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

[mathmari]

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:

 

[I like Serena]

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 $\phi'(y)=1.5\,\frac{\text{minute}}{\text{km}}$.

 

[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 $\phi'(y)=1.5\,\frac{\text{minute}}{\text{km}}$.

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:

 

[I like Serena]

How about the mean value theorem instead?

 

[mathmari]

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?

 

[I like Serena]

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 $\phi(80\,\text{km})=120\,\text{minutes}$?

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:

 

[mathmari]

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*}

$\phi$ 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:

 

[I like Serena]

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.

 

[mathmari]

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:

 

[I like Serena]

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.

 

[mathmari]

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:

 

[I like Serena]

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.

 

[mathmari]

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:

 

[I like Serena]

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.

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.

 

[mathmari]

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:

 

[I like Serena]

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.

 

[mathmari]

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: