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 = x^{3} . Find the rate of change of the volume with respect to x when x = 5 units.
Solution :
v = x^{3}
dv/dx = 3x^{2}
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) = πr^{2}
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
tan 45 = x/5 1 = x/5 x = 5 |
tan θ = x/5 x = 5 tan θ |
differentiating x = 5 tan θ
dx/dt = 5 sec^{2}θ (dθ/dt)
dx/dt = 5 sec^{2}45 (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.
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