Question

what is the closest point of x^4+y^4=1 to x+8y=24?
(similar to one of the other question that has been written here)

Answer 1

x= (24-8y)
now plug in x

make the function 1 a function of x

Answer 2

differentiate it to find vals of x
get the points and find the distances

Answer 3

sry mistake

Answer 4

\[\sqrt[4]{1-y ^{4}} - (24-8y)\]

Answer 5

no

(24-8y)^4 + y^4 = 1

Answer 6

aha u wanna find the point that is for on both of these equation?

Answer 7

that is on both of these equation*

Answer 8

Differentiate the fourth-degree equation implicitly and solve for the derivative to get \[y'=-x ^{3}/y ^{3}.\]
This is the slope of the line tangent to the curve at
(x,y).. The slope of the normal line to the curve there is
\[y ^{3}/x ^{3}.\]
The line has slope -1/8, and any line perpendicular to it has slope 8, the same slope as the normal line to the curve.
So we have\[y ^{3}/x ^{3}=8,\]
simplifying to y= 2x. Then
\[x ^{4}+(2x)^{4}=1\rightarrow x= 1/\sqrt[4]{17}.\] So
\[y = 2/\sqrt[4]{17},\]and the point on the curve nearest the line is
\[(1/\sqrt[4]{17},2/\sqrt[4]{17}).\]

Answer 9

thanks very heplful now i can't understand the way that u found y′=−x3/y3.

Answer 10

\[4x ^{3} + 4y ^{3}y' = 0\]
after implicit differentiation is done. Solve for y'.

Answer 11

I see thanks abtrehearn

Answer 12

:^)