Rates involving two or more variables (1 Viewer)

[luvotomy]

hello =)

can someone please show me how to work out the questions below, it would be greatly appreciated:

1) 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.

2) A chute drops sand at a constant rate of 8mcubed per minute. 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 2dp, when its height is 2m.

 

[gordon.tan2008]

1. Use pythagoras since 'being overhead' means perpendicular. let hypotenuse be s or something while the other length of the triangle be x taken x is horizontal (length 2 is given) . From the given information dx/dt = 500. s is the distance between obs and plane.
To find ds/dt use the chain rule ds/dt= ds/dx x dx/dt
From pythagoras theorem s = (x^2+4)^1/2 so ds/dx = x (X^2+4)^-1/2
so ds/dt = [x ( x^2+4)^-1/2] x 500 since dx/dt is given.
from the information given you let s = 5 and by pythagoras x = (5^2-2^2)^1/2 = 21^1/2
therefore ds/dt = 100(21)^1/2km/h

2. Assume the cone is a right angled cone where the height drawn through it to the centre of the circle base is right angled. Let the height be h and radius be r. You are given h = 2r so r=h/2
You need to find dh/dt for rate of change of height and the given information is dv/dt = 8 (sand being poured in)
Chain Rule: dh/dt = dh/dv x dv/dt = 1/(dv/dh) x dv/dt (reciprocal)
The volume of a cone is given by 1/3(pi)r^2h and since r=h/2 the volume is h^3(pi)/12
dh/dt = h^2(pi)/4 x 8 and when height is 2 then dh/dt = 8(pi) approximately 25.13m/s

So hard without diagram and notation

 

[vds700]

hello =)

can someone please show me how to work out the questions below, it would be greatly appreciated:

1) 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.

2) A chute drops sand at a constant rate of 8mcubed per minute. 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 2dp, when its height is 2m.

1) Have a look at my diagram firsdt. Dont laugh, I did it with paint lol, the blue box is representing the plane which is x metres horizontally from the viewer and a metres directly fro m the viewer.

dx/dt = 500 km/h
also by pythagoras' theorem, x² + 4 = a²
a = sqrt(x² + 4)
x = sqrt(a² - 4)
da/dx = x/sqrt(x² + 4)
when a = 5, x = sqrt 21

da/dt = da/dx.dx/dt
=500x/sqrt(x² + 4)
=500 sqrt21/5
= 100 sqrt21 km/h

 

[vds700]

1. Use pythagoras since 'being overhead' means perpendicular. let hypotenuse be s or something while the other length of the triangle be x taken x is horizontal (length 2 is given) . From the given information dx/dt = 500. s is the distance between obs and plane.
To find ds/dt use the chain rule ds/dt= ds/dx x dx/dt
From pythagoras theorem s = (x^2+4)^1/2 so ds/dx = x (X^2+4)^-1/2
so ds/dt = [x ( x^2+4)^-1/2] x 500 since dx/dt is given.
from the information given you let s = 5 and by pythagoras x = (5^2-2^2)^1/2 = 21^1/2
therefore ds/dt = 100(21)^1/2km/h

2. Assume the cone is a right angled cone where the height drawn through it to the centre of the circle base is right angled. Let the height be h and radius be r. You are given h = 2r so r=h/2
You need to find dh/dt for rate of change of height and the given information is dv/dt = 8 (sand being poured in)
Chain Rule: dh/dt = dh/dv x dv/dt = 1/(dv/dh) x dv/dt (reciprocal)
The volume of a cone is given by 1/3(pi)r^2h and since r=h/2 the volume is h^3(pi)/12
dh/dt = h^2(pi)/4 x 8 and when height is 2 then dh/dt = 8(pi) approximately 25.13m/s

So hard without diagram and notation

lol u beat me to it though i did the diagram

 

[tommykins]

hello =)

can someone please show me how to work out the questions below, it would be greatly appreciated:

1) 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.

Draw up a triangle.

2) A chute drops sand at a constant rate of 8mcubed per minute. 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 2dp, when its height is 2m.[/QUOTE]
2) you're trying to find dh/dt when h = 2

We have dv/dt = 8

chain rule means we need dv/dt * dh/dv

Volume of cone = 1/3.pi.r^2.h but since we're given h = 2r, sub r = h/2 into the volume of cone.

differentiate to find dv/dh, but make it the recipricol to fnid dh/dv.

All a matter of subbing in by then.

 

[luvotomy]

Thanks alot guys