Find equations of the tangent line and normal line to the curve at the given point.

Question

y= x + √x, (1,2)

Answer 1

To find the equation of a line we need to know a point on that line and the slope of that line (point slope form)

y - y1 = m*(x - x1)

(x1, y1) is the point on the line

m is the slope of the line

The point is given, the only missing quantity is the slope. The derivative of the function gives the slope of the tangent line at a given point.

Computation of derivative:

y = x + x^(1/2)

y' = 1 + (1/2)*x^(-1/2)

y'(1) = 1 + (1/2)*1^(-1/2) = 1 + 1/2 = 3/2

Finding the equation of the tangent line:

m = 3/2, which is the slope of the tangent line found from the derivative

The point that the tangent line passes through is (1,2)

Using the point slope form of a line we get

y - 2 = (3/2)*(x - 1)

y = 3/2*x + 1/2

Finding the equation of the normal line:

The slope of the normal line is the negative reciprocal of the slope of the tangent line because the two lines are perpendicular

m = -2/3

The point that the normal line passes through is (1,2)

Using the point slope form of a line we get

y - 2 = (-2/3)*(x - 1)

y = -2/3*x + 4/3