Write the equation of lines tangent and normal to the following function at (0, π). To find derivative, use implicit differentiation.

x^2cos^2y - siny = 0

Note: I forgot the ^2 for cos on the previous question. Sorry.

x = 0, y = pi

-x^2 2 cos y sin y dy +cos^2y 2 x dx - cos y dy = 0

2 cos^2 y x dx=cosy (2x^2siny+1)dy

dy/dx = slope m

m = 2 x cos y/(2x^2siny+1)
at (0,pi)
m = 2 * 0 = 0 Oh my :) horizontal there
so what is y at x = 0 and y = pi
LOL, zero
so along the x axis

I mean parallel to x axis and y = pi

for part 2, normal is up the y axis
see
http://www.wolframalpha.com/widgets/view.jsp?id=91851988c40ebbe236f5561e167c9ab8

Just copy your function into there:

x^2cos^2y - siny = 0

look at the solution close to (0,0)

To find the equation of the line tangent and normal to the function at the point (0, π), we first need to find the derivative of the function using implicit differentiation.

Step 1: Implicit Differentiation
Let's differentiate the given equation with respect to x.

To differentiate x^2cos^2y - siny = 0, we apply the rules of differentiation, treating y as a function of x:

d/dx (x^2cos^2y) - d/dx (siny) = d/dx (0)

Using the chain rule, the derivative of cos^2y with respect to x is:

-2x * cos^2y * dy/dx

The derivative of siny with respect to x is:

-cosy * dy/dx

Differentiating the constant term 0 gives us:

0

So, the differentiated equation becomes:

2x * cos^2y * dy/dx - cosy * dy/dx = 0

Now, solve for dy/dx:

dy/dx (2x * cos^2y - cosy) = 0

dy/dx = 0 / (2x * cos^2y - cosy)

Step 2: Finding the derivative at (0, π)
Now, substitute x = 0 and y = π into the derivative to find the slope of the tangent line at the point (0, π):

dy/dx = 0 / (2(0) * cos^2(π) - cos(π))

Since cos(π) = -1, the denominator becomes:

dy/dx = 0 / (0 - (-1))

dy/dx = 0 / (-1)

dy/dx = 0

So, the slope of the tangent line at (0, π) is 0.

Step 3: Equation of the tangent line
The equation of a line can be written in the form y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line.

Substituting the values (0, π) and dy/dx = 0 into the equation, we get:

y - π = 0(x - 0)

y - π = 0

y = π

Therefore, the equation of the tangent line to the function at (0, π) is y = π.

Step 4: Equation of the normal line
The normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line's slope.

Since the slope of the tangent line is 0, the slope of the normal line is undefined (vertical line). Therefore, the equation of the normal line to the function at (0, π) is x = 0.

To summarize:
- The equation of the tangent line to the function at (0, π) is y = π.
- The equation of the normal line to the function at (0, π) is x = 0.