\) and not a proper subset?

is it wrong to say that our circle is a proper subset of \(\displaystyle \mathbb{R}^2\)?

i'm caring about the notation because i study them carefully and wanna be correct when i write them in future questions.

i think the speed and the velcity is confuse me a lot in the last questions i posted. i don't know it precisely and it is like the speed we don't care about the direction. i'm confused when you included \(\displaystyle \boldsymbol{\gamma}_0\) to our curve \(\displaystyle \boldsymbol{\gamma}\)

what do you mean by the zero in \(\displaystyle \boldsymbol{\gamma}_0\)? i'm just confused by the notation

Please post the EXACT wording of the problem.

 

[fresh_42]

i think if i write the curve as \(\displaystyle A\) or \(\displaystyle \bold{r}\) or \(\displaystyle \bold{r}(t)\) or \(\displaystyle \boldsymbol{\gamma}(t)\) i'm not change the properties of the curve right?

No, you can use a letter whatever you want. However, there are standards that help reading and writing a lot. [imath] \mathbf{r} [/imath] is usually a radius, and we do not have a radius as a variable here. We have [imath] \mathbf{r} =\mathbf{r}(t)=1[/imath]. If you choose [imath] r [/imath] as your path name it will be a matter of time until you get confused with the constant radius in the example.

[imath] A [/imath] is even worse since it is normally used as the notation of an area.

if i study curves they use interchangebly \(\displaystyle \bold{r}(t)\) and \(\displaystyle \boldsymbol{\gamma}(t)\). i think \(\displaystyle \boldsymbol{\gamma}(t)\) is the best notation for curves

... which we do have here. A curve (a circle) that we follow from timestamp [imath] t=0 [/imath] to timestamp [imath] t=2\pi. [/imath]

If I use [imath] t\mapsto \gamma(t) [/imath] as a general path, and [imath] \gamma(t)=(x(t),y(t))=(\cos(t),\sin(t)) [/imath] I can calculate everything and still have the letters [imath] r [/imath] and [imath] A [/imath] in hand, in case we consider different circles with arbitrary or even changing radii, or want to calculate the area [imath] A=\pi. [/imath] It makes no sense to create confusion.

you're using the notation of sphere \(\displaystyle \mathbb{S}\) for our curve. why? why it's a subset of \(\displaystyle \mathbb{R}^2\) and not a proper subset?

It is a circle of radius one. Its official name is unit circle or unit sphere, and its official notation is [imath] \mathbb{S}^1. [/imath] The power one denotes the fact that it is one-dimensional, although embedded in the two-dimensional Euclidean plane [imath] \mathbb{R}^2 .[/imath]

I always use [imath] \subseteq [/imath] because a) it is a shortcut on my keyboard and b) I do not want to even think about equality. Everybody knows that a circle isn't the entire plane. [imath] \subset [/imath] or [imath] \subsetneq [/imath] would pay way too much attention to that fact: why did he emphasize this triviality, are there examples where equality holds, etc. All senseless questions that could come up if I emphasized the strict inequality. [imath] \subseteq [/imath] is simply my default notation.

is it wrong to say that our circle is a proper subset of \(\displaystyle \mathbb{R}^2\)?

Of course not, but why? It's trivial.

i'm caring about the notation because i study them carefully and wanna be correct when i write them in future questions.

Notations are arbitrary, but conventions are not. Your exercise considers two different curves, one of which happens to be a circle. However, the shape is completely irrelevant to the questions about speed. Speed is a local quantity: speed to a certain moment in time! Again, the speed on the circle happens to be constant because of a famous formula involving the sine and the cosine of an angle (here a time). This is the result of a computation, not a given fact! The other curve is [imath] t\mapsto (0,\sin(t)) [/imath] which doesn't have a constant speed. Btw., the speed around the circle may be constant, but the velocity is not!

