
Differentiation 4-61
Also,
i.e.,
1 1 1
1
100
1
120
11
600
600
54 54
1 2
R R R
R
= +
= +
=
= = . .
We have to find dR/dt.
By differentiating Eq. (1), we get
− = − −
⇒ = +
1 1 1
1 1
2
1
2
1
2
2
2
2
1
2
1
2
2
2
R
dR
dt
R
dR
dt
R
dR
dt
dR
dt
R
R
dR
dt
R
dR
dt
= + −
= −
( . )
( )
( . )
( )
( . )54 54
1
100
0 5
1
120
0 6
2
2 2
0.27267 ΩΩ/min.
Therefore, R is decreasing at a rate of 0.27267 W/min.
Example 4.58 A point P on the parabola y
2
= 4ax is moving with a velocity of v ft/s at time t. Find the
rate of increase of coordinates x and y at time t.
Solution: Given the equation of the parabola is y
2
= 4ax.
Let P(x, y) be any point on the parabola.
We have to find dx and dy.
Differentiating ...