Expert: Scotto - 3/13/2009

Question
(x2/a2)+(y2/b2)=1 is the equation for an ellipse with the verticies (+-a,0)(0,+-b) and a slope m =-0.5 in the point (1,1).
a) Find a and b
b)Suppose there was an error in measurement of the slope m of at most 0.01. Use differentials to approximate the maximum error incurrec when calculating the ellipse's area. what is the maximum relative error?
c)suppose that the verticies of the ellipse from part a start moving. If the slope in the point (1,1) decreases at a rate of .5 units/second, at what rate is the area of the ellipse changing after one second?

Thank you!

Answer
For an ellipse, the equation is really suppose to be
(x/a)² + (y/b)² = 1.

For (1,1) to be on the graph, we need (1/a)² + (1/b)² = 1.
This can be transformed into 1/a² + 1/b² = 1.
We can then say the 1/a² = 1 - 1/b² = (b²-1)/b².
From here, we can ignore the middle up above and invert the
expressions on the left and right, giving a² = b²/(b²-1).

Now we know that the slope at (1,1) is -0.5.  
Using our equation and taking the derivative implicitly gives
2x/a² + 2yy'/b² = 0.  This says that y'(2y/b²) = - 2x/a².
Mulitplying both sides by the inverse of 2y/b² gives the answer
y' = -xb²/(ya²).  Since we are at the point (1,1), this is
-0.5 = -b²/a², which goes to a² = 2b².

Putting this back in the equation found at the end of the paragagraph above the last would say that 2b² = b²/(b²-1).
Multiply by the inverse of the left side and get 2(b²-1) = 1.
This converts to b²-1 = 0.5, which goes to b² = 1.5.  So we know that b=±√1.5.

Now since from the pargraph before, which ended with a² = 2b², we can put b in here and find out that a² = 2(1.5) = 3.  This means that a=±√3.

a) The variables a and b are given in the equation
(x/a)² + (y/b)² = 1.
When x is 0, (y/b)² = 1, so y² = b², so y = ±b.
When y is 0, (x/a)² = 1, so x² = a², so x = ±a.
Those are the points on the a and b axis where x and y reach the maximum and minimum.

b) The area of an ellipse is almost like a circle.
A cirlce's area is πr².  An ellipse's area is almost the same.
The variable a and b are for the short and long radius of the ellipse (or long and short).  So the area is πab.

Note that if a=√3 and b=√1.5, then (1,1) is on the graph, for
(x/a)² + (y/b)² would then be 1/3 + 1/2.25, which is the same as
3/9 + 4/9.  We can see that this is 7/9.

c) We know that the area is abπ.   The derivative with respect to both and b varying is a’bπ + ab’π = π(a’b +b’a).  
Now if  we put in a=√3 and b = √1.5 with a’ and b’ = 0.1, we get
0.1*π(√3+√1.5) =0.15π√2, which is approximately 0.067.