Find an equation for the tangent and normal to the curve x^2 + y^2 = 9 at x=1
To find the equation of the tangent and normal to a curve at a specific point, we need to use calculus.
First, let's find the derivative of the curve x^2 + y^2 = 9 with respect to x. To do this, we need to implicitly differentiate the equation.
Differentiating both sides of the equation with respect to x:
2x + 2y * (dy/dx) = 0
Now, let's find the slope of the tangent line at the point (1, y).
Substitute x = 1 into the equation:
2(1) + 2y * (dy/dx) = 0
Simplifying, we have:
2 + 2y * (dy/dx) = 0
Now, let's find the value of dy/dx at x = 1. To do this, we need to substitute y = ±√(9 - x^2) into the equation:
2 + 2y * (dy/dx) = 0
Putting y = √(9 - x^2):
2 + 2√(9 - x^2) * (dy/dx) = 0
And putting y = -√(9 - x^2):
2 + 2(-√(9 - x^2)) * (dy/dx) = 0
We now have the value of dy/dx at x = 1.
Next, let's find the equation of the tangent line. We know the slope of the tangent line is equal to dy/dx at the point (1, y) and this point lies on the line. So, we have the slope and a point, which allows us to use the point-slope form of a line:
y - y1 = m(x - x1)
Using x1 = 1, y1 = ±√(9 - 1^2) and m = dy/dx, we can substitute these values into the equation to find the equation of the tangent line.
Similarly, we can find the equation of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, we can substitute -1/m (negative reciprocal of dy/dx) as the slope in the point-slope form equation to find the equation of the normal line.
By substituting the values and simplifying, we can find the equations of the tangent and normal lines to the curve x^2 + y^2 = 9 at x = 1.