I have a question from the Book of Leithold. I have solved it but  i need a review for this problem. This is the question: Find an equation of the tangent to the curve $y=\frac{1}{3}x^2+2$ which is perpendicular to the line $x-y=0$. The statement in Portuguese is the same in English. the response of the book is $4x+4y-11=0$. What´s wrong with my solution?
May you see the question in this link. http://pt.scribd.com/doc/94701954/Questao-23-Leithold
 A: You reached $y=-x+\frac{11}{4}$. That is equivalent to the answer $4x+4y-11=0$ given in the book. Just multiply through by $4$ and bring everything over to the left-hand side. 
I had a bit of a hard time with your question. The main problem is that you typed the question incorrectly. At this time, your question reads, in part, 
"Find an equation of the tangent to the curve $y=\frac{1}{3}x^2+2$ which is perpendicular to the line $x-y=0$."
However, in your solution (in translation) the question reads
"Find an equation of the tangent to the curve $y=-\frac{1}{3}x^2+2$ which is perpendicular to the line $x-y=0$."
And indeed the answer corresponds to the choice $y=-\frac{1}{3}x^2+2$.
What you should have done:  Note that the slope of the line $x-y=0$ is $1$. Therefore the slope of any line perpendicular to $x-y=0$ is $-\frac{1}{1}$, which is $-1$. (If a line has $\ell$ has slope $m\ne 0$, then the slope of any line perpendicular to $\ell$ is $-\frac{1}{m}$.)
Now we calculate the slope of the tangent line to $y=-\frac{1}{3}x^2+2$ at the general point with coordinate $x$. We have $\frac{dy}{dx}=-\frac{2}{3}x$.
Thus we want $-\frac{2}{3}x=-1$, giving $x=\frac{3}{2}$. 
If $x=\frac{3}{2}$, then the corresponding $y$ is $\frac{5}{4}$. Thus our tangent line has equation 
$$y-\frac{5}{4}=(-1)\left(x-\frac{3}{2}\right).$$
This simplifies to your $y=-x+\frac{11}{4}$, and also to the answer in the book. 
Remark: There are a number of mistakes in the handwritten solution that ou link to. These happen to cancel! For example, you end up setting $-\frac{2}{3}x$ equal to $1$ (not $-1$). Thus you end up with the wrong value of $x$. Later, a little sign slip cancels the mistake. I would urge you to look in detail at the solution given above.
