Given that x²cos y-sin y=0 ,(0,π):
a)verfiy that given point is on the curve.
b)use implicit differentiation to find the slope of the above curve at the given point.
c)find the equation for tangent and normal to the curve at that point.
(a) trés simples, non?
(b)
2x cosy - x^2 siny y' - cosy y' = 0
y' = (2x cosy)/(cosy + x^2 siny)
So, at (0,π), y'=0
(c) so, the tangent is a horizontal line, and the normal is a vertical line through (0,π).
To verify that the point (0, π) lies on the curve x²cos(y) - sin(y) = 0, we need to substitute the values of x and y into the equation and see if it satisfies the equation.
a) Substituting x = 0 and y = π into the equation:
(0)²cos(π) - sin(π) = 0 - 0 = 0.
Since the equation evaluates to 0 when x = 0 and y = π, we can conclude that the given point (0, π) lies on the curve x²cos(y) - sin(y) = 0.
b) Now, let's use implicit differentiation to find the slope of the curve at the given point.
Differentiate both sides of the equation with respect to x, treating y as a function of x:
d/dx(x²cos(y) - sin(y)) = d/dx(0)
2xcos(y) - x²sin(y) * dy/dx - cos(y) * dy/dx = 0
Since we are interested in finding the slope at the point (0, π), we substitute x = 0 and y = π into the equation:
2(0)cos(π) - (0)²sin(π) * dy/dx - cos(π) * dy/dx = 0
- dy/dx - cos(π) * dy/dx = 0
-(1 + cos(π)) * dy/dx = 0
-2 * dy/dx = 0
dy/dx = 0
Therefore, the slope of the curve at the point (0, π) is 0.
c) To find the equation of the tangent and normal lines to the curve at the point (0, π), we need to find the equations of these lines in point-slope form.
The equation of a line in point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line.
Using the given point (0, π) and the slope dy/dx = 0, we can write the equations for the tangent and normal lines:
Tangent line:
Using the point-slope form, we have:
y - π = 0(x - 0)
y - π = 0
y = π
Therefore, the equation of the tangent line to the curve at the point (0, π) is y = π.
Normal line:
Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, we have:
m(normal) = -1/m(tangent) = -1/0 = undefined.
An undefined slope indicates that the normal line is vertical. Since the tangent line at the point (0, π) is horizontal (y = π), the normal line will be a vertical line passing through the same point.
Therefore, the equation of the normal line to the curve at the point (0, π) is x = 0.