What is wrong with this derivation of the distance $\pi(a+b)$ around an ellipse? (From Strang's Calculus) I am self-studying Gil Strang's text Calculus. Exercise 30 in Section 8.2 reads:

The area of an ellipse is $\pi ab$. The area of a strip around it (width $\Delta$) is $\pi(a + \Delta)(b + \Delta) - \pi ab ≈ \pi(a + b)\Delta$. The distance around the ellipse seems to be $\pi(a + b)$. But this distance is impossible to find—what is wrong?

I am struggling to figure this out. It really seems that if one takes the limit of the area of the strip over $\Delta$ as $\Delta$ goes to zero, you get $\pi(a + b)$ as the distance around the ellipse.
Please help!
 A: I believe @heropup's comment is the key to understanding this puzzle. Let's however first apply this argument to a circle and also to a rectangle, to add to the confusion and then see what (if any) difference there is compared to the ellipse.

Given a circle of radius $r$, consider a circle of constant width $\delta$ from this circle; i.e a circle of radius $r+ \delta$. Then the difference in areas is $A(\delta) = \pi (r+ \delta)^2 - \pi r^2 = 2 \pi r \delta + \pi \delta^2$. Dividing by $\delta$ and taking the limit $\delta \to 0$ (i.e computing $A'(0)$) is given by
\begin{align}
A'(0) = 2 \pi r
\end{align}
which is exactly the circumference of the circle. 

Let's now try this for a rectangle of side lengths $a,b$. Let's consider now a rectangle which is a constant width $\delta$ apart. Note that this means the new rectangle will have dimensions $a+ 2 \delta$ and $b + 2 \delta$. Then the difference in areas is the area of the "slit", given by
\begin{align}
A(\delta) &= (a+ 2\delta)(b+2 \delta) - ab \\
&= 2 \delta(a+b) + 4 \delta^2 
\end{align}
This implies, $A'(0) = 2(a+b)$, which is once again the perimeter of the rectangle.

So why does this not work for the ellipse? Well we need to be more cautious of what we mean by "take a new figure of constant width $\delta$ apart". This idea was intuitively obvious for circles and rectangles that we didn't even have to think about what we were doing.
In the argument you presented, what you considered was two ellipses, the first ellipse $E_{a,b}$ having semi-axes of dimensions $a$ and $b$, while the second ellipse $E_{a+\delta, b + \delta}$ has dimensions $a+ \delta$ and $b + \delta$. If you think in terms of parametrizations, they are parametrized, for $t \in [0, 2\pi]$ as
\begin{align}
t \mapsto (a \cos t, b \sin t)
\end{align}
 and
\begin{align}
t \mapsto \left( (a+ \delta) \cos t, (b+ \delta) \sin t \right)
\end{align}
respectively. For any given $t \in [0, 2\pi]$ the following statement is true:

The distance between the points $(a \cos t, b \sin t)$ and $\left( (a+ \delta) \cos t, (b+ \delta) \sin t \right)$ is $|\delta|$.

However, if $t \notin \{0, \pi/2, \pi, 3 \pi/2, 2 \pi...\}$ (i.e the angle $t$ doesn't lie on one of the coordinate axes) then what happens is that if you're initially at the point $(a \cos t, b \sin t)$, and you draw the straight line which is normal to the ellipse $E_{a,b}$ at this point, then this normal line will intersect the ellipse $E_{a+\delta, b + \delta}$ at a point DIFFERENT from $\left( (a+ \delta) \cos t, (b+ \delta) \sin t \right)$ (draw a picture and this sentence will be very clear). 
This shows that the width is not a constant $\delta$ around the ellipse.
For the circle/rectangle on the other hand, the normal line to the first circle/rectangle will intersect the second circle/rectangle at the point you expect, hence they have a constant width. 
