Show that there is a 1-km section that he covers in 1.5 minutes

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In summary: But why do we need the continuity of the inverse function?To apply the mean value theorem.Otherwise the function could jump so that there isn't a point where we drive 1.5 minute/km....which wouldn't make sense.
  • #1
mathmari
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Hey! :giggle:

A cyclist, constantly driving forward, covers a distance of $80$ km in exactly $2$ hours. Show that there is a $1$-kilometer section of this journey that he covers in exactly $1.5$ minutes.
Hint: Continuity of the inverse function.

So we have to define a function $\phi :T\rightarrow K, \ x\mapsto \frac{120x}{80}$, where $T$ is the set of time and $K$ is the set of kilometer. Or can we not just define the function like that?

Then we want to show that there is a $y$ such that $\phi (y)=1,5$ and that $y=1$, right?

:unsure:
 
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  • #2
Hey mathmari!

I think you mean a function $\phi: K \to T$, where $K$ is the set of distances and $T$ is the set of times, with the boundary conditions $\phi(0)=0$ and $\phi(80\,\text{km})=120\,\text{minutes}$, and it is an increasing function.
We don't know what that function is. The function you gave is just an example of what it could be. 🤔

Then we would want to show that there is an $y$ such that $\phi'(y)=1.5\,\frac{\text{minute}}{\text{km}}$. 🤔
 
  • #3
Klaas van Aarsen said:
I think you mean a function $\phi: K \to T$, where $K$ is the set of distances and $T$ is the set of times, with the boundary conditions $\phi(0)=0$ and $\phi(80\,\text{km})=120\,\text{minutes}$, and it is an increasing function.
We don't know what that function is. The function you gave is just an example of what it could be. 🤔

Then we would want to show that there is an $y$ such that $\phi'(y)=1.5\,\frac{\text{minute}}{\text{km}}$. 🤔

Ahh! :geek:

It holds that : If $f(x)$ is continuous on its domain, and a injective function, then $f^{-1}$ is a function that is also continuous on its domain.
But do we have information to show that?
Given the values $\phi(0)=0$ and $\phi(80)=120$ it seems like IVT, but I don't think that it helps us here..

Could you give me a hint?

:unsure:
 
  • #4
How about the mean value theorem instead? 🤔
 
  • #5
Klaas van Aarsen said:
How about the mean value theorem instead? 🤔

Then we get that there is a c in $(0,120)$ such that $\phi '(c) =\frac{\phi(120)-\phi(0)}{120-0}=\frac{80}{120}$.

How is this relates to the inverse function?
 
  • #6
mathmari said:
Then we get that there is a c in $(0,120)$ such that $\phi '(c) =\frac{\phi(120)-\phi(0)}{120-0}=\frac{80}{120}$.

Don't we have $c$ in $(0,\,80\,\text{km})$ and $\phi(80\,\text{km})=120\,\text{minutes}$? 🤔

mathmari said:
How is this relates to the inverse function?
$\phi$ is the inverse function of the usual distance as function of time, isn't it? :unsure:
 
  • #7
Klaas van Aarsen said:
Don't we have $c$ in $(0,\,80\,\text{km})$ and $\phi(80\,\text{km})=120\,\text{minutes}$? 🤔

Oh yes... We have \begin{equation*}\phi '(c)={\frac {\phi (80)-\phi (0)}{80-0}}={\frac {120-0}{80-0}}={\frac {120}{80}}={\frac {120}{80}}=1.5\end{equation*}
Klaas van Aarsen said:
$\phi$ is the inverse function of the usual distance as function of time, isn't it? :unsure:

Yes. But we have that the derivative at $x_0$ is equal to $1.5$. :unsure:
 
  • #8
mathmari said:
Yes. But we have that the derivative at $x_0$ is equal to $1.5$.
Indeed. So we're using the continuity of the inverse of the distance function, combined with the fact that it's differentiable. 🤔
 
  • #9
Klaas van Aarsen said:
Indeed. So we're using the continuity of the inverse of the distance function, combined with the fact that it's differentiable. 🤔

I am confused about that right now. So do we use the formula of the derivative of the inverse function? :unsure:
 
  • #10
mathmari said:
I am confused about that right now. So do we use the formula of the derivative of the inverse function?
What do you mean by the formula of the derivative? 🤔

We're using the formula of the mean value theorem. 🤔
 
  • #11
Klaas van Aarsen said:
What do you mean by the formula of the derivative? 🤔

We're using the formula of the mean value theorem. 🤔

Yes, by the mean value theorem we have shown that there is a $y$ such that $\phi'(y)=1.5$.

But what does this mean? I am confused that we have now the derivative of $\phi$ :unsure:
 
  • #12
mathmari said:
Yes, by the mean value theorem we have shown that there is a $y$ such that $\phi'(y)=1.5$.

But what does this mean? I am confused that we have now the derivative of $\phi$
It means that there is a location in the trajectory that we traverse with $1.5$ minutes per km.
Put otherwise, there is a section of 1 km that takes 1.5 minutes. 🤔
 
  • #13
Klaas van Aarsen said:
It means that there is a location in the trajectory that we traverse with $1.5$ minutes per km.
Put otherwise, there is a section of 1 km that takes 1.5 minutes. 🤔

How do we see that the section is of length 1 km ?

But why do we need the continuity of the inverse function?

:unsure:
 
  • #14
mathmari said:
How do we see that the section is of length 1 km ?

Because that is how we choose our function $\phi$.
The derivative is supposed to represent the time it takes to complete a section of 1 km as function of distance. 🤔

mathmari said:
But why do we need the continuity of the inverse function?
To apply the mean value theorem.
Otherwise the function could jump so that there isn't a point where we drive 1.5 minute/km. 🤔
 
Last edited:
  • #15
Klaas van Aarsen said:
To apply the mean value theorem.
Otherwise the function could jump so that there isn't a point where we drive 1.5 minute/km. 🤔

So we can use the derivative, i.e. we know that $\phi$ is differentiable because of the continuity of the inverse function ? :unsure:
 
  • #16
mathmari said:
So we can use the derivative, i.e. we know that $\phi$ is differentiable because of the continuity of the inverse function ?
It is given that we continuously drive forward.
That means that the distance as function of time is continuously differentiable everywhere with a positive derivative (positive velocity).
The inverse function theorem tells us that the inverse of the distance function (which we called $\phi$) is then also continuously differentiable everywhere. 🤔
 
  • #17
Klaas van Aarsen said:
It is given that we continuously drive forward.
That means that the distance as function of time is continuously differentiable everywhere with a positive derivative (positive velocity).
The inverse function theorem tells us that the inverse of the distance function (which we called $\phi$) is then also continuously differentiable everywhere. 🤔

I got it now! Thanks! :geek:
 

Related to Show that there is a 1-km section that he covers in 1.5 minutes

1. How do you determine the speed of the person in the given scenario?

The speed of the person can be calculated by dividing the distance traveled (1 km) by the time taken (1.5 minutes). This will give the speed in kilometers per minute. To convert it to a more commonly used unit, such as kilometers per hour, simply multiply the result by 60.

2. Can you explain the concept of speed in more detail?

Speed is a measure of how fast an object is moving. It is calculated by dividing the distance traveled by the time taken. It is typically measured in units such as kilometers per hour, miles per hour, or meters per second.

3. Is it possible for the person to cover the 1-km section in a shorter amount of time?

Yes, it is possible for the person to cover the 1-km section in a shorter amount of time. The given time of 1.5 minutes is just an example and the actual time taken will depend on the person's speed.

4. How accurate is this calculation of speed?

The accuracy of this calculation depends on the accuracy of the distance and time measurements. If the measurements are precise, then the calculated speed will also be precise. However, factors such as external forces (e.g. wind resistance) and human error can affect the accuracy of the calculation.

5. Can this calculation of speed be applied to all types of movement?

No, this calculation of speed is specifically for linear motion where the object is moving in a straight line. It cannot be applied to other types of movement such as circular or rotational motion.

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