* SOLUTION 13 :* Begin with
*x*^{2} + *xy* + *y*^{2} = 1 . Differentiate both sides of the equation, getting

*D* ( *x*^{2} + *xy* + *y*^{2} ) = *D* ( 1 ) ,

2*x* + ( *xy*' + (1)*y* ) + 2 *y y*' = 0 ,

so that (Now solve for *y*' .)

*xy*' + 2 *y y*' = - 2*x* - *y* ,

(Factor out *y*' .)

*y*' [ *x* + 2*y* ] = - 2 *x* - *y* ,

and the first derivative as a function of *x* and *y* is

(Equation 1)

.

To find *y*'' , differentiate both sides of this equation, getting

.

Use Equation 1 to substitute for *y*' , getting

(Get a common denominator in the numerator and simplify the expression.)

.

This answer can be simplified even further. Note that the original equation is

*x*^{2} + *xy* + *y*^{2} = 1 ,

so that

(Equation 2)

*x*^{2} + *y*^{2} = 1 - *xy* .

Use Equation 2 to substitute into the equation for *y*'' , getting

,

and the second derivative as a function of *x* and *y* is

.

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* SOLUTION 14 :* Begin with
*x*^{2/3} + *y*^{2/3} = 8 . Differentiate both sides of the equation, getting

*D* ( *x*^{2/3} + *y*^{2/3} ) = *D* ( 8 ) ,

*D* ( *x*^{2/3} ) + *D* ( *y*^{2/3} ) = *D* ( 8 ) ,

(Remember to use the chain rule on
*D* ( *y*^{2/3} ) .)

(2/3)*x*^{-1/3} + (2/3)*y*^{-1/3} *y*' = 0 ,

so that (Now solve for *y*' .)

(2/3)*y*^{-1/3} *y*' = - (2/3)*x*^{-1/3} ,

,

and

,

Since lines tangent to the graph will have slope $ -1 $ , set *y*' = -1 , getting

,

- *y*^{1/3} = -*x*^{1/3} ,

*y*^{1/3} = *x*^{1/3} ,

( *y*^{1/3} )^{3} = ( *x*^{1/3} )^{3} ,

or

*y* = *x* .

Substitue this into the ORIGINAL equation
*x*^{2/3} + *y*^{2/3} = 8 , getting

*x*^{2/3} + (*x*)^{2/3} = 8 ,

2 *x*^{2/3} = 8 ,

*x*^{2/3} = 4 ,

( *x*^{2/3} )^{3} = 4^{3} ,

*x*^{2} = 64 ,

or

.

If *x*=8 , then *y*=8 , and the tangent line passing through the point (8, 8) has slope -1 .

If *x*=-8 , then *y*=-8 , and the tangent line passing through the point (-8, -8) has slope -1 .

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* SOLUTION 15 :* Since the equation
*x*^{2} - *xy* + *y*^{2} = 3 represents an ellipse, the largest and smallest values of *y* will occur at the highest and lowest points of the ellipse. This is where tangent lines to the graph are horizontal, i.e., where the first derivative *y*'=0 . The largest and smallest values of *x* will occur at the right-most and left-most points of the ellipse. This is where tangent lines to the graph are vertical, i.e., where the first derivative *y*' does not exist.

Begin by taking the derivative of both sides of the equation, getting

*D* ( *x*^{2} - *xy* + *y*^{2} ) = *D* ( 3 ) ,

*D* ( *x*^{2} ) - *D* ( *xy* ) + *D* ( *y*^{2} ) = *D* ( 3 ) ,

(Remember to use the chain rule on
*D* ( *y*^{2} ) .)

2*x* - ( *xy*' + (1)*y* ) + 2 *y y*' = 0 ,

so that (Now solve for *y*' .)

2*x* - *xy*' - *y* + 2 *y y*' = 0 ,

2 *y y*'- *xy*' = *y* - 2*x* ,

(Factor out *y*' .)

*y*' [ 2*y* - *x* ] = *y* - 2*x* ,

and

(Equation 1)

.

If the first derivative *y*'=0 , then (If
, then *A*=0 .)

*y* - 2*x* = 0 ,

so that

*y* = 2*x* .

Substituting this into the original equation
*x*^{2} - *xy* + *y*^{2} = 3 leads to

*x*^{2} - *x* (2*x*) + (2*x*)^{2} = 3 ,

*x*^{2} - 2*x*^{2} + 4*x*^{2} = 3 ,

3*x*^{2} = 3 ,

*x*^{2} = 1 ,

and

.

Thus, the maximum value of *y* occurs when *x*=1 and *y*=2 , i.e., at the point

(1, 2) .

The minimum value of *x* occurs when *x*=-1 and *y*=-2 , i.e., at the point

(-1, -2) .

