Expert: Scotto - 3/21/2009

Question
1. Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=16 at the point (8,1). The equation of this tangent line can be written in the form y=mx=b where m is: ______ and where b is: ________

2. find the slope of the tangent line to the curve
-3x^2-4xy-2y^3=-12 at the point (-2,2) .

3. Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at the point (3,-1).

4. If 3x^2+3x+xy=4 and y(4)=-14, find y'(4) by implicit differentiation.

5. If (x^2)/(49) + (y^2)/(16)+1 and y(2)=3.83326, find y'(2) by implicit differentiation.  

Answer
1. Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=16 at the point (8,1). The equation of this tangent line can be written in the form y=mx=b where m is: ______ and where b is: ________

Doing implicit differentiation on the equation given,
we get (using the product rule and assuming y is the function and x is the variable)
x(3y²)y’ + y^3 + xy’ + y = 0.

First, lets verify that (8,1) is on the curve:  
(8)(1) + 8(1) = 16 – yep, it’s there.

Solving for y’ gives us y’(3xy² + x)=-(y^3 + y),
or y’ = -(y^3 + y)/(3xy² + x).
Given that x is 8 and y is 1, we get
y’ = -(1 + 1)/(3*8 + 8) = -1/16.
We have a point and a slope, so the line is (y-y0) = m(x-x0).
Here, we have y – 1 = -(x-8)/16.
Working that out gives y – 1 = -x/16 – ½, or y = -x/16 + ½.


2. Find the slope of the tangent line to the curve
-3x^2 - 4xy - 2y^3 = -12 at the point (-2,2) .

I’ll start this problem by verifying the (-2,2) is on the curve.
-3(4) – 4(-2)2 – 2(8) = -12 + 16 –12 = -12: yeah, its there.

OK, differentiating gives us –6x –4xy’ – 4y –6y²y’ = 0.
Again, lets solve for y’  To start with factor it out and move the other terms over.

You get y’(-4x-6y²) = 6x + 4y.  Doing the division gives us
y’ = (6x+4y)/(-4x - 6y²) = -(3x+2y)/(2x+3y²).
It sure looks like something would cancel, but that’s not it.

Putting in (-2,2) gives y’ = -(-6+4)/(-4+12) = 2/8 = 1/4.
Putting these into the point-slope form gives us
y-2 = (1/4)(x+2).  Multiplying this out gives us
y = x/4 + 1/2 + 2 = x/4 + 5/2 = (x+10)/4.
All of the linear equations in the line above are right.
It just depends on which one you need.

Yet this was silly – all the question asked for was the slope,
and I could have ended there: its 1/4.


3. Find the slope of the tangent line to the curve (a lemniscate)
2(x^2 + y^2)² = 25(x^2 - y^2) at the point (3,-1).

Let’s start off by looking at what the questions asking for:
the slope.  All we have to do is find y’ and put in x and y.

First, lets take the derivative of that lemniscale.
4(x² + y²)(2x + 2yy’) = 50x – 50yy’.  That wasn’t so bad.

We now have to multiply things out some.
That gives 8x(x² + y²) + 8yy’(x² + y²) = 50x – 50yy’.
Putting y’ on one side gives
8yy’(x² + y²) + 50yy’ = 50x - 8x(x² + y²).
Factoring out y’ gives y’(8y(x² + y²) + 50y) = 50x - 8x(x² + y²).
Doing division gives y’ = (50x - 8x(x² + y²).)/(8y(x² + y²) + 50y).
Now I’m thinking about finding the line,
but all we need is the slope.

Take y’ = (50x - 8x(x² + y²))/(8y(x² + y²) + 50y)
and put in (3,-1) for (x,y).
That gives (150 – 24(10))/(8(10)) = -90/80 = -9/8.


4. If 3x² + 3x + xy = 4 and y(4) = -14,
find y'(4) by implicit differentiation.

Implicit differentiation gives us
6x + 3 + {yeah, a product rule } xy’ + y = 0.
So xy’ = -(y + 6x + 3), then y’ = -(y + 6x + 4)/x.

Given x is 4 and y is 14, y’ = -(14 + 24 + 4)/4 = -21/2 = -10 ½.


4. If 3x² + 3x + xy = 4 and y(4)=-14,
find y'(4) by implicit differentiation.
Implicit differentiation gives us
6x + 3 + {yeah, a product rule } xy’ + y = 0.
So xy’ = -(y + 6x + 3), then y’ = -(y + 6x + 4)/x.

Given x is 4 and y is 14, y’ = -(14 + 24 + 4)/4 = -21/2 = -10 ½.


5. I will assume the equation was suppose to be
x²/49 + y²/16 = 1 and y(2)=3.83326,
find y'(2) by implicit differentiation.

This is (x/7)² + (y/4)² = 1.  Again, lets check it out.
If x is 2, we have 4/49 + y²/16 = 1, so y² = 45/49.
This says that y = √(780/16) = √195 / 2 = 6.98212,
which is not what was given.

Without the proper equation, I can’t solve this problem.
Basically, the solution is similar to the other four problems.
You have the proper equation, so just go through the steps.