Question

Consider a line tangent to the curve x^(2/3)+y^(2/3)=1. Show that the distance between the x-intercept and the y-intercept of this line is always 1.

Answer 1

Are you sure about your equation?

Answer 2

Your equation is fine. Stay tuned for the answer.

Answer 3

yes i'm sure.it is \[x^{2/3}+y \left\{ 2/3\right\}\]

Answer 4

Is it
\[
x^{2/3} + y^{2/3}=1
\]

Answer 5

I have trig solution for it

Answer 6

yes it is. what is your solution ?

Answer 7

Put
\[
x(t)=\cos ^3(t)\\
y(t)=\sin^3(t)
\]

Answer 8

Let c be a value of t where we draw the tangent. The slop at c is
\[
\frac{y'(c)}{x'(c)}=-\tan(c)
\]

Answer 9

You do the computation

Answer 10

The equation of the tangent at t=c is
\[
y- y(c) = -\tan(c)(x-x(c))
\]

Answer 11

Put x=0 and find the y-intercept b to be
\[
b=\sin ^3(c)+\sin (c) \cos ^2(c)
\]

Answer 12

Put y=0 and find the x-intercept a to be
\[
a=\cot (c) \left(\sin ^3(c)+\sin (c) \cos ^2(c)\right)
\]

Answer 13

Compute
\[
a^2 + b^2
\]
and you find that is equal to 1

Answer 14

What is your background in Math?

Answer 15

i have a question. how do you find the slope that is -tan(c)

Answer 16

I'm first year student in univercity

Answer 17

I will send you a solution that does not involve trig

Answer 18

Is this part of your implicit differentiation subject?

Answer 19

yes it is. i couldn't understand the subject well