Teachers and Students > Universities

Derivative - Max and Min application!

<< < (2/6) > >>

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:

--- Quote from: sgvaibhav on January 08, 2010, 01:42:08 pm ---i did not understand the method here...

--- End quote ---

Use of gradients of tangents and perpendicular lines...  :-\ Nope?

Navigation

[0] Message Index

[#] Next page

[*] Previous page