$m=\frac{-3}{\sqrt{2}}$

$nm=-1\therefore n=\frac{\sqrt{2}}{3}$

$\frac{\sqrt{2}}{3}\left(\frac{\pi }{4}\right)+d=sin\frac{\pi }{4}\therefore d=\frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{3}\frac{\pi }{4}\simeq 0.34$

$y=0.47x+0.34$ . The curve and the normal line are shown in the following figure:

$f\left(x\right)=x^3,x_0=1,\frac{df}{dx}{\left|\right.}_1=3x^2{\left|\right.}_1=3\therefore m=3$ and $n=-\frac{1}{3}$ Now use the property that the normal line $y=nx+d$ and the curve $y=x^3$ pass through the point $\left(1,1\right)$ to find $d$ and so obtain the equation of the normal line:

$1=n+d\therefore d=1-n=1+\frac{1}{3}=\frac{4}{3}$ . Thus the equation of the normal line is $y=-\frac{1}{3}x+\frac{4}{3}.$ The curve and the normal line through $\left(1,1\right)$ are shown below:

1. 1. $f\left(x\right)=x^4+2x^2\frac{df}{dx}=4x^3+4x,\frac{df}{dx}{\left|\right.}_{x=1}=8$
      
      tangent line $y=8x+c$ . This passes through $\left(1,3\right)$ so    $y=8x-5$

      normal line $y=-\frac{1}{8}x+\underset{̣}{.}$ This passes through $\left(1,3\right)$ so $y=-\frac{1}{8}x+\frac{25}{8}.$

   2. $f\left(x\right)=\sqrt{1-x^2}\frac{df}{dx}=\frac{-x}{\sqrt{1-x^2}}\frac{df}{dx}{\left|\right.}_{x=\frac{\sqrt{2}}{2}}=-1$
      
      tangent line $y=-x+c$ . This passes through $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ so $y=-x+\sqrt{2}$
      
      normal line $y=x+d$ . This passes through $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ so $y=x$ .
      
      At $\left(1,0\right)$ the tangent line is $x=1$ and the gradient is infinite (the line is vertical), and the normal line is $y=0$ .
   3. $f\left(x\right)=x^{\frac{1}{2}}\frac{df}{dx}=\frac{1}{2}{x}^{-\frac{1}{2}}\frac{df}{dx}{\left|\right.}_{x=1}=\frac{1}{2}$
      
      tangent line: $y=\frac{1}{2}x+c$ . This passes through $\left(1,1\right)$ so $y=\frac{1}{2}x+\frac{1}{2}$
      
      normal line: $y=-2x+d$ . This passes through $\left(1,1\right)$ so $y=-2x+3$ .
2. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular.

   Point of intersection: $e^{-x}=e^{ax}$ i.e. $-x=ax\therefore x=0$ ( $a=-1$ not sensible)

   The tangent line to $y=e^{ax}$ is  $y=mx+c$  where $m=a{e}^{ax}{\left|\right.}_{x=0}=a$

   The tangent line to $y=e^{-x}$ is  $y=kx+g$  where $k=-e^{-x}{\left|\right.}_{x=0}=-1$

   These two lines are perpendicular if $a\left(-1\right)=-1\text{i.e.}a=1.$