Answer:
1) Thus point (3,4) is represented as (5,53[tex]^{o}[/tex]) in polar format.
2) Polar form is [tex]r=tan(\theta )sec(\theta)[/tex]
Step-by-step explanation:
Any point (x,y) can be represented in polar format (r,θ) as
[tex]r=\sqrt{x^{2}+y^{2}}[/tex]
[tex]\theta =tan^{-1}(\frac{y}{x})[/tex]
Using the above formula
The point (3,4) is represented in polar format as
[tex]r=\sqrt{3^{2}+4^{2}}\\\\r=\sqrt{9+16}=5\\\\\theta =tan^{-1}(\frac{4}{3})=53^{o}[/tex]
Thus point (3,4) is represented as (5,53[tex]^{o}[/tex])
2)
The given curve is [tex]y=x^{2}[/tex]
to convert it to polar form put
[tex]x=rcos(\theta )\\\\y=rsin(\theta )[/tex]
Thus the curve becomes
[tex]rsin(\theta )=r^{2}\times cos^{2}(\theta )\\\\r=\frac{sin(\theta )}{cos^{2}(\theta )}\\\\r=tan(\theta )sec(\theta)[/tex]
