FINDING RATE OF CHANGE FROM A WORD PROBLEM

If θ is the angle made by the tangent to the curve

y  =  f(x)

at the point (x, y) , then the slope of the curve at (x, y) is f'(x) = tanθ,

where θ is measured in the anti clock wise direction from the X -axis. Note that, f ′(x) is also denoted by

dy/dx 

and also called instantaneous rate of change.

Problem 1 :

If the volume of a cube of side length x is v = x³ . Find the rate of change of the volume with respect to x when x = 5 units.

Solution :

v  =  x³

dv/dx  =  3x²

dv/dx at x  = 5  ==>  3(5)²

=  75 units.

Problem 2 :

If the mass m(x) (in kilograms) of a thin rod of length x (in meters) is given by, m(x) = √3x then what is the rate of change of mass with respect to the length when it is x = 3 and x = 27 meters.

Solution :

m(x) = √3√x

Rate of change of mass  m'(x)  =  √3(1/2√x)

m'(x)  =  (√3/2√x)

When x  =  3

m'(3)  =  (√3/2√3)  ==>  1/2 kg/m

When x  =  27

m'(27)  =  (√3/2√27)  ==>  1/6 kg/m

Problem 3 :

A stone is dropped into a pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate at 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?

Solution :

Area of ripple(A)  =  πr²

dr/dt  =  2 cm/sec

r  =  5

Supposed to find dA/dt

dA/dt  =  2πr (dr/dt)

dA/dt  =  2π(5)(2)

dA/dt  =  20π

Problem 4 :

A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shore line. How fast is the beam moving along the shore line when it makes an angle of 45° with the shore?

Solution :

Rate of revolution by the beacon :

1 revolution  =  10 seconds

360 degree  =  10 seconds

1 sec  =  2π/10

Angular velocity (d θ/dt)  =  2π/10  ==>  π/5

In the triangle,

tan θ  =  Opposite side/ Adjacent side

differentiating x  =  5 tan θ

dx/dt  =  5 sec²θ (dθ/dt)

dx/dt  =  5 sec²45 (π/5)

dx/dt  =  5 (√2)² (π/5)

dx/dt  =  2π km/sec

So, the beam is moving at the rate of 2π km/sec.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.