This means the circle itself becomes irrelevant (except for that formula) and is the reason I spoke of [imath] \gamma [/imath] instead. No radius [imath] \boldsymbol r [/imath] and no area [imath] \boldsymbol A, [/imath] only Pythagoras.

i think the speed and the velcity is confuse me a lot in the last questions i posted. i don't know it precisely and it is like the speed we don't care about the direction. i'm confused when you included \(\displaystyle \boldsymbol{\gamma}_0\) to our curve \(\displaystyle \boldsymbol{\gamma}\)

I only introduced [imath] \gamma_0 [/imath] to distinguish it from the circle that I used [imath] \gamma [/imath] for. Two curves, two names. Otherwise, we will get a mess. The geometry doesn't play a role for the calculations.

Physicists avoid the word speed at all. They do not use it. It is a velocity, which is a vector, not a number. It is the vector you follow when you lose traction on a road, and its length, i.e. how far you will be cast into the wilderness. This length is commonly named speed, but physicists speak of the norm or the length of the velocity vector instead. With only one exception: the speed of light. But that is because of historical reasons.

what do you mean by the zero in \(\displaystyle \boldsymbol{\gamma}_0\)? i'm just confused by the notation

I meant that I needed something to distinguish between [imath] t\mapsto \gamma(t)=(\cos(t),\sin(t)) [/imath] and [imath] t\mapsto \gamma_0(t)=(0,\sin(t)) .[/imath] I could have chosen any other letter instead, but many of them have their own conventions and we are still talking about curves, or paths along curves.

 

[fresh_42]

If you cannot use the formula for the velocity, yet, then we need an argument why it is the derivative.

The velocity can also be defined as the marginal gain of distance at a marginal amount of time

[math] \displaystyle{\lim_{ h \to 0} \dfrac{\gamma(t+h)-\gamma(t)}{h}=\lim_{ h \to 0} \dfrac{(\cos(t+h),\sin(t+h))-(\cos(t),\sin(t))}{h}=\lim_{ h \to 0}\dfrac{(\cos(t+h)-\cos(t)\, , \,\sin(t+h)-\sin(t))}{h} } [/math]
Now apply the addition theorems for cosine and sine and try to determine the limit.

If you have done this, then compute the length of this vector.

If limits are also not available, then consider
[math] \gamma(t+h)-\gamma(t)=\begin{pmatrix}\cos(t+h)-\cos(t)\\ \sin(t+h)-\sin(t)\end{pmatrix}= \begin{pmatrix}\cos(t)(\cos(h)-1)-\sin(t)\sin(h)\\ \sin(t)(\cos(h)-1)+\cos(t)\sin(h)\end{pmatrix} [/math]and
[math] \dfrac{\gamma(t+h)-\gamma(t)}{h}=\begin{pmatrix}\cos(t)\cdot \dfrac{\cos(h)-1}{h}-\sin(t)\cdot \dfrac{\sin(h)}{h}\\[10pt] \sin(t)\cdot\dfrac{\cos(h)-1}{h}+\cos(t)\cdot\dfrac{\sin(h)}{h} \end{pmatrix} [/math]Now figure out where those quotients tend to if we make [imath] h [/imath] smaller and smaller, i.e. if [imath] h [/imath] tend to zero.

 

[logistic_guy]

No, you can use a letter whatever you want. However, there are standards that help reading and writing a lot. [imath] \mathbf{r} [/imath] is usually a radius, and we do not have a radius as a variable here. We have [imath] \mathbf{r} =\mathbf{r}(t)=1[/imath]. If you choose [imath] r [/imath] as your path name it will be a matter of time until you get confused with the constant radius in the example.

[imath] A [/imath] is even worse since it is normally used as the notation of an area.



... which we do have here. A curve (a circle) that we follow from timestamp [imath] t=0 [/imath] to timestamp [imath] t=2\pi. [/imath]

If I use [imath] t\mapsto \gamma(t) [/imath] as a general path, and [imath] \gamma(t)=(x(t),y(t))=(\cos(t),\sin(t)) [/imath] I can calculate everything and still have the letters [imath] r [/imath] and [imath] A [/imath] in hand, in case we consider different circles with arbitrary or even changing radii, or want to calculate the area [imath] A=\pi. [/imath] It makes no sense to create confusion.