If the first derivative *y*' does not exist, then the denominator in Equation 1 equals zero (Why?), i.e.,

2*y* - *x* = 0 ,

so that

*x* = 2*y* .

Substituting this into the original equation
*x*^{2} - *xy* + *y*^{2} = 3 leads to

(2*y*)^{2} - (2*y*) *y* + *y*^{2} = 3 ,

4 *y*^{2} - 2*y*^{2} + *y*^{2} = 3 ,

3*y*^{2} = 3 ,

*y*^{2} = 1 ,

and

.

Thus, the maximum value of *x* occurs when *y*=1 and *x*=2 , i.e., at the point

(2, 1) .

The minimum value of *x* occurs when *y*=-1 and *x*=-2 , which occurs at the point

(-2, -1) .

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* SOLUTION 16 :* Begin with
(*x*^{2}+*y*^{2})^{2} = 2*x*^{2}-2*y*^{2} . Differentiate both sides of the equation, getting

*D* (*x*^{2}+*y*^{2})^{2} = *D* ( 2*x*^{2}-2*y*^{2} ) ,

*D* (*x*^{2}+*y*^{2})^{2} = *D* ( 2*x*^{2} ) - *D* ( 2*y*^{2} ) ,

(Remember to use the chain rule on *D* ( *y*^{2} ) and
*D* ( 2*y*^{2} ) .)

2 (*x*^{2}+*y*^{2}) *D* (*x*^{2}+*y*^{2}) = 4*x* - 4 *y y*' ,

2 (*x*^{2}+*y*^{2}) (2*x*+2*yy*') = 4*x* - 4 *y y*' ,

so that (Now solve for *y*' .)

4*x* (*x*^{2}+*y*^{2}) + 4*y*(*x*^{2}+*y*^{2}) *y*' = 4*x* - 4 *y y*' ,

4 *y y*' + 4*y*(*x*^{2}+*y*^{2}) *y*' = 4*x* - 4*x* (*x*^{2}+*y*^{2}) ,

4 [ *y y*' + (*x*^{2}*y*+*y*^{3}) *y*'] = 4 [ *x* - (*x*^{3}+*xy*^{2}) ] ,

(Divide out 4 .)

*y y*' + (*x*^{2}*y*+*y*^{3}) *y*' = *x* - *x*^{3}-*xy*^{2} ,

(Factor out *y*' .)

*y*' [ *y* + (*x*^{2}*y*+*y*^{3}) ] = *x* - *x*^{3}-*xy*^{2} ,

and

(Equation 1)

.

If *y*'=0 , then (If
, then *A*=0 .)

*x* - *x*^{3}-*xy*^{2} = 0 ,

*x* (1 - *x*^{2}-*y*^{2}) = 0 ,

and

(**)

*x* = 0 or
1 - *x*^{2}-*y*^{2} = 0 .

If *x*=0 in the original equation
(*x*^{2}+*y*^{2})^{2} = 2*x*^{2}-2*y*^{2} , then

(0+*y*^{2})^{2} = 0-2*y*^{2} ,

*y*^{4} + 2*y*^{2} = 0 ,

*y*^{2} ( *y*^{2} + 1 ) = 0 ,

and

*y*=0 .

Note, however, if *x*=0 and *y*=0 are substituted into Equation 1, we get the indeterminate form " 0/0 " . Is *y*'=0 at the point (0, 0) , i.e., does
*y*'(0, 0) = 0 ? We will show that the value of the derivative is NOT zero at (0, 0) . Assume that *y*' is continuous at (0, 0) , and that
*y*'(0, 0) = 0 . Thus,

(Equation 2)

.

But we also have that

(The limit on the right-hand side takes the indeterminate form " 0/0 " , so use L'Hopital's Rule, remembering that *y* is an impicit function of *x* .)

(Evaluate the limit, and use Equation 2.)

.

Since the numerator approaches 1 and the denominator approaches 0 , this limit DOES NOT EXIST. But this contradicts Equation 2. Thus, our assumption in Equation 2 is false, and
for *x*=0 . (NOTE: It can be shown using polar coordinates that the slope of the graph as it passes through the origin (twice !) is +1 and -1 . Returning to (**), we next have that

1 - *x*^{2}-*y*^{2} = 0 ,

or

*y*^{2} = 1 - *x*^{2} .

Substitute this into the original equation
(*x*^{2}+*y*^{2})^{2} = 2*x*^{2}-2*y*^{2} , getting

( *x*^{2}+(1 - *x*^{2}) )^{2} = 2*x*^{2}-2(1 - *x*^{2}) ,

1 = 2*x*^{2} - 2 + 2 *x*^{2} ,

3 = 4*x*^{2} ,

*x*^{2} = 3/4 ,

or

.

If
, then
. Thus *y*'=0 at the following points :

,

,

,

and

.

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