1. use implicit differentiation:
cosr-cot θ=e^-r θ
-sin r dr/dθ - (-csc^2 θ) = e^-r (1) + θ(e^-r)(-dr/dθ)
-sin r dr/dθ + θ(e^-r)(dr/dθ) = e^-r - csc^2 θ
dr/dθ (-sin r + θ(e^-r) ) = e^-r - csc^2 θ
dr/dθ = (e^-r - csc^2 θ) / ( -sin r + θ(e^-r) )
check my steps
#2 Again, you have to differentiate implicitly.
xy +2x-y=0
x(dy/dx) + y(1) + 2 - dy/dx = 0
dy/dx (x - 1) = -y - 2
dy/dx = (y+2)/(1-x) , I switched signs for top and bottom
slope of -4x - 2y = 10 is -2
so the slope of the normal is 1/2
then (y+2)/(1-x) = 1/2
2y + 4 = 1 -x
x = -2y-3
sub that into xy +2x-y=0
y(-2y-3) + 2(-2y-3) - y = 0
-2y^2 - 3y - 4y - 6 - y = 0
2y^2 + 8y + 6 = 0
y^2 + 4y + 3 = 0
(y+1)(y+3) = 0
y = -1 or y = -3
so find their matching x's
You now have 2 points where the normal has a slope of 1/2
Just find those equations.
#4, just a straight application of the product rule. You MUST know how to do that.
1. Find dr/d θ for cosr-cot θ=e^-r θ
2. Find the normal equation to the curve xy +2x-y=0 that are parallel to the line -4x - 2y = 10.
I know for this I'd have to find the derivative of the curve given. If I rearrange the equation it's (y=xy+2x.... y=2.. If I were to find the derivative right?)
4. Find first derivative for y=4t(-2t^2-10)^5
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