It is a circle of radius one. Its official name is unit circle or unit sphere, and its official notation is [imath] \mathbb{S}^1. [/imath] The power one denotes the fact that it is one-dimensional, although embedded in the two-dimensional Euclidean plane [imath] \mathbb{R}^2 .[/imath]

I always use [imath] \subseteq [/imath] because a) it is a shortcut on my keyboard and b) I do not want to even think about equality. Everybody knows that a circle isn't the entire plane. [imath] \subset [/imath] or [imath] \subsetneq [/imath] would pay way too much attention to that fact: why did he emphasize this triviality, are there examples where equality holds, etc. All senseless questions that could come up if I emphasized the strict inequality. [imath] \subseteq [/imath] is simply my default notation.


Of course not, but why? It's trivial.


Notations are arbitrary, but conventions are not. Your exercise considers two different curves, one of which happens to be a circle. However, the shape is completely irrelevant to the questions about speed. Speed is a local quantity: speed to a certain moment in time! Again, the speed on the circle happens to be constant because of a famous formula involving the sine and the cosine of an angle (here a time). This is the result of a computation, not a given fact! The other curve is [imath] t\mapsto (0,\sin(t)) [/imath] which doesn't have a constant speed. Btw., the speed around the circle may be constant, but the velocity is not!

This means the circle itself becomes irrelevant (except for that formula) and is the reason I spoke of [imath] \gamma [/imath] instead. No radius [imath] \boldsymbol r [/imath] and no area [imath] \boldsymbol A, [/imath] only Pythagoras.




I only introduced [imath] \gamma_0 [/imath] to distinguish it from the circle that I used [imath] \gamma [/imath] for. Two curves, two names. Otherwise, we will get a mess. The geometry doesn't play a role for the calculations.

Physicists avoid the word speed at all. They do not use it. It is a velocity, which is a vector, not a number. It is the vector you follow when you lose traction on a road, and its length, i.e. how far you will be cast into the wilderness. This length is commonly named speed, but physicists speak of the norm or the length of the velocity vector instead. With only one exception: the speed of light. But that is because of historical reasons.



I meant that I needed something to distinguish between [imath] t\mapsto \gamma(t)=(\cos(t),\sin(t)) [/imath] and [imath] t\mapsto \gamma_0(t)=(0,\sin(t)) .[/imath] I could have chosen any other letter instead, but many of them have their own conventions and we are still talking about curves, or paths along curves.

thank

i think i'm understanding. you explain everything i want nicely.

since i can name my new curve anything i'll name it eta \(\displaystyle \boldsymbol{\eta}(t) = (0, \sin t)\)

i think this new curve represent y-axis inside \(\displaystyle \mathbb{S}^1\). it's a vertical line of length 2 or better to say \(\displaystyle -1 < y < 1\)

my \(\displaystyle y\) here is \(\displaystyle y(t) = \sin t\)

the velocity is \(\displaystyle \boldsymbol{\eta}'(t) = (0, \cos t)\) give me \(\displaystyle y'(t) = \cos t\)

i've the velocity of the new curve but i don't how to make calculation support the speed of \(\displaystyle 0.8\) unit per second

if i do speed = \(\displaystyle \vert|\boldsymbol{\eta}'(t)\vert| = \vert|\cos t \vert|\)

If you cannot use the formula for the velocity, yet, then we need an argument why it is the derivative.

The velocity can also be defined as the marginal gain of distance at a marginal amount of time

[math] \displaystyle{\lim_{ h \to 0} \dfrac{\gamma(t+h)-\gamma(t)}{h}=\lim_{ h \to 0} \dfrac{(\cos(t+h),\sin(t+h))-(\cos(t),\sin(t))}{h}=\lim_{ h \to 0}\dfrac{(\cos(t+h)-\cos(t)\, , \,\sin(t+h)-\sin(t))}{h} } [/math]
Now apply the addition theorems for cosine and sine and try to determine the limit.

If you have done this, then compute the length of this vector.

i see this formula a lot in calculsu and i still don't understand why they use this complicated formula when the derivative is just \(\displaystyle \frac{dy}{dt}\). it's very confusing. tell me frankly how many years do you spent to understand this monster?

