SOLUTION: 1. Let In(x^2 - 3y) = x - y - 1define a differentiable function y of x. Find an equation of the line tangent to the graph of the equation at the point (2,1) 2. Water is poured i

 
Let ln(x^2 - 3y) = x - y - 1define a differentiable 
function y of x. Find an equation of the line tangent 
to the graph of the equation at the point (2,1) 


                ln(x² - 3y) = x - y - 1

Differentiate each term implicitly (i.e., without solving 
for y)

On the left use d[ln(u)]/dx = u'/u


                   2x - 3y'
                 ---------- = 1 - y' - 0
                   x² - 3y

                   2x - 3y'
                 ---------- = 1 - y'
                   x² - 3y

Substitute (x.y) = (2,1)

               2(2) - 3y'
             -------------- = 1 - y'
               (2)² - 3(1)

                    4 - 3y'
                  --------- = 1 - y'
                     4 - 3

                    4 - 3y'
                  --------- = 1 - y'
                       1

                    4 - 3y' = 1 - y'

                       -2y' = -3

                         y' = 3/2

So the slope m of the tangent line is m = 3/2

So the tangent line has slope m = 3/2 and passes 
through the point (2,1) 

So we use the point-slope form:

                     y - y1 = m(x - x1)

                      y - 1 = (3/2)(x - 2)

                      y - 1 = (3/2)x - 3

                          y = (3/2)x - 2 

Or if you want it in the standard form of a line:

                   3x - 2y = 4 

----------------------                   


2. Water is poured into a conical cup at the rate of 2/3 
cubic inches per second. If the cup is 6 inches tall and 
the top of the cup has a radius of 2 inches, how fast is 
the water level rising when the water is 4 inches deep?

Let's assume this is its largest cross-section,

 _______________
 \   2" |  2"  /
  \     |     / 
   \    |6"  / 
    \   |   /
     \  |  / 
      \ | /    
       \|/      
             

Now we'll freeze it at some unknown arbitrary water
level, x.  Let the radius of the circle which is the
circle of the surface of the water level at x be y.
     
 _______________
 \   2" |  2"  /
  \     |     / 
   \    |6"  / 
    \___|_y_/_
     \  |  / |
      \ |x/  |x   
       \|/   |   
             ¯

The volume of the water at height x and radius of
surface circle y is 

V = pr²h/3

V = py²x/3

This has 3 unknowns, so we need to reduce the number
of unknowns to 2 by finding a way to relate two of them.
By similar triangles,

y/x = 2/6

So this gives y = x/3

Substituting that in

V = py²x/3

gives

V = p(x/3)²x/3

V = p(x²/9)x/3

V = px³/27

V = (p/27)x³

Now take the derivative with respect to time t
Don't forget the chain rule, i.e., to take the
derivative of the "inside", when the "inside"
is not what you are taking the derivative with
respect to.

dV/dt = 3(p/27)x²·dx/xt

dV/dt = (p/9)x²·dx/dt

We are given that dV/dt = 2/3 in³/min 

So we substitute that and freeze it at x=4,
which means to substitute those in:

2/3 = (p/9)4²·dx/dt

2/3 = (16p/9)·dx/dt)

3/(8p) = dx/dt

dx/dt = .1193662073 in/min

Edwin