Differentiation 2

( 1 ) V = ds / dt = 6t2

So the velocity at t = 4,

6 ( 4 )2 = 96m /s



( 2 ) y = x1/2

dy / dx = ( 1 / 2 )x -1/2

So when the slope is 1 / 6,

( 1 / 2 )x – 1/2 = 1 / 6

x1/2 = 3

x = 9

Put x = 9 back to the equation,

y = 91/2 = 3

So the point is ( 9 , 3 ).


2007-08-05 00:16:17 補充：
For question 2, the slope is found by evaluating dy / dx.

2007-08-05 00:43:03 補充：
Sorry that for Q1, there should not be the unit m / s as the displacement is s but not s metres and the time is t but not t seconds.

1.v
= ds/dt
= 3(2t^2)
= 6t^2

At t=4, v = 6(4^2) = 96.

2. Gradient of the tangent to the graph is 1/6 --->
dy/dx = 1/6
1/2 * x^(-1/2) = 1/6
1 / [x^(1/2)] = 1/3
3 = x^(1/2)
x = 9

Substitute x=9 into y=x^(1/2),
y = 9^(1/2) = 3

So the point is (9,3)