Rate of change questions :\ (1 Viewer)

[coeyz]

1.) A 4m long ladder is leaning against a wall. Its base is slipping away from the wall at a constant rate of 2m/s. At what rate, to 2 decimal places, will the top of the aldder be slipping down the wall when the base is 1m out from the wall? (ans: moving down at the rate of 0.52m/s)

2.) A chute drops sand at a constant rate of 8m^3 /min. As the sand falls to the ground it forms a cone shape such that the height of the cone is twice its radius. Find the rate at which the height of the sand will be changing, when its height is 2m. (ans: 2.55m/min)

thanks guyss

 

[Drongoski]

Q1)



i.e. ladder is going down at 0.516 ... m/sec


I'll let someone else have a go at Q2

 

[jet]

 

[jet]

Q1)



i.e. ladder is going down at 0.516 ... m/sec


I'll let someone else have a go at Q2

It is really preferable not to use implicit differentiation, as it is more a 4 unit thing. You are better off taking the positive square root, as you know y is a distance, and that can be differentiated explicitly.

 

[coeyz]

thankyou guys, can you also do this for me?

An observer sees a plane directly overhead, 2km above the ground. The plane is moving horizontally at a constant speed of 500km/h. Find the rate at which the distance will be increasing between the observer and the plane when the distance between them is 5km

ans: 458km/h

thanks again~~

 

[coeyz]

oh i can do it now~ thankss anyway guys

 

[coeyz]

oh last question please

The volume of water in a tidal pool is given by the formula V = 2cos(3pi/x),
where x is the depth of water in the pool. Find the rate at which the depth of the pool will be increasing when the volume in the pool is increasing at the rate of 12m^3/h and the depth is 1.2m

V = 2cos(3pi/x)
dv/dx = 6pi/x^2 sin(3pi/x)

dv/dt = 12

then,

dx/dt = dx/dv . dv/dt
= 1 / [ 6pi/x^2 sin(3pi/x) ] . 12
sub x = 1.2,
then i got the ans dx/dt is 6.71m/h




but the ans is 0.92m/h
which bit did i do wrong
thanks a lot

 

[Drongoski]

It is really preferable not to use implicit differentiation, as it is more a 4 unit thing. You are better off taking the positive square root, as you know y is a distance, and that can be differentiated explicitly.

Good point. Thanks.

 

[Drongoski]

oh last question please

The volume of water in a tidal pool is given by the formula V = 2cos(3pi/x),
where x is the depth of water in the pool. Find the rate at which the depth of the pool will be increasing when the volume in the pool is increasing at the rate of 12m^3/h and the depth is 1.2m

V = 2cos(3pi/x)
dv/dx = 6pi/x^2 sin(3pi/x)

dv/dt = 12

then,

dx/dt = dx/dv . dv/dt
= 1 / [ 6pi/x^2 sin(3pi/x) ] . 12
sub x = 1.2,
then i got the ans dx/dt is 6.71m/h




but the ans is 0.92m/h
which bit did i do wrong
thanks a lot

I remember doing this same question a few days ago & getting it wrong because I forgot the argument of the sin( ) is in radians, not degrees. I think that explains it.

 

[oly1991]

I remember doing this same question a few days ago & getting it wrong because I forgot the argument of the sin( ) is in radians, not degrees. I think that explains it.

yeh this is the link to the thread

http://community.boredofstudies.org/12/mathematics/208205/calculus-physical-world-q.html

 

[Drongoski]

yeh this is the link to the thread

http://community.boredofstudies.org/12/mathematics/208205/calculus-physical-world-q.html

Thank you, oly1991.

 

[coeyz]

oiccc.. thanks guys