# Integral and a slope problem

• May 19th 2008, 04:11 PM
333arizona
Integral and a slope problem
find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*

also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated. :)
• May 19th 2008, 04:21 PM
Mathstud28
Quote:

Originally Posted by 333arizona
find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*

also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated. :)

$\displaystyle \int\tan^2(x)+\sin(x)dx$

$\displaystyle \tan^2(x)+1=\sec^2(x)\Rightarrow\tan^2(x)=\sec^2(x )-1$

So $\displaystyle \int\tan^2(x)+\sin(x)dx=\int\sec^2(x)-1+\sin(x)dx=\tan(x)-x-\cos(x)+C$
• May 19th 2008, 04:32 PM
Mathstud28
Quote:

Originally Posted by 333arizona
find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*

also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated. :)

$\displaystyle 3x^2+2xy+6y^2-3x-8y=0$

differntiating implicity we get

$\displaystyle 6x+2[xy'+y]+12y\cdot{y'}-3-8y'=0$

Substituting $\displaystyle x=1$ and $\displaystyle y=1$

we get $\displaystyle 6(1)+2[y'+1]+12\cdot{y'}-3-8y'=0\Rightarrow{5+6y'=0}\Rightarrow{y'=\frac{-5}{6}}$