Rate of change calculus (1 Viewer)

[pLuvia]

John Observes a helicopter is 260 m from him. The helicopter is flying horizontally towards him at a rate of 30m/s at an altitude of 100 m. How fast is the helicopter approaching him?

Answer: 27⁹/₁₃

 

[*fkr]

im shit at maxima and minima but i figured it out using pythagoras though its not exactely 27/9/13

 

[klaw]

im shit at maxima and minima but i figured it out using pythagoras though its not exactely 27/9/13

We're not finding the side, we're finding the rate that the helicopter is approaching him

 

[klaw]

Let r be horizontal distance and x be distance from John to the helicopter

dr/dt=30 m/s
x^2=100^2+r^2
dx/dr=r/sqrt(100^2+r^2)
.:dx/dt=30r/sqrt(100^2+r^2)
When x=260, r=240

When r=240, dx/dt= 360/13=answer

 

[pLuvia]

thanks klaw

 

[pLuvia]

The volume (Vm³) of water in a given vessel is given by V=2x²+2x, where x is the depth of the water in metres. If water is entering the vessel at a constant of 0.03m³/s, find the rate at which the depth of water is increasing when x=3/4 m. Express your answer in m/min

 

[PLooB]

The volume (Vm³) of water in a given vessel is given by V=2x²+2x, where x is the depth of the water in metres. If water is entering the vessel at a constant of 0.03m³/s, find the rate at which the depth of water is increasing when x=3/4 m. Express your answer in m/min

V=2x²+2x

dV/dx = 4x+2

dV/dt= 0.03

we need dx/dt

so dx/dt = dV/dt x dx/dV = 0.03 x 1/(4x+2) = 0.03/(4x+2)

so when the depth is 0.75m (x=0.75) the rate at which the level is increasing is:

dx/dt = 0.03/[4(0.75)+2]= 0.006m/s x 60 = 0.36m/min

The deth of the water is increasing at 0.36 metres per minute.

Haven't done these in a long time so I'm not too sure.

 

[pLuvia]

aa thanks mate

 

[pLuvia]

In figure, AB = 3cm, AC = 6cm, BC = xcm and angle A = @
a) Express x² in terms of @
b) If @ increases at the rate of 1/3 radian/sec, find the rate of change of x with respect to time when @ = pi/3

 

[webby234]

a)

x^2 = 6^2 +3^2 - 2 x 6 x 3 cos@

x^2 = 45 - 36 cos@

 

[webby234]

b) dx/dt = dx/d@ x d@/dt

d@/dt = 1/3

x = (45-36cos@)^1/2
dx/d@ = 1/2 x 36sin@ (45-36cos@)^(-1/2)

dx/dt = 6sin@/(sqrt(45-36cos@)

At pi/3

3sqrt3/sqrt(45-18)

= 3sqrt3/3sqrt3
= 1