Expert: Alon Mandes - 11/29/2008

Question
QUESTION: Working some practice problems over the Thanksgiving break! Fun times eh? This one has me stumped: A particle is moving along the curve given by (xy^3/(1+y^2))=8/5. Assume that the x-coordinate is decreasing at a rate of 6 cm/sec when the particle is at the point (1,2). a) At what rate is the -coordinate of the point changing at that instant?

ANSWER: Hello Heather, sometimes a math break is needed during holidays..
Let's review a very helpful rule :
d/dt { f(x,y) } = [∂f/∂x][dx/dt]+[∂f/∂y][dy/dt]
=[∂f/∂x]x'(t)+[∂f/∂y]y'(y). Where " ∂ " resemble the partial derivative of the 2-variables function. Now, of course this rule can
be written in implicit function as well..here it is :
dy/dt=[dy/dx][dx/dt]
y'(t)=y'(x)*x'(t).
So, now all we have to find is y'(x) for the curve xy^3/(1+y^2)=8/5.

Let's solve the problem according to the 2nd method :
d/dx { xy^3/(1+y^2)=8/5 } gives us :
d/dx { xy^3/(1+y^2) } = d/dx { 8/5 } which also is :
d/dx { xy^3/(1+y^2) } = 0.
To perform the left part we have to keep in mind that :
d/dx {x} = 1   &   d/dx {y} = y'.
Ok then, let's oprate the d/dx operator :
y^3/(1+y^2) + x[3y^2*y'(1+y^2)-2y*y'*y^3]/(1+y^2)^4 = 0
y^3/(1+y^2) + xy'[3y^2*(1+y^2)-2y^4]/(1+y^2)^4 = 0
y'(x,y)= { y^3/(1+y^2) } / { x[3y^2*(1+y^2)-2y^4]/(1+y^2)^4 }.
y'(1,2)= { 2^3/(1+2^2) } / { 1[3*2^2*(1+2^2)-2*2^4]/(1+2^2)^4 }.
y'(1,2)= { 8/5 } / { 1[12*5-2*16]/(5)^4 }.
y'(1,2)= { 8/5 } / { 28/625 } = 150/7 = 35.71
OK, now we claim that :
y'(t)=y'(x)*x'(t) {At the point (1,2)}. So,
y'(t)=35.71*6
y'(t)=214.28. & this value is the y-coordinate rate change.

Alon.


---------- FOLLOW-UP ----------

QUESTION: WOW! thanks for the quick response. I got lost when you started taking the derivative of the left side. I may just be used to doing it another way. The image attached will show what I mean. Once I have the derivative can I just solve for dy/dt when x=1 and y=2? Where do we use the fact that dx/dt = -6?

Answer
Heather, the calculation you attached was perfectly kind of correct
with 2 mistakes :
1st : [f/g]'=[f'g-g'x]/g^2.
2nd : It should be dy/dx instead of dy/dt, because you are deriving
with respect to x & not with respect to t.
So, if you do so, you will get : y'=Function(x,y). substitute the
point (1,2) & you will have y' as a number. Then, you will have to
find y'(t) from : y'(t)=y'(x,y)*x'(t)={number}*6.
I apologize, I had a mistake in my previous calculation : It should
be (1+y^2)^2 & not (1+y^2)^4.
& another thing : This sort of derivatives is called "Implicit",
because y is not a function of x..but its from the kind of
G(x,y)=8/5. So, the method here is to derive both sides..In which
case, the derivative of constant will become zero, & that can give
you a way to separate y' as a function of x & y.

Any more question, I'm here.

Alon.