Question
Find the slope of the line tangent to the curve xsin(3y)−cosy=−π at the point (π,π2) .
Answers
Answered by
GPT 3.5
To find the slope of the line tangent to the curve at the point (π,π²), we need to find the derivative of the curve with respect to x and then evaluate it at the given point.
First, let's rewrite the equation of the curve in terms of y:
xsin(3y) - cosy = -π
Subtract xsin(3y) from both sides:
-cosy = -π - xsin(3y)
Divide both sides by -1 to isolate cosy:
cosy = π + xsin(3y)
Now, differentiate both sides of the equation with respect to x using the chain rule:
-sin(y) * dy/dx = sin(3y) + xsin(3y)' = sin(3y) + 3xsin(3y)' = sin(3y) + 3x(3cos(3y)) = sin(3y) + 9xcos(3y)
Since we want to find the slope at the point (π,π²), substitute π for x and π² for y:
-sin(π²) * dy/dx|_(π,π²) = sin(3π²) + 9πcos(3π²)
To evaluate this expression, we need to know the values of sin(π²) and cos(3π²). Since π² is not a special angle, we will need to use a calculator or a mathematical software to find their values.
Therefore, the slope of the line tangent to the curve at the point (π,π²) is given by:
-m = sin(3π²) + 9πcos(3π²) / sin(π²)
First, let's rewrite the equation of the curve in terms of y:
xsin(3y) - cosy = -π
Subtract xsin(3y) from both sides:
-cosy = -π - xsin(3y)
Divide both sides by -1 to isolate cosy:
cosy = π + xsin(3y)
Now, differentiate both sides of the equation with respect to x using the chain rule:
-sin(y) * dy/dx = sin(3y) + xsin(3y)' = sin(3y) + 3xsin(3y)' = sin(3y) + 3x(3cos(3y)) = sin(3y) + 9xcos(3y)
Since we want to find the slope at the point (π,π²), substitute π for x and π² for y:
-sin(π²) * dy/dx|_(π,π²) = sin(3π²) + 9πcos(3π²)
To evaluate this expression, we need to know the values of sin(π²) and cos(3π²). Since π² is not a special angle, we will need to use a calculator or a mathematical software to find their values.
Therefore, the slope of the line tangent to the curve at the point (π,π²) is given by:
-m = sin(3π²) + 9πcos(3π²) / sin(π²)