Answer:
1) Thus point (3,4) is represented as (5,53
) in polar format.
2) Polar form is
![r=tan(\theta )sec(\theta)](https://img.qammunity.org/2020/formulas/mathematics/college/fzd77c1yn8udonvhik9yyijbrggh0i6m1z.png)
Explanation:
Any point (x,y) can be represented in polar format (r,θ) as
![r=\sqrt{x^(2)+y^(2)}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wrjyv8rhm0x0tl1crdqbsi2nz6hetrf5as.png)
![\theta =tan^(-1)((y)/(x))](https://img.qammunity.org/2020/formulas/mathematics/college/4skxcuvzh62clujtax8b3d3sgnlc1c1e0p.png)
Using the above formula
The point (3,4) is represented in polar format as
![r=\sqrt{3^(2)+4^(2)}\\\\r=√(9+16)=5\\\\\theta =tan^(-1)((4)/(3))=53^(o)](https://img.qammunity.org/2020/formulas/mathematics/college/clz2e4wpm9tyzyfqz6g6j13pc8l4u2iw70.png)
Thus point (3,4) is represented as (5,53
)
2)
The given curve is
![y=x^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qfh1zkisge7wtl99lc0zrsdplbk57w91cd.png)
to convert it to polar form put
![x=rcos(\theta )\\\\y=rsin(\theta )](https://img.qammunity.org/2020/formulas/mathematics/college/jnty6v1qb54or76jxpzumdy66vtgav7kyo.png)
Thus the curve becomes
![rsin(\theta )=r^(2)* cos^(2)(\theta )\\\\r=(sin(\theta ))/(cos^(2)(\theta ))\\\\r=tan(\theta )sec(\theta)](https://img.qammunity.org/2020/formulas/mathematics/college/999c5l79o9mvy0wu0bzzpjtpy2l35ub85y.png)