Find equations for the lines that are tangent and normal to the graph of y = sinx + 3 at x = pi

Now, the slope of the tangent line at any point (x,y) on the curve is just y'(x) = cosx

At x=pi, y' = -1 and y=3

So, we want the line through (pi,3) with slope -1.

y-3 = -(x-pi) = pi-x
y = -x + 3+pi

The slope of the normal line is -(1/-1) = 1

y-3 = (x-pi)
y = x + 3-pi

Well, aren't we tangentially dropping into the world of calculus? Let's dive right in!

To find the tangent line at a specific point on a curve, we need two things: the slope of the curve at that point and a point on the curve.

The slope of the curve at x = pi can be found by taking the derivative of y = sin(x) + 3 with respect to x. The derivative of sin(x) is cos(x), so the derivative of y = sin(x) + 3 is y' = cos(x).

Plugging in x = pi will give us the slope at that point: y' = cos(pi) = -1.

Now, let's find the y-coordinate of the point on the curve at x = pi. Plugging in x = pi into y = sin(x) + 3, we get y = sin(pi) + 3 = 0 + 3 = 3. So our point is (pi, 3).

Using the point-slope form of a line (y - y₁ = m(x - x₁)), we can find the equation of the tangent line:
y - 3 = -1(x - pi)
y - 3 = -x + pi
y = -x + pi + 3

Now, the normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the slope of the tangent line (which is 1 in this case).

Using the point-slope form again, we can find the equation of the normal line:
y - 3 = 1(x - pi)
y - 3 = x - pi
y = x - pi + 3

So there you have it! The equations for the tangent and normal lines to the curve y = sin(x) + 3 at x = pi are:
Tangent line: y = -x + pi + 3
Normal line: y = x - pi + 3

Now, go forth and tangentially amaze your friends with your newfound line-profiling skills!

To find the equations for the lines that are tangent and normal to the graph of y = sin(x) + 3 at x = π, we need to first find the slope of the tangent line at that point.

Step 1: Find the derivative of y = sin(x) + 3.
Taking the derivative of sin(x) gives us cos(x), and the derivative of a constant is 0.
Therefore, the derivative of y = sin(x) + 3 is dy/dx = cos(x).

Step 2: Evaluate the derivative at x = π.
Substitute π into the derivative expression, dy/dx = cos(x).
dy/dx = cos(π) = -1.

Step 3: Find the equation for the tangent line using the point-slope form.
Using the point-slope form of an equation, y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope, we can substitute the values we have to form the equation.
At x = π, the point on the graph is (π, sin(π) + 3) = (π, 3).
Therefore, the equation for the tangent line is y - 3 = -1(x - π).
Simplifying, we get y - 3 = -x + π, or y = -x + (π + 3).

Step 4: Find the equation for the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is -1, the slope of the normal line will be 1.
Using the point-slope form again, we can write the equation for the normal line.
The point of tangency is (π, 3). Therefore, the equation for the normal line is y - 3 = 1(x - π).
Simplifying, we get y - 3 = x - π, or y = x + (3 - π).

So, the equation for the tangent line is y = -x + (π + 3), and the equation for the normal line is y = x + (3 - π).

To find the equations of the lines that are tangent and normal to the graph of y = sinx + 3 at x = pi, we need to determine the slope of the tangent line and the normal line at that point.

1. Tangent line:
To find the slope of the tangent line, we need to take the derivative of the given function with respect to x. Differentiating y = sinx + 3, we get:
dy/dx = cosx

Evaluate the derivative at x = pi:
dy/dx(pi) = cos(pi) = -1

So, the slope of the tangent line at x = pi is -1.

Using the point-slope form of a line, we can write the equation of the tangent line:
y - y₁ = m(x - x₁)

Substitute x₁ = pi, y₁ = sin(pi) + 3 = 3, and m = -1 into the equation:
y - 3 = -1(x - pi)
y - 3 = -x + pi
y = -x + pi + 3
y = -x + 3 + pi (simplify if desired)

Therefore, the equation of the tangent line is y = -x + 3 + pi.

2. Normal line:
To find the slope of the normal line, we can use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line at x = pi is 1.

Using the same point-slope form of a line, we can write the equation of the normal line:
y - y₁ = m(x - x₁)

Substitute x₁ = pi, y₁ = sin(pi) + 3 = 3, and m = 1 into the equation:
y - 3 = 1(x - pi)
y - 3 = x - pi
y = x - pi + 3
y = x - (pi - 3) (simplify if desired)

Therefore, the equation of the normal line is y = x - (pi - 3).

In summary, the equation of the tangent line is y = -x + 3 + pi, and the equation of the normal line is y = x - (pi - 3).