Physical Applications of Calculus, 2 3unit questions need help! (1 Viewer)

[currysauce]

1. A kite 20m high, is blown along horizontally at 3m/s by the wind. At what rate is the string being let out when it is 40m long?



2. A chute drops sand at constant rate of 8m^3 per 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, to 2.d.p, when its height is 2m.

 

[Pace_T]

mate thats 2 unit questions
i done them last nite
lol
anyway i will copy from my workbook

 

[currysauce]

oops, so it is... ahh well

 

[Pace_T]

1. A kite 20m high, is blown along horizontally at 3m/s by the wind. At what rate is the string being let out when it is 40m long?


you've got a triangle

ground, straight up is 20m
the horizontal line is 3ms^-1 (let this be x)
the hypoteneuse is 40m (let this b y)

dx/dy = 3

dy/dt = dy/dx * dx/dt

now, y^2 = x^2 + 20^2
y = sqrt(x^2 + 20^2)
dy/dx = x/sqrt(x^2 + 20^2)

since x = sqrt1200 (pyfagoras theorem)
dy/dx = 0.866025403
.'. dy/dt = 0.866025403*3
=2.6ms^-1

 

[Pace_T]

2. A chute drops sand at constant rate of 8m^3 per 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, to 2.d.p, when its height is 2m.


ok, you know that h = 2r, r = 1/2*h

we need to find dh/dt
dh/dt = dv/dt * dh/dv

dv/dt = 8 (given)

dh/dv = 1/(dv/dh)
now,
v = 1/3pi*r^3h
but h = 2r
so v = (1/3*pi*(1/2h)^2)*h (i subbed in r = .5h)
v = 1/12*pi*h^3

dv/dh = 1/4*pi*h^2

dh/dv = 1/(1/4*pi*h^2)

since h = 2, dh/dv = 1/pi

now, dh/dt = 8/pi
= 2.546
= 2.55mm^-1 (2dp.)

 

[currysauce]

thankyou good sir