1. ## Derivatives

1st one: differentiate y=(x^3/2)+(48/x)

2nd one: The fixed point A has coordinates (8, -6,5) and the variable point P has coordinates (t,t2t).
a. Show that AP^2=6t^2-24t+125
b. Hence find the value of t for which the distance AP is least.
c. Determine this least distance.
thank u

2. Originally Posted by thereddemon

1st one: differentiate y=(x^3/2)+(48/x)

2nd one: The fixed point A has coordinates (8, -6,5) and the variable point P has coordinates (t,t2t).
a. Show that AP^2=6t^2-24t+125
b. Hence find the value of t for which the distance AP is least.
c. Determine this least distance.
thank u
$y = \frac{x^3}{2} + 48 x^{-1}$
$y' = \frac{3x^2}{2} - 48 x^{-2}$

For the second

$AP^2 = (t - 8)^2+(t+6)^2+(2t-5)^2$ expand this.

For the second part, take the derivative and set this to zero. I'm sure you can do the rest.

3. wow fast response! thank u so much! u rock