calculusgeometry
This is a longer question that I am having trouble with. The part I can't do is c, where you must find the area of the triangle formed by a normal and tangent. My problem is that I can't visualise the problem and am unable to find the lengths of the sides of the triangle. I am not sure whether information from the previous questions is necessary, so in doubt I am leaving in the previous questions, but the one I need help with is c.
Thanks
Following rschwieb's advice:
Note that
$\ \ \ \bullet$ $\color{darkgreen}{x=0}$ is the line coinciding with the $y$-axis.
$\ \ \ \bullet$ $\color{darkblue}{x+y=0}$ has standard form $y=-x$; so, its graph is the line of slope $-1$ passing through the origin.
$\ \ \ \bullet$ $\color{maroon}{y=x-1}$ is the equation of the line of slope $1$ with $y$-intercept $(0,-1)$.
The graphs of these lines are shown below:
The triangle in question is shaded light green (on my display) above. To find the coordinates of the vertices of this triangle, you have to solve systems of equations.
Apparently there is a vertex around the point $(1/2,-1/2)$. Let's check that. Here, we need to find the intersection of the lines $\color{darkblue}{x+y=0}$ and $\color{maroon}{y=x-1}$. I'll leave it to you to solve this system. It indeed has the solution $(1/2,-1/2)$.
Once you have the coordinates of all three vertices, computing the area of the triangle should be easy. For instance, you can take the base to be $1$ (along the $y$-axis) and the corresponding height to be $1/2$.
In polar coordinates the equation of the folium is $$ r=\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta} $$ For the area of the region $A_1$ the limits of integration for $\theta$ are: $$ 0\le \theta \le \arctan \left( \frac {\sqrt[3]{2}}{\sqrt[3]{4}}\right)=\arctan \sqrt[3]{\frac{1}{2}}=\alpha $$ and the area is given by: $$ A_1=\frac{1}{2}\int_0^ \alpha r^2 d\theta=\frac{1}{2}\int_0^ \alpha \left[\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta}\right]^2 d\theta $$ with the substitution $u=1+\tan^3 \theta$ we have: $$ du=3\tan^2 \theta \sec^2 \theta d \theta $$ and the limits of integration becomes: $$ 1\le u \le 1+\frac{1}{2} $$ So the area is $$ A_1=\frac{3}{2}\int_1^{\frac{3}{2}} \frac{du}{u^2}=\frac{1}{2} $$
Best Answer
Since $y'= 2x-\frac{1}{2}x^3$, the coefficient of tangent at $P$ is $k_t=\frac{3}{2}$, so the coefficient of normal at $P$ is $k_n=-\frac{2}{3}$. So the tangent at $P$ has the equation $y=\frac{3}{2}(x-1)+2$ and the normal $y=-\frac{2}{3}(x-1)+2$. Thus $T=(-\frac{1}{3},0)$ and $N=(0,\frac{8}{3})$. Now you have $\vec{TP}=(\frac{4}{3},2)$ and $\vec{NP}=(1,-\frac{2}{3})$, so $TP=\frac{2\sqrt{13}}{3}$, $NP=\frac{\sqrt{13}}{3}$, and since the angle at $P$ is right, we have that the area is $\frac{13}{9}$.