Question
Find the slope of the line tangent to the curve xsin(3y)−cosy=−π at the point (π,π2) . (1 point) Responses −3π − 3 π −1 − 1 0 0 1 1 3π
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GPT 3.5
To find the slope of the line tangent to the curve, we need to first find the derivative of the curve and then evaluate it at the given point.
Taking the derivative of the curve xsin(3y) - cosy = -π with respect to x, we get:
sin(3y) + x(3cos(3y))(dy/dx) + sin(y)(dy/dx) = 0
Now we need to solve for dy/dx.
To do this, we can rearrange the equation and solve for dy/dx:
(3cos(3y))(dy/dx) + sin(y)(dy/dx) = -sin(3y)
(dy/dx)(3cos(3y) + sin(y)) = -sin(3y)
dy/dx = -sin(3y) / (3cos(3y) + sin(y))
Now we can evaluate this derivative at the point (π,π^2):
dy/dx = -sin(3π^2) / (3cos(3π^2) + sin(π^2))
Since sin(3π^2) = 0 and sin(π^2) = sin(π) = 0, we have:
dy/dx = 0 / (3cos(3π^2) + 0) = 0
Therefore, the slope of the line tangent to the curve at the point (π,π^2) is 0.
Taking the derivative of the curve xsin(3y) - cosy = -π with respect to x, we get:
sin(3y) + x(3cos(3y))(dy/dx) + sin(y)(dy/dx) = 0
Now we need to solve for dy/dx.
To do this, we can rearrange the equation and solve for dy/dx:
(3cos(3y))(dy/dx) + sin(y)(dy/dx) = -sin(3y)
(dy/dx)(3cos(3y) + sin(y)) = -sin(3y)
dy/dx = -sin(3y) / (3cos(3y) + sin(y))
Now we can evaluate this derivative at the point (π,π^2):
dy/dx = -sin(3π^2) / (3cos(3π^2) + sin(π^2))
Since sin(3π^2) = 0 and sin(π^2) = sin(π) = 0, we have:
dy/dx = 0 / (3cos(3π^2) + 0) = 0
Therefore, the slope of the line tangent to the curve at the point (π,π^2) is 0.