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
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. To find dr/d θ for the equation cosr - cot θ = e^-r θ, we can differentiate both sides with respect to θ using the chain rule and product rule.
Let's start by differentiating the left side:
d/dθ (cosr) - d/dθ (cot θ) = d/dθ (e^-r θ)
To differentiate cosr, we use the chain rule:
d/dθ (cosr) = -sinr * dr/dθ
To differentiate cot θ, we use the quotient rule:
d/dθ (cot θ) = -csc^2 θ * d/dθ (θ)
Simplifying, we get:
-sinr * dr/dθ + csc^2 θ = -r e^-r θ
Now, let's solve for dr/dθ by isolating the derivative term:
dr/dθ = (-r e^-r θ + csc^2 θ) / (-sinr)
2. To find the normal equation to the curve xy + 2x - y = 0 that is parallel to the line -4x - 2y = 10, we'll first find the derivative of the curve.
The given equation is: xy + 2x - y = 0
If we rearrange the equation, it becomes:
y = xy + 2x
To find the derivative of y with respect to x, we'll differentiate both sides with respect to x:
d/dx (y) = d/dx (xy + 2x)
To differentiate the right side, we'll use the product rule:
d/dx (xy + 2x) = x * d/dx (y) + y * d/dx (x) + 2
Since we're looking for the normal equation, we want the slope of the curve to be perpendicular to the slope of the given line. The slope of the given line is -4/2 = -2.
Therefore, for the normal equation, we want the derivative to have a slope of 1/2 (the negative reciprocal of -2).
Hence, we can set the derivative equal to 1/2 and solve for y in terms of x:
x * d/dx (y) + y * d/dx (x) + 2 = 1/2
Simplifying, we get:
x * dy/dx + y + 2 = 1/2
Now, we can rearrange this equation to find the normal equation.
3. To find the first derivative of y = 4t(-2t^2-10)^5, we can use the power rule and the chain rule.
Let's start by applying the power rule to the expression (-2t^2-10)^5:
(5) * (-2t^2-10)^4 * d/dt (-2t^2-10)
To find the derivative of (-2t^2-10), we can apply the sum rule:
d/dt (-2t^2-10) = d/dt (-2t^2) + d/dt (-10)
For the first term, we use the power rule:
d/dt (-2t^2) = (-2) * (2t) = -4t
The derivative of a constant term is zero: d/dt (-10) = 0
Substituting these derivatives back into the first expression, we get:
(5) * (-2t^2-10)^4 * (-4t) = -20t(-2t^2-10)^4
Hence, the first derivative of y = 4t(-2t^2-10)^5 is -20t(-2t^2-10)^4.