 

[fresh_42]

thank

i think i'm understanding. you explain everything i want nicely.

since i can name my new curve anything i'll name it eta \(\displaystyle \boldsymbol{\eta}(t) = (0, \sin t)\)

i think this new curve represent y-axis inside \(\displaystyle \mathbb{S}^1\). it's a vertical line of length 2 or better to say \(\displaystyle -1 < y < 1\)

my \(\displaystyle y\) here is \(\displaystyle y(t) = \sin t\)

the velocity is \(\displaystyle \boldsymbol{\eta}'(t) = (0, \cos t)\) give me \(\displaystyle y'(t) = \cos t\)

i've the velocity of the new curve but i don't how to make calculation support the speed of \(\displaystyle 0.8\) unit per second

if i do speed = \(\displaystyle \vert|\boldsymbol{\eta}'(t)\vert| = \vert|\cos t \vert|\)


i see this formula a lot in calculsu and i still don't understand why they use this complicated formula when the derivative is just \(\displaystyle \frac{dy}{dt}\). it's very confusing. tell me frankly how many years do you spent to understand this monster?

See my edit of the previous post.

The "monster" is not so wild. It is basically the slope of secants (= intersects a curve at two points) where the distance between the two intersection points gets smaller and smaller until it vanishes and the secants become a tangent (= touches a curve at one point) and the limit is the direction of that tangent.





In the picture, we have [imath] \gamma(t)=(t, 0.2 \cdot t^2) [/imath]. Sorry, I used a picture that I already had, so [imath] x=t [/imath] and we are interested in the velocity in [imath] p=(-5,5) [/imath] at time [imath] t=-5 [/imath] or location [imath] x=-5, [/imath] depends on how we describe it. Anyway, we have [imath] t=x [/imath] here.

[math] \dfrac{\gamma(t+h)-\gamma(t)}{h}=\dfrac{1}{h}\cdot \left(\begin{pmatrix}-5+h \\ 0.2\cdot (-5+h)^2\end{pmatrix}-\begin{pmatrix}-5\\ 0.2\cdot (-5)^2\end{pmatrix}\right)=\dfrac{1}{h}\begin{pmatrix}h\\ 0.2\cdot (-10h+h^2)\end{pmatrix}=\begin{pmatrix}1 \\ -2+h\end{pmatrix}\stackrel{h\to 0}{\longrightarrow }\begin{pmatrix}1\\- 2 \end{pmatrix} [/math]
This is the direction of the tangent line, the velocity vector at [imath] (-5,5) .[/imath] The slope of a straight is [imath] \dfrac{\Delta y}{\Delta x}=\dfrac{y_A-y_B}{x_A-x_B}=\dfrac{y_2-y_1}{x_2-x_1} [/imath] (I listed a few notations since I don't know which one you are used to if you calculate how steep a straight is.)

Since we have [imath] (1,-2), [/imath] it means [imath] 1 [/imath] step in [imath] +x [/imath] direction requires [imath] 2 [/imath] steps in [imath] -y [/imath] direction, i.e. [imath] \dfrac{-2}{1}=-2 [/imath] is the slope of the tangent, and the velocity at this point if we slide down the parabola linearly (no acceleration!) to the [imath] x [/imath]-axis. The speed is [imath] \sqrt{1^2+(-2)^2}=\sqrt{5} [/imath] per second, or whatever the unit on the [imath] x=t [/imath]-axis is.

I don't want to scare you, but you should see what they have made out of this simple concept of tangents if you really want to see monsters. The entire physics is about this limit!

 

[logistic_guy]

See my edit of the previous post.

The "monster" is not so wild. It is basically the slope of secants (= intersects a curve at two points) where the distance between the two intersection points gets smaller and smaller until it vanishes and the secants become a tangent (= touches a curve at one point) and the limit is the direction of that tangent.


