find the lines that are

tangent and normal to th ecurve at the point given

1. 2xy+piesiny=2pie (1, pie/2)

2. x^2cos^2y-siny=0 (0, pie)

To find the lines that are tangent and normal to a curve at a given point, we can use calculus by finding the derivative of the curve and evaluating it at the given point.

1. Curve: 2xy + πesiny = 2π

To find the derivative of the curve, we differentiate each term with respect to x:

d/dx(2xy) + d/dx(πesiny) = d/dx(2π)

2y + 2xdy/dx + πe*cosy*dy/dx = 0

Now, we can evaluate the derivative at the given point (1, π/2):

2(π/2) + 2(1) *dy/dx + πe*cos(π/2)*dy/dx = 0

π + 2dy/dx + πe*0 *dy/dx = 0

π + 2dy/dx = 0

dy/dx = -π/2

Therefore, the derivative (slope) of the curve at the point (1, π/2) is -π/2.

To find the equation of the tangent line, we use the point-slope form:

y - y1 = m(x - x1)

Using the point (1, π/2) and the slope -π/2:

y - π/2 = (-π/2)(x - 1)

Simplifying the equation, we get:

y = (-π/2)x - π/2 + π/2

y = (-π/2)x

So, the tangent line to the curve at the point (1, π/2) is y = (-π/2)x.

To find the equation of the normal line, we use the fact that the product of the slopes of two perpendicular lines is -1.

The slope of the normal line can be found by taking the negative reciprocal of the slope of the tangent line:

m(normal) = -1/(-π/2) = 2/π

Using the point (1, π/2) and the slope 2/π:

y - π/2 = (2/π)(x - 1)

Simplifying the equation, we get:

y = (2/π)x + π/2 - π/2

y = (2/π)x

So, the normal line to the curve at the point (1, π/2) is y = (2/π)x.

2. Curve: x^2*cos^2y - siny = 0

To find the derivative of the curve, we differentiate each term with respect to x:

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

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

Now, we can evaluate the derivative at the given point (0, π):

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

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

dy/dx + dy/dx = 0

2dy/dx = 0

dy/dx = 0

Therefore, the derivative (slope) of the curve at the point (0, π) is 0.

Since the slope is 0, the equation of the tangent line is simply the equation of the horizontal line passing through the point (0, π):

y = π

So, the tangent line to the curve at the point (0, π) is y = π.

To find the equation of the normal line, we again use the fact that the product of the slopes of two perpendicular lines is -1.

The slope of the normal line can be found by taking the negative reciprocal of the slope of the tangent line (since the tangent line is horizontal, its slope is 0):

m(normal) = -1/0 (undefined)

Since the slope is undefined, the equation of the normal line will be vertical and of the form x = a, where a is the x-coordinate of the point. In this case, x = 0.

So, the normal line to the curve at the point (0, π) is x = 0.