Expert: Paul Klarreich - 5/9/2007

Question
Function y=f(x) is defined by 5sqrt(x)+xy+y^3=11?
a). find an expression for f'(x)=dy/dx in terms of x and y.
b). find the equation of the line that is tangent to the graph of y=f(x) at the point (0.25,2)
c). use the tangent line from part (b) to estimate f(0.6).
d). write an equation whose solution is the exact value of f(0.6).
to the nearest thousandth, what is f(0.6)?
e). would it be appropriate to use the tangent line from part (b) to estimate f(-0.1)? explain
can you please explain this problem to me
thanks

Answer
Questioner:   bhavika
Category:  Calculus
Private:  No
 
Subject:  calculus
Question:  Function y=f(x) is defined by 5sqrt(x)+xy+y^3=11?
a). find an expression for f'(x)=dy/dx in terms of x and y.
b). find the equation of the line that is tangent to the graph of y=f(x) at the point (0.25,2)
c). use the tangent line from part (b) to estimate f(0.6).
d). write an equation whose solution is the exact value of f(0.6).
to the nearest thousandth, what is f(0.6)?
e). would it be appropriate to use the tangent line from part (b) to estimate f(-0.1)? explain
can you please explain this problem to me
thanks
................................
Hi, Bhavika,

Your f(x) is defined by   5 sqrt(x) + xy + y^3 = 11

a) Use Implicit Differentiation:

First term: 5 sqrt(x) = 5x^1/2
D(5x^1/2) = 5(1/2)x^-1/2 = 5/2sqrt(x)

Second term: xy.  Use the product rule:

D(xy) = x(dy/dx) + y(1)

Third term:  y^3.  Use the chain rule:

D(y^3) = 3y^2 dy/dx

Fourth term:  11.  Derivative is zero.

The differentiated equation is:

5/2sqrt(x) +  x dy/dx + y + 3y^2 dy/dx = 0

Solve for dy/dx:

 x dy/dx  + 3y^2 dy/dx = - 5/2sqrt(x) - y

 dy/dx(x + 3y^2) = - 5/2sqrt(x) - y
                         5
 dy/dx(x + 3y^2) = - --------- - y
                     2 sqrt(x)

                     5 + 2y sqrt(x)
 dy/dx(x + 3y^2) = - --------------
                       2 sqrt(x)

               5 + 2y sqrt(x)
 dy/dx = - ---------------------
            (x + 3y^2) 2 sqrt(x)
....................
b). find the equation of the line that is tangent to the graph of y=f(x) at

the point (0.25,2)

Find the slope from the above value of dy/dx:

               5 + 2(2) sqrt(1/4)
 dy/dx = - ---------------------
            (1/4 + 3(4)) 2 sqrt(1/4)

               5 + 2
 dy/dx = - ------------------
            (1/4 + 12) (1)

                7
 dy/dx = - ------------
            (1/4 + 12)

                28
 dy/dx = - ------------
            (1 + 48)

            28       4
 dy/dx = - ---- = - ---
            49       7

Now the line has this equation:

y - 2 = -4/7(x - 1/4)
       -4x + 1
y - 2 = -------
          7

    -4x + 1
y  = ------- + 2
       7
    -4x + 15
y  = ---------
       7
..............................
c). use the tangent line from part (b) to estimate f(0.6).

Substitute  x = 0.6 into the answer:

    -4(0.6) + 15
y  = -------------
       7

    -2.4 + 15
y  = ----------
         7

    12.6
y  = ----- = 1.8
      7
................................................
d). write an equation whose solution is the exact value of f(0.6).
to the nearest thousandth, what is f(0.6)?

This would be the original equation, with  x = 0.6:

5 sqrt(0.6) + (0.6)y + y^3 = 11

I don't see an easy way to solve this equation.  Let's leave that aside for now.

....................................
e). would it be appropriate to use the tangent line from part (b) to estimate f(-0.1)? explain

'Appropriate' means 'Would you get a decent answer?'  I think the answer to that is NO.  If the graph between  x = 0.25 and  x = -0.1  is well-behaved, then the answer is YES.

But in between  x = 0.25 and x = -0.1 lies  x = 0.  And the value of dy/dx at  x = 0 would be:

               5 + 2y sqrt(0)
 dy/dx = - ---------------------
            (0 + 3y^2) 2 sqrt(0)

which has a zero denominator, so is undefined.  So the graph is not well-behaved between your two points and the result will be unpredictable.