View attachment 38486


In the picture, we have [imath] \gamma(t)=(t, 0.2 \cdot t^2) [/imath]. Sorry, I used a picture that I already had, so [imath] x=t [/imath] and we are interested in the velocity in [imath] p=(-5,5) [/imath] at time [imath] t=-5 [/imath] or location [imath] x=-5, [/imath] depends on how we describe it. Anyway, we have [imath] t=x [/imath] here.

[math] \dfrac{\gamma(t+h)-\gamma(t)}{h}=\dfrac{1}{h}\cdot \left(\begin{pmatrix}-5+h \\ 0.2\cdot (-5+h)^2\end{pmatrix}-\begin{pmatrix}-5\\ 0.2\cdot (-5)^2\end{pmatrix}\right)=\dfrac{1}{h}\begin{pmatrix}h\\ 0.2\cdot (-10h+h^2)\end{pmatrix}=\begin{pmatrix}1 \\ -2+h\end{pmatrix}\stackrel{h\to 0}{\longrightarrow }\begin{pmatrix}1\\- 2 \end{pmatrix} [/math]
This is the direction of the tangent line, the velocity vector at [imath] (-5,5) .[/imath] The slope of a straight is [imath] \dfrac{\Delta y}{\Delta x}=\dfrac{y_A-y_B}{x_A-x_B}=\dfrac{y_2-y_1}{x_2-x_1} [/imath] (I listed a few notations since I don't know which one you are used to if you calculate how steep a straight is.)

Since we have [imath] (1,-2), [/imath] it means [imath] 1 [/imath] step in [imath] +x [/imath] direction requires [imath] 2 [/imath] steps in [imath] -y [/imath] direction, i.e. [imath] \dfrac{-2}{1}=-2 [/imath] is the slope of the tangent, and the velocity at this point if we slide down the parabola linearly (no acceleration!) to the [imath] x [/imath]-axis. The speed is [imath] \sqrt{1^2+(-2)^2}=\sqrt{5} [/imath] per second, or whatever the unit on the [imath] x=t [/imath]-axis is.

I don't want to scare you, but you should see what they have made out of this simple concept of tangents if you really want to see monsters. The entire physics is about this limit!

one month ago when i see graph i run away because it's confusing. i spend a lot of time in the last few weeks to understand graphs. i think i can understand your parabola graph. it come with this equation usually \(\displaystyle y = x^2\). i still don't understand the purpose of other numbers. i'll do a lot of algebra this month and the next month to master it.

give me some time to read and understand your edit post and this post. hold on fresh_42, i'll get back to you soon

 

[fresh_42]

give me some time to read and understand your edit post and this post. hold on fresh_42, i'll get back to you soon

Take all the time you need, make your own examples (not too complicated), and sketch some situations [imath] \displaystyle{\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h} }[/imath]. So much depends on it: 95% of calculus, everything in physics, and therefore the world as we currently understand it. It is a simple quotient and the limit only means the transition from secant to tangent. This concept is really important.

It all began with Leibniz and Newton at the end of the seventeenth century. That was 350 years ago and people developed a lot of mathematics ever since. But in the end, it is this tiny quotient that rules the world. We approximate complicated equations (blue function) by linear approximations and calculate with them instead. However, a linear approximation is nothing else than that (green) tangent in the picture.

 

[logistic_guy]

i'm back

i'm spent time to open reading my calculus book. i practice to solve \(\displaystyle y = x\) the derivative \(\displaystyle \frac{dy}{dx} = 1\)

for reason i don't understand the derivstive can also be writen with limit

\(\displaystyle \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\) the notation don't make sense but it give correct result

i'll find the derivative at point \(\displaystyle (-5,-5)\)

\(\displaystyle \frac{dy}{dx} = \lim_{h \to 0} \frac{-5 + h - -5}{h} = \lim_{h \to 0} \frac{-5 + h + 5}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)

with eta notation \(\displaystyle \boldsymbol{\eta}(t) = (t,t)\)

