The graph of r = 7/ (3 cos(theta) + 2 sin(theta)) is a line. What is the slope of the line? How do you do this!!!
r = 7/(3cosθ + 2sinθ)
r(3cosθ+2sinθ) = 7
3rcosθ + 2rsinθ = 7
3x+2y = 7
I guess you can handle it from there, yeah?
It says that -2/3 is wrong
no doubt. Did you try -3/2 ?
3x+2y = 7
2y = -3x + 7
y = -3/2 x + 7/2
Whoops bad math:) Thanks!
To find the slope of the line represented by the polar equation r = 7/(3cos(θ) + 2sin(θ)), we can use the fact that the slope of a line in polar coordinates is given by the derivative of r with respect to θ.
To calculate the derivative, we need to express r in terms of Cartesian coordinates. A conversion formula for r in terms of x and y can be used:
r = sqrt(x^2 + y^2)
Substituting x = rcos(θ) and y = rsin(θ) into the equation, we get:
sqrt(x^2 + y^2) = 7/(3cos(θ) + 2sin(θ))
Squaring both sides of the equation to eliminate the square root, we have:
x^2 + y^2 = 49/(9cos^2(θ) + 12cos(θ)sin(θ) + 4sin^2(θ))
Substituting x = rcos(θ) and y = rsin(θ) again, we get:
r^2 = 49/(9cos^2(θ) + 12cos(θ)sin(θ) + 4sin^2(θ))
Now, we can differentiate both sides of the equation with respect to θ to find the derivative. Since r is a function of θ, we use the chain rule to differentiate r^2 on the left-hand side.
2rr' = 49 * [-9sing^2(θ) + 9cos^2(θ) - 6cos(θ)sin(θ)] / (9cos^2(θ) + 12cos(θ)sin(θ) + 4sin^2(θ))^2
Simplifying the right-hand side of the equation:
2rr' = 49 * [9cos^2(θ) - 9sin^2(θ) - 6cos(θ)sin(θ)] / (9cos^2(θ) + 12cos(θ)sin(θ) + 4sin^2(θ))^2
Simplifying further:
r' = 49 * [9cos^2(θ) - 9sin^2(θ) - 6cos(θ)sin(θ)] / (2r * (9cos^2(θ) + 12cos(θ)sin(θ) + 4sin^2(θ))^2)
The slope of the line in polar coordinates is r'.
We can see that the value of r' is dependent on θ. Since the slope is not a constant value and varies with different values of θ, the graph of this polar equation does not represent a straight line.
Therefore, it is not possible to determine the slope of the line represented by the polar equation r = 7/(3cos(θ) + 2sin(θ)).