Question

A particle moves along the curve below. y = √ 3 + x 3 y=3+x3 As it reaches the point ( 1 , 2 ) (1,2), the y y-coordinate is increasing at a rate of 3 cm/s 3 cm/s. How fast is the x x-coordinate of the point changing at that instant?

Answer 1

Answer:

The x-coordinate of the point changing at ¼cm/s

Step-by-step explanation:

Given

y = √(3 + x³)

Point (1,2)

Increment Rate = dy/dt = 3cm/s

To calculate how fast is the x-coordinate of the point changing at that instant?

First, we calculate dy/dx

if y = √(3 + x³)

dy/dx = 3x²/(2√(3 + x³))

At (x,y) = (1,2)

dy/dx = 3(1)²/(2√(3 + 1³))

dy/dx = 3/2√4

dy/dx = 3/(2*2)

dy/dx = ¾

Then we calculate dx/dt

dx/dt = dy/dt ÷ dy/dx

Where dy/dx = ¾ and dy/dt = 3

dx/dt = ¾ ÷ 3

dx/dt = ¾ * ⅓

dx/dt = ¼cm/s

The x-coordinate of the point changing at ¼cm/s