Derivative - Max and Min application!

[SGVaibhav]

Question asked in the picture below!

THanks a lot for help!!

[Ghost Of Highbury]

I dont know how to find the closest point.

However, about the formula, Both ur suggested formulas are correct.

This is because its *squared*

lets say if x =10

first formula : (18-10)^2 = 64

second formula : (10-18)^2 = 64

hope that helped

[Alpha]

I can't view the pic. 

[SGVaibhav]

alpha, its weird that u cannot see the pic

astar, i am getting different answers everytime i solve this,
can u please solve it with some working.

i dont understand what to find out after getting the derivative

[Alpha]

I really can't see... I've tried all that I could. 

[SGVaibhav]

try now
____
where ru astar   im dying here !!
lol im so dependent on him..

[Alpha]

Closest to the point >>> Line joining (18, 0) to the required point must be perpendicular to the point on the curve.

>>> Grad. at point on curve = Inverse sign, take reciprocal of grad. at point on curve

Let point be (x, y).

dy/dx = 2x

(x, y)               (18, 0)

(y - 0)/ ( x - 18) = - 1/ 2x

y = - 1/ 2x (x- 18)

On line which passes th. (x, y), y = x^2 (eq. of curve).

x^ 2 = -1/ 2x (x- 18)

2x^3 = -x + 18

Can you complete the rest...?

Am not sure of the answer, it's a trial only. 

[SGVaibhav]

i did not understand the method here...

[astarmathsandphysics]

(y-0)^2=(0-y)^2 so both are the same.
d^2=(x-18)^2+(y-0)^2 and y=x^2
d^2=x^2-36x+324+x^4
differentiate to get 2x-36+4x^3=0
2x^3+x-18=0
(x-2)(2x^2+4x+9)=0
x=2 cos the quadratic has no solutions.
and y=2^2=4

[Alpha]

i did not understand the method here...

Use of gradients of tangents and perpendicular lines...  Nope?

[SGVaibhav]

i finally decided to ask my sir for help, but he is on vacation to malaysia.
i will ask my friends, because they use some different method.

[Alpha]

Sorry, was in a hurry...

Okay, lemme try to explain what I used.

Point on curve closest to (18, 0).

In general, when you're required to find the closest distance, you look for a line perpendicular to your point.

If I give you a set of parallel lines, the closest distance between them is the perpendicular height.

That's what I used. The tangent to the point (you are required to find) on the curve must be perpendicular to the line joining the point required (on the curve) and (18, 0).

I tried to draw it... Don't laugh, did that too in a hurry.  

[SGVaibhav]

lol thanks for the effort , the drawing is good

[Alpha]

lol thanks for the effort , the drawing is good

Come-well!

[Alpha]

Hey you got the answer??