HW6 Solutions Section 3.5

This is the solutions to homework 6. For now there are only solutions. But I will slowly add errors to avoid.

Problems with a strike through them have not yet been answered.
3.5 #2a,2b,2c, 7,12,24,26,52
3.6 #6,11,12,16,20,24,30,40,48,50
3.9 #4,30

3.5 #2a
$8x+18y\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{4x}{9y}$

Point Loss:
Failing to simplify $\frac{8}{18}$

3.5 #2b

Error:
$\sqrt{4-\frac{4}{9}x^2}\neq\sqrt{4}-\sqrt{\frac{4}{9}x^2}$

3.5 #2c

3.5 #7
$2x+(y+x\frac{dy}{dx})-2y\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{2x+y}{2y-x}$

3.5 #12
$1=\cos\left(xy^2\right)\left(y^2+x\left(2y\frac{dy}{dx}\right)\right)$
$\frac{dy}{dx}=\frac{1}{2xy}\left(\frac{1}{\cos\left(xy^2\right)}-y^2\right)$

3.5 #24
$\frac{dy}{dx}\sec(x)+y(\sec(x)\tan(x))=\tan(y)+x\left(\sec(y)\frac{dy}{dx}\right)$
$\frac{dy}{dx}(\sec(x)-x\sec(y)=\tan(y)-y\sec(x)\tan(x)$
$\frac{dy}{dx}=\frac{\tan(y)-y\sec(x)\tan(x)}{\sec(x)-x\sec(y)}$

3.5 #26
$2x+(2y+2xy^{\prime})-2yy^{\prime}+1=0$
At (x,y)=(1,2)
$2+(4+2y^{\prime})-4y^{\prime}+1=0$
$y^{\prime}=\frac{7}{2}$
$y-2=\frac{7}{2}(x-1)$

Errors:
$\frac{d}{dx}(2xy)\neq 2x\frac{dy}{dx}$
Correction: $\frac{d}{dx}(2xy)=2y+2x\frac{dy}{dx}$
$\frac{d}{dx}(2xy)\neq 2\frac{dy}{dx}$
Correction: $\frac{d}{dx}(2xy)=2y+2x\frac{dy}{dx}$
$\frac{d}{dx}(2xy)\neq y+x\frac{dy}{dx}$
Correction: $\frac{d}{dx}(2xy)=2y+2x\frac{dy}{dx}$
Forgetting to differentiate the fourth term, $x$.
Forgetting to carry a term from one line to the next.
Sign errors.

Point Loss:
Answering half the problem.

3.5 #57
$\frac{d}{dx}\left(\cos^{-1}(x)\right)=\mbox{?}$
Let $y=\cos^{-1}(x)$.
$\frac{dy}{dx}=\mbox{?}$
$y=\cos^{-1}(x)$
Note that $y\in\mbox{range of }\cos^{-1}(x)=[0,\pi]$
$\cos(y)=x$
Take the derivative with respect to x on both sides.
$-\sin(y)\frac{dy}{dx}=1$
But $\sin(y)=\sin\left(\cos^{-1}(x)\right)=\sqrt{1-x^2}$
Where we can either use triangles or the trigonometric identity: $\sin^2(\theta)+\cos^2(\theta)=1$ to obtain $|\sin(y)|=\sqrt{1-\cos^2(y)}$
Note that because $y\in[0,\pi]$, we have $\sin(y)\in[0,1]$ and so $|\sin(y)|=\sin(y)$
So $\frac{d}{dx}\left(\cos^{-1}(x)\right)=\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}$