\(\displaystyle \frac{\boldsymbol{\eta}(t + h) - \boldsymbol{\eta}(t)}{h} = \frac{1}{h} \left(\begin{pmatrix}-5+h \\ -5+h\end{pmatrix}-\begin{pmatrix}-5\\ -5\end{pmatrix}\right) = \frac{1}{h} \begin{pmatrix}h \\ h\end{pmatrix}\stackrel{h\to 0}{\longrightarrow }\begin{pmatrix}1\\1 \end{pmatrix}\)

it mean \(\displaystyle 1 \) step in \(\displaystyle +x \) direction require \(\displaystyle 1 \) step in \(\displaystyle +y \) direction, i.e. \(\displaystyle \dfrac{1}{1}=1\) is the slope of the tangent

i steal some of your idea to complete the concept

this limit approach is like asking kid and blond girl what is \(\displaystyle 1 + 1\)

the kid say \(\displaystyle 1 + 1 = 2\)

the blond girl say \(\displaystyle 1 + 1 = 9*1000 + 15 - \frac{18}{3} + \sqrt{100} - 13 - 9000 - 10 + 6= 2\)

 

[fresh_42]

Sounds good.

I know a similar joke:

A: "What is seven squared?"
B: 49.
A: "How did you know so fast?"
B: Easy. I calculated 5 times 10 and subtracted one.

You can make it a bit more complicated by considering [imath] x\mapsto c\cdot x [/imath] with a fixed number [imath] c [/imath] or even [imath] x\mapsto c\cdot x^n. [/imath] The sine functions you began this thread with are a bit more complicated (post #6).

a) [imath] \displaystyle{ \lim_{h \to 0} \dfrac{\sin (h)}{h} = 1.} [/imath]

The sine of an angle [imath] h [/imath] is the quotient of the opposite cathetus and the hypotenuse, which has length one in our example. The opposite cathetus shrinks if the angle shrinks and this at nearly the same speed the closer we get to zero. Thus, we are left with [imath] 1/1=1. [/imath]

b) [imath] \displaystyle{ \lim_{h \to 0} \dfrac{\cos (h)-1}{h} = 0.} [/imath]

The cosine of an angle [imath] h [/imath] is the quotient of the adjacent cathetus and the hypotenuse, which has length one in our example.
The adjacent cathetus becomes the hypotenuse by a shrinking angle, i.e. the quotient becomes one, and [imath] \cos(h) -1 [/imath] becomes zero faster than the division by [imath] h [/imath] could compensate, so that the value of the limit is zero.

This could be a possible explanation you have been asked for in the problem statement. But be careful! I cheated! Firstly, I asked
https://www.wolframalpha.com/input?i=lim+(h+to+0)+(sin+h+/+h)= and https://www.wolframalpha.com/input?i=lim+(h+to+0)+((-1+++cos+h)+/+h)= about the result, and secondly, I know where to look up the formulas

[math]\begin{array}{lll} \dfrac{\sin(h)}{h}&=\dfrac{1}{h}\left(\dfrac{h}{1}-\dfrac{h^3}{1\cdot 2\cdot 3}+\dfrac{h^5}{1\cdot 2\cdot 3\cdot 4\cdot 5}\mp\ldots\right) =\dfrac{1}{1}-\dfrac{h^2}{1\cdot 2\cdot 3}+\dfrac{h^4}{1\cdot 2\cdot 3\cdot 4\cdot 5}\mp\ldots \stackrel{h\to 0}{\longrightarrow }1\\[12pt] \dfrac{\cos(h)-1}{h}&=-\dfrac{1}{h}+\dfrac{1}{h}\left(1-\dfrac{h^2}{1\cdot 2}+\dfrac{h^4}{1\cdot 2\cdot 3\cdot 4}\mp\ldots\right) =-\dfrac{h}{1\cdot 2}+\dfrac{h^3}{1\cdot 2\cdot 3\cdot 4}\mp\ldots \stackrel{h\to 0}{\longrightarrow }0 \end{array}[/math]
It all depends on how fast the opposite cathetus shrinks if the angle shrinks, or how fast one minus the length of the adjacent cathetus shrinks if the angle shrinks. The division by [imath] h [/imath] makes the quotient increase if we shrink [imath] h. [/imath] WolframAlpha and those formulas answered the question of how fast. My verbal argument tried to hide this weakness. It is therefore important to actually calculate the limit - like you did in your example - rather than only describe it.

