Y 3 + X Y + X 2 = 0and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn't find a valid answer.Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.

Question

Y    3    +  X  Y  +      X    2    =  0and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn&#039;t find a valid answer.Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.

xgirlrogueim · Accepted Answer

Let   F  =      y    3    +  x  y  +      x    2  . Then      F    x    ′    =  y  +  2  x      F    y    ′    =  3      y    2    +  x            d      y              d      x        =  −            F      x      ′              F      y      ′        =  −            y      +      2      x              3              y        2            +      x      We know   x  =  10 and we need the corresponding y which is the solution of       y    3    +  10  y  +  100  =  0 which is cubic equation that you can solve using Cardano. This equation has a single real root given by  y  =                    10                  (                      2055                    −          45          )                    3              3              2                  /                3              −            10              2                  /                3                            3                  (                      2055                    −          45          )                    3        ≃  −  3.93003If you do not use Cardano, only a numerical method could be used. The simplest would be Newton, which, starting from a &quot;reason able&quot; guess       y    0   will update it according to      y          n      +      1        =      y    n    −            f      (              y        n            )                      f        ′            (              y        n            )      with   f  (  y  )  =      y    3    +  10  y  +  100. By inspection, you could see that there is a root close to −4; so, let us select       y    0    =  −  4. Soo, Newton iterates will be −3.93103,−3.93003.

planhetkk · Accepted Answer

Y    3    +  X  Y  +      X    2    =  0  3      Y    2    d  Y  +  Y  d  X  +  X  d  Y  +  2  X  d  X  =  0  (  3      Y    2    +  X  )  d  Y  =  −  (  Y  +  2  X  )  d  X            d      Y              d      X        =  −            Y      +      2      X              3              Y        2            +      X      Are you sure that you need more ? If yes, you will have to solve       Y    3    +  X  Y  +      X    2    =  0 for   Y and put it back into the above formula.

Answered question

Answer & Explanation

New Questions in Calculus 2