Given a curve, say y^3 = x^2, and a value, say 4.
How can I find a point in which the curve will have slope 4?
2 answers
I understand the equation for slope in this case would be: dy/dx = 2x/3y. But how can you find a point that satisfies both 2x/3y and y^3 = x^2?
for your given example , dy/dx = 2x/(3y^2)
so you set that equal to 4
2x/(3y^2) = 4
2x = 12y^2
x = 6y^2
now plug that back into the original y^3 = x^2
y^3 = 36y^4
36y^4 - y^3 = 0
y^3(36y - 1) = 0
y = 0 or y = 1/36
if y = 0, the x = 0 <------ ****
if y = 1/36, then x = 6(1/36)^2 = 1/216
However, the derivative is indeterminate at (0,0), it would be 0/0
take a point very very close to the origin.
let y = .001
then x^2 = .001^3 , using my calculator ....
x = appr .000031622 and
dy/dx = 2x/(3y^2) = appr 666,667 appears to be approaching infinitiy
for the other point (1/216,1/36)
dy/dx = (1/108) / (3(1/1296)
= (1/108)(1296/3) = 4 , as needed.
so you set that equal to 4
2x/(3y^2) = 4
2x = 12y^2
x = 6y^2
now plug that back into the original y^3 = x^2
y^3 = 36y^4
36y^4 - y^3 = 0
y^3(36y - 1) = 0
y = 0 or y = 1/36
if y = 0, the x = 0 <------ ****
if y = 1/36, then x = 6(1/36)^2 = 1/216
However, the derivative is indeterminate at (0,0), it would be 0/0
take a point very very close to the origin.
let y = .001
then x^2 = .001^3 , using my calculator ....
x = appr .000031622 and
dy/dx = 2x/(3y^2) = appr 666,667 appears to be approaching infinitiy
for the other point (1/216,1/36)
dy/dx = (1/108) / (3(1/1296)
= (1/108)(1296/3) = 4 , as needed.