Question

What is the slope of st.line xcosa+ysina=p? ( Find by using derivative)​

Answer 1

Assume that
`a` and
`p` are constants. The slope of the line will be equal to

• 
  `\displaystyle -(cos((a)))/(sin((a))) = \cot{(a)}` if
  `sin(a) \\e 0`;
• Infinity if
  `sin(a) = 0`.

Explanation:

Rewrite the expression of the line to express
`y` in terms of
`x` and the constants.

Substract
`x\cdot cos((a))` from both sides of the equation:

`y sin((a)) = p - xcos((a))`.

In case
`sin(a) \\e 0`, divide both sides with
`sin(a)`:

`\displaystyle y = - (cos((a)))/(sin((a)))\cdot x+ (p)/(sin((a)))`.

Take the first derivative of both sides with respect to
`x`.
`(p)/(sin((a)))` is a constant, so its first derivative will be zero.

`\displaystyle (dy)/(dx) = - (cos((a)))/(sin((a)))`.

`\displaystyle (dy)/(dx)` is the slope of this line. The slope of this line is therefore

`\displaystyle - (cos((a)))/(sin((a))) = -\cot{(a)}`.

In case
`sin(a) = 0`, the equation of this line becomes:

`y sin((a)) = p - xcos((a))`.

`xcos((a)) = p`.

`\displaystyle x = (p)/(cos((a)))`,

which is the equation of a vertical line that goes through the point
`\displaystyle \left(0, (p)/(cos((a)))\right)`. The slope of this line will be infinity.