Also, I spoke of linear approximations and that they are used instead of the function. This is of course only true close to the point where we calculated the tangent. Differentiation is a local phenomenon. Sure, there are functions that are everywhere differentiable, as [imath] f(x)=0.2 \cdot x^2 [/imath] in the example, but a tangent, a differentiation at a single point, [imath] p=(-5,5) [/imath] as in the example, is local. The tangent only approximates the curve [imath] (x,f(x)) [/imath] around [imath] p. [/imath] The further we go away from [imath] p [/imath] the bigger the distance between tangent and function and the tangent can no longer be used as an approximation.
[math] f'(x)=(0.2\cdot x^2)'=0.4\cdot x [/math] defines a tangent of my parabola everywhere, but the green one that can be used at [imath] p=(-5,5) [/imath] has a slope of [imath] f'(-5)=0.4 \cdot (-5)=-2. [/imath] All others are worse up to really bad approximations at [imath] p. [/imath]

 

[logistic_guy]

Sounds good.

thank

I know a similar joke:

A: "What is seven squared?"
B: 49.
A: "How did you know so fast?"
B: Easy. I calculated 5 times 10 and subtracted one.

nice joke

You can make it a bit more complicated by considering [imath] x\mapsto c\cdot x [/imath] with a fixed number [imath] c [/imath] or even [imath] x\mapsto c\cdot x^n. [/imath] The sine functions you began this thread with are a bit more complicated (post #6).

limit with \(\displaystyle x\) not scaring me. i can hadle it easily, but weird function of trigometry difficult. i didn't pay attention in calculus class, now i've to deal with the sequences

a) [imath] \displaystyle{ \lim_{h \to 0} \dfrac{\sin (h)}{h} = 1.} [/imath]

b) [imath] \displaystyle{ \lim_{h \to 0} \dfrac{\cos (h)-1}{h} = 0.} [/imath]

this i understand now. it's clear what happen in (post #6)

This could be a possible explanation you have been asked for in the problem statement. But be careful! I cheated! Firstly, I asked
https://www.wolframalpha.com/input?i=lim+(h+to+0)+(sin+h+/+h)= and https://www.wolframalpha.com/input?i=lim+(h+to+0)+((-1+++cos+h)+/+h)= about the result, and secondly, I know where to look up the formulas

first time i know i can do limit in wolframit don't show steps

[math]\begin{array}{lll} \dfrac{\sin(h)}{h}&=\dfrac{1}{h}\left(\dfrac{h}{1}-\dfrac{h^3}{1\cdot 2\cdot 3}+\dfrac{h^5}{1\cdot 2\cdot 3\cdot 4\cdot 5}\mp\ldots\right) =\dfrac{1}{1}-\dfrac{h^2}{1\cdot 2\cdot 3}+\dfrac{h^4}{1\cdot 2\cdot 3\cdot 4\cdot 5}\mp\ldots \stackrel{h\to 0}{\longrightarrow }1\\[12pt] \dfrac{\cos(h)-1}{h}&=-\dfrac{1}{h}+\dfrac{1}{h}\left(1-\dfrac{h^2}{1\cdot 2}+\dfrac{h^4}{1\cdot 2\cdot 3\cdot 4}\mp\ldots\right) =-\dfrac{h}{1\cdot 2}+\dfrac{h^3}{1\cdot 2\cdot 3\cdot 4}\mp\ldots \stackrel{h\to 0}{\longrightarrow }0 \end{array}[/math]

this i don't understand

It all depends on how fast the opposite cathetus shrinks if the angle shrinks, or how fast one minus the length of the adjacent cathetus shrinks if the angle shrinks. The division by [imath] h [/imath] makes the quotient increase if we shrink [imath] h. [/imath] WolframAlpha and those formulas answered the question of how fast. My verbal argument tried to hide this weakness. It is therefore important to actually calculate the limit - like you did in your example - rather than only describe it.

