rates question (1 Viewer)

Satiric · Apr 12, 2009

i need help with this question..

this is from exercise 22.4 Q15 couchman & jones

A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm cubed. Water is poured in at a constant rate of 5 Litres per minute. Find the rate at which the water level is rising when the depth is 30 cm. (solved)

okay i have another question guys..
its similar and easier..but i cant get the hang of it :S
anyways the question is
The cross-section of a trough l metre long is in the shape of an isoseles triangle of base measurement of 2a metre and height b metre. Water leaks from the trough at a constant rate of c metre cubed/min. FInd the rate at which the water level is falling when the depth of the water is b/2 metre.

Thanks in advance

jet · Apr 12, 2009

Are you sure that is the whole question?

GUSSSSSSSSSSSSS · Apr 12, 2009

^^ i agree

does it give perhaps the angle at which the sides of the trough are inclined??? cos otherwise you would have anotha pronumeral in there..

Satiric · Apr 12, 2009

yep. Thats the whole question word for word.
its impossible :S

jet · Apr 12, 2009

Actually, it makes sense now... Just imagine the right angle at the apex of the triangle. Hence, the angles are 45 degrees.
$\text{Let the base of the triangle formed by the water in the trough be } b \\ \therefore \text{ half the length of the base can be calculated using simple trig:} \\ tan45^{\circ} = \frac{x}{\frac{b}{2}} \\ b = 2x \\ \text{Hence the area of the triangle } A = \frac{1}{2}bh \\ = \frac{1}{2}(2x)(x) \\ = x^2 cm^2 \\ \therefore \text{ the volume of the water } V = x^2 \times 200 \\ = 200x^2 cm^3 \\ \text{Now, } \frac{dV}{dt} = 5000 ml \, min^{-1} \\ \text{but, } \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \\ 5000 = 400x \frac{dx}{dt} \\ \text{when } x = 30, 5000 = 12000\frac{dx}{dt} \\ \frac{dx}{dt} = 0.42667 cm \, min^{-1}$

The reason i converted the dv/dt to mL was because the volume was expressed in cubic centimetres, and 1 cubic centimetre = 1 mL

GUSSSSSSSSSSSSS · Apr 12, 2009

jetblack2007 said:
Actually, it makes sense now... Just imagine the right angle at the apex of the triangle. Hence, the angles are 45 degrees.
$\text{Let the base of the triangle formed by the water in the trough be } b \\ \therefore \text{ half the length of the base can be calculated using simple trig:} \\ tan45^{\circ} = \frac{x}{\frac{b}{2}} \\ b = 2x \\ \text{Hence the area of the triangle } A = \frac{1}{2}bh \\ = \frac{1}{2}(2x)(x) \\ = x^2 cm^2 \\ \therefore \text{ the volume of the water } V = x^2 \times 200 \\ = 200x^2 cm^3 \\ \text{Now, } \frac{dV}{dt} = 5000 ml \, min^{-1} \\ \text{but, } \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \\ 5000 = 400x \frac{dx}{dt} \\ \text{when } x = 30, 5000 = 12000\frac{dx}{dt} \\ \frac{dx}{dt} = 0.42667 cm \, min^{-1}$

The reason i converted the dv/dt to mL was because the volume was expressed in cubic centimetres, and 1 cubic centimetre = 1 mL

shit completely missed where it said "right angle"

oh well

you win lol

EvoRevolution · Apr 12, 2009

yea lol this i had troubles with this question aswell, the only one i didnt get.
But i didn't wht the F___ a water trough was so yea.

GUSSSSSSSSSSSSS · Apr 12, 2009

EvoRevolution said:
yea lol this i had troubles with this question aswell, the only one i didnt get.
But i didn't wht the F___ a water trough was so yea.

AHAHAHAHA

but it all works out well..

