Calculus Question

Question

Let C denote the curve x^2/3+y^2/3 = 1. Let P denote a point on the curve C in the first quadrant. Let L denote the normal line to the curve C at the point P. If L passes through the point (1/2,0) find the gradient of the line L. Give your answer correct to two decimal places.

My tutor hinted us that x^2/3+y^2/3 = 1 should be replaced with cos³(t) + sin³(t) = 1, but I have no idea how to get there. Please assist me in understanding this question. Thank you.

Answer 1

Nice answer by Yefim. Here is another way to reach that final equation in terms of u.

Realize that there are two ways to calculate the slope of L (in terms of x₀)

One is by setting up the slope between two points (x₀, y₀) and (1/2, 0).

m₁ = y₀/(x₀-1/2)

The 2nd method is to find y' using implicit differentiation. That does result in -y^1/3/x^1/3.

So, the slope of L will be the negative reciprocal, x^1/3/y^1/3.

Another formula for the slope of L is therefore x₀^1/3/y₀^1/3.

And y₀ = (1-x₀^2/3)^3/2.

Set the two slopes equal to each other replacing y₀ with its equivalent in terms of x₀. Solve for x₀, then determine y_0. Use either slope formula to find slope of L.

Here is a graph depicting this.

desmos.com/calculator/6dutjh9q8d

Answer 2

Let P(x₀, y₀) is tangent point. So x₀^2/3 + y₀^2/3 = 1. Now slope of normal line L m = - 1/y'.

Wer differentiate implicitly equation of C: 2/3x^-1/3 + 2/3y^-1/3y' = 0, y' = - x^-1/3/y^-1/3 = - y^1/3/x^1/3.

So slope(gradient) of line L at point P is equel m = x₀^1/3/y₀^1/3 and equation of line L: y = y₀ + :x₀^1/3/y₀^1/3(x - x₀). Becouse point (1/2, 0) belong to to line L we have: 0 = y₀ + 1/2x₀^1/3/y₀^1/3 - x₀^4/3/y₀^1/3,

or 2y₀^4/3 + x₀^1/3 - 2x₀^4/3 = 0 and x₀^2/3 + y₀^2/3 = 1, y₀^2/3 = 1 - x₀^2/3.

2(1 - x₀^2/3)² + x₀^1/3 - 2x₀^4/3 = 0. After squaring we get: 2 - 4x₀^2/3 + x₀^1/3 = 0.

This is quadratic equation for x₀^1/3 = u; We get 4u² - u - 2 = 0, u = (1 ± √(33))/8. We ned only positive (because we in 1st quadrant). u = (1 + √(33))/8, x₀^1/3 = (1 + √(33))/8 = 0,84307.

Then y₀^1/3= √ (1 - 0.84307²) = 0.53780.

So gradient(slope) of line: m = 0.84307/0.53780 = 1.57