Also, I spoke of linear approximations and that they are used instead of the function. This is of course only true close to the point where we calculated the tangent. Differentiation is a local phenomenon. Sure, there are functions that are everywhere differentiable, as [imath] f(x)=0.2 \cdot x^2 [/imath] in the example, but a tangent, a differentiation at a single point, [imath] p=(-5,5) [/imath] as in the example, is local. The tangent only approximates the curve [imath] (x,f(x)) [/imath] around [imath] p. [/imath] The further we go away from [imath] p [/imath] the bigger the distance between tangent and function and the tangent can no longer be used as an approximation.
[math] f'(x)=(0.2\cdot x^2)'=0.4\cdot x [/math] defines a tangent of my parabola everywhere, but the green one that can be used at [imath] p=(-5,5) [/imath] has a slope of [imath] f'(-5)=0.4 \cdot (-5)=-2. [/imath] All others are worse up to really bad approximations at [imath] p. [/imath]

i still didn't understand this shrink angle and approximation game. if you let me find the derivative normally i say this is my game

 

[mario99]

here is the question

The formula \(\displaystyle A = (\cos t, \sin t)\) represents an object moving counterclockwise at a speed of \(\displaystyle 1\) unit per second on the unit circle. At point \(\displaystyle A\) Jo says \(\displaystyle (-\sin t, \cos t)\) is a vector that describes the object's velocity. Let \(\displaystyle P = (0,\sin t)\) be the projection of \(\displaystyle A\) on the y-axis. As time passes, notice that \(\displaystyle P\) moves with varying speed up and down this axis. When \(\displaystyle P\) passes \(\displaystyle (0,0.6)\), its speed is \(\displaystyle 0.8\) unit per second. Make calculations that support this statement.

Alternative way! You can show that the speed is [imath]0.8[/imath] unit per second at [imath]0.6[/imath] unit without explicitly finding the derivative. If the object moves from [imath]0.6[/imath] unit to [imath]0.6001[/imath] unit, it basically did not move. From this idea, you can find the velocity.

Let [imath]v[/imath] be the velocity and [imath]s[/imath] be the speed.


[imath]\displaystyle v = \frac{\Delta y}{\Delta t}\text{j} = \frac{0.6001 - 0.6}{\sin^{-1}0.6001 - \sin^{-1}0.6}\text{j} = \frac{0.001}{0.64475 - 0.64350}\text{j} = \frac{0.001}{0.00125}\text{j} = 0.8\text{j}[/imath] unit per second [imath] \ \ \ \ \ \ \ \ (\text{j}[/imath] means up)


[imath]\displaystyle s = \Vert v \Vert = \Vert 0.8\text{j} \Vert = 0.8 [/imath] unit per second

 

[jonah2.0]

Beer induced query follows.

thank

i think if i write the curve as \(\displaystyle A\) or \(\displaystyle \bold{r}\) or \(\displaystyle \bold{r}(t)\) or \(\displaystyle \boldsymbol{\gamma}(t)\) i'm not change the properties of the curve right?

if i study curves they use interchangebly \(\displaystyle \bold{r}(t)\) and \(\displaystyle \boldsymbol{\gamma}(t)\). i think \(\displaystyle \boldsymbol{\gamma}(t)\) is the best notation for curves

you're using the notation of sphere \(\displaystyle \mathbb{S}\) for our curve. why? why it's a subset of \(\displaystyle \mathbb{R}^2\) and not a proper subset?

is it wrong to say that our circle is a proper subset of \(\displaystyle \mathbb{R}^2\)?

i'm caring about the notation because i study them carefully and wanna be correct when i write them in future questions.

i think the speed and the velcity is confuse me a lot in the last questions i posted. i don't know it precisely and it is like the speed we don't care about the direction. i'm confused when you included \(\displaystyle \boldsymbol{\gamma}_0\) to our curve \(\displaystyle \boldsymbol{\gamma}\)

what do you mean by the zero in \(\displaystyle \boldsymbol{\gamma}_0\)? i'm just confused by the notation


View attachment 38485

What is the ISBN of your textbook?