. Given that x²cos y_sin y=0,(0,π).
A. Verify that the given points on the curve.
B.use implicit differention to find the slope of the above curve at the given point.
C.find the equation of tangent and normal to the curve at that.
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To verify that the given points (0,π) lie on the curve, you need to substitute these values into the equation and check if it satisfies the equation.
A. Verify the given points on the curve:
For the given equation x²cos(y) - sin(y) = 0, substitute x = 0 and y = π into the equation:
(0)²cos(π) - sin(π) = 0
0 - 0 = 0
Since the equation is satisfied, the point (0,π) lies on the curve.
B. Using implicit differentiation to find the slope of the curve:
Differentiate both sides of the equation with respect to x, treating y as a function of x, and apply the chain rule when necessary.
Differentiating x²cos(y) - sin(y) = 0 with respect to x:
2x * cos(y) - x² * sin(y) * dy/dx - cos(y) * dy/dx - sin(y) = 0
Now, we can solve for dy/dx, the slope of the curve:
dy/dx = (2x * cos(y) - sin(y)) / (x² * sin(y) + cos(y))
C. Find the equation of the tangent and normal to the curve at the given point:
To find the equation of the tangent line, we need both the slope dy/dx and the point (0,π). Using the slope dy/dx we found in step B, substitute x = 0 and y = π into the equation.
Tangent equation using point-slope form:
y - y₁ = m(x - x₁)
Tangent equation at point (0,π):
y - π = [(2 * 0 * cos(π) - sin(π)) / (0² * sin(π) + cos(π))] * (x - 0)
Simplifying, we get the equation of the tangent line:
y - π = -sin(π) * x
To find the equation of the normal line, we need the slope of the normal line, which is the negative reciprocal of the slope of the tangent line at the given point (0,π). So the slope of the normal line is 1.
Normal equation at point (0,π):
y - π = 1(x - 0)
Simplifying, we get the equation of the normal line:
y - π = x - 0
Therefore, the equation of the tangent line is y = -sin(π)x + π, and the equation of the normal line is y = x + π.