Question

Find the point where the normal to y=x+x^1/2+1 at (4, 7) crosses the y- axis.

Answer 1

To find the normal to the curve at a given point, we need to determine the derivative of the curve at that point and then find the negative reciprocal of it.

First, let's find the derivative of the curve y = x + x^(1/2) + 1 by differentiating each term separately:

dy/dx = d/dx(x) + d/dx(x^(1/2)) + d/dx(1)
= 1 + (1/2)(x^(-1/2)) + 0
= 1 + 1/(2√x)

Next, let's find the derivative at the point (4, 7) by substituting x = 4 into the derivative we just found:

dy/dx = 1 + 1/(2√4)
= 1 + 1/4
= 5/4

Now, let's find the negative reciprocal of the derivative:

m = -1/(dy/dx)
= -1/(5/4)
= -4/5

So the slope of the normal at the point (4, 7) is -4/5.

To find the point where the normal crosses the y-axis, we take the given point (4, 7) and use the point-slope form of a line:

y - y1 = m(x - x1)

Plugging in the values (x1, y1) = (4, 7) and m = -4/5:

y - 7 = -4/5(x - 4)

Now, let's solve for y when x = 0:

y - 7 = -4/5(-4)

y - 7 = 16/5

y = 16/5 + 7

y = 16/5 + 35/5

y = 51/5

Therefore, the point where the normal to the curve y = x + x^(1/2) + 1 at (4, 7) crosses the y-axis is (0, 51/5).