Satiric · Apr 12, 2009

jetblack2007 said:
Actually, it makes sense now... Just imagine the right angle at the apex of the triangle. Hence, the angles are 45 degrees.
$\text{Let the base of the triangle formed by the water in the trough be } b \\ \therefore \text{ half the length of the base can be calculated using simple trig:} \\ tan45^{\circ} = \frac{x}{\frac{b}{2}} \\ b = 2x \\ \text{Hence the area of the triangle } A = \frac{1}{2}bh \\ = \frac{1}{2}(2x)(x) \\ = x^2 cm^2 \\ \therefore \text{ the volume of the water } V = x^2 \times 200 \\ = 200x^2 cm^3 \\ \text{Now, } \frac{dV}{dt} = 5000 ml \, min^{-1} \\ \text{but, } \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \\ 5000 = 400x \frac{dx}{dt} \\ \text{when } x = 30, 5000 = 12000\frac{dx}{dt} \\ \frac{dx}{dt} = 0.42667 cm \, min^{-1}$

The reason i converted the dv/dt to mL was because the volume was expressed in cubic centimetres, and 1 cubic centimetre = 1 mL

holy crap.
thanks a lot man. =D

Drongoski · Apr 12, 2009

I thought you just posted a new problem. I typed out the solution (with great difficulty for me) and when I pressed the button to send it says invalid thread. what a wasted effort.

I'm assuming top of trough is 2a wide (question ambiguous to me)

$Let\ x\ metre\ be\ height fr \ bottom\\ \\\therefore \ \ V(x)\ \ =\ \ 1 \times \frac{1}{2}\times \frac{2ax}{b}\times x\ \ =\ \ \frac{ax^{2}}{b}\\ \\ \therefore \ \ \frac{\mathrm{d} V}{\mathrm{d} x}\ \ =\ \ \frac{2ax}{b}\ \ \ and\ \ \frac{\mathrm{d} V}{\mathrm{d} t}\ \ =\ \ -\ c\\ \\ Now\ \ \ \ \frac{\mathrm{d} V}{\mathrm{d} t}\ \ =\ \ \frac{\mathrm{d} V}{\mathrm{d} x}\times \frac{\mathrm{d} x}{\mathrm{d} t}\\ \\\therefore \ \ -\ c\ \ =\ \ \frac{2ax}{b}\times \frac{\mathrm{d} x}{\mathrm{d} t}\\ \\ \therefore \ \ \frac{\mathrm{d} x}{\mathrm{d} t}\ \ =\ \ \frac{-bc}{2ax} \ \ =\ \ \frac{-bc}{2a(\frac{b}{2})}\ \ =\ \ -\frac{c}{a}$

Satiric: did u not post a 2nd rates problem not long ago to which above is my attempted solution.

Satiric · Apr 13, 2009

Drongoski said:
I thought you just posted a new problem. I typed out the solution (with great difficulty for me) and when I pressed the button to send it says invalid thread. what a wasted effort.

I'm assuming top of trough is 2a wide (question ambiguous to me)

$Let\ x\ metre\ be\ height fr \ bottom\\ \\\therefore \ \ V(x)\ \ =\ \ 1 \times \frac{1}{2}\times \frac{2ax}{b}\times x\ \ =\ \ \frac{ax^{2}}{b}\\ \\ \therefore \ \ \frac{\mathrm{d} V}{\mathrm{d} x}\ \ =\ \ \frac{2ax}{b}\ \ \ and\ \ \frac{\mathrm{d} V}{\mathrm{d} t}\ \ =\ \ -\ c\\ \\ Now\ \ \ \ \frac{\mathrm{d} V}{\mathrm{d} t}\ \ =\ \ \frac{\mathrm{d} V}{\mathrm{d} x}\times \frac{\mathrm{d} x}{\mathrm{d} t}\\ \\\therefore \ \ -\ c\ \ =\ \ \frac{2ax}{b}\times \frac{\mathrm{d} x}{\mathrm{d} t}\\ \\ \therefore \ \ \frac{\mathrm{d} x}{\mathrm{d} t}\ \ =\ \ \frac{-bc}{2ax} \ \ =\ \ \frac{-bc}{2a(\frac{b}{2})}\ \ =\ \ -\frac{c}{a}$

Satiric: did u not post a 2nd rates problem not long ago to which above is my attempted solution.

yea i did. sorry about that.
I thought i had it..when i didnt :S
anyways thanks a lot.
can u plz explain how you got 2ax/b in the first line? Im really stupid :S
thanks

Drongoski · Apr 13, 2009

Imagine the full isosceles triangle base 2a; go x metres from bottom up the perpendicular bisector of the triangle. draw a line parallel to the base (the top of the trough). Then the smaller lower triangle is similar to the original cross=sectional isosceles triangle. Therefore, since similar triangles are proportional, x/(its base) = b/2a so thst base of smaller triangle is 2ax/b and the area of triangle = half x base x height = (1/2) . (2ax/b) .x = ax^2/b and for its volume multiply by length = 1 metre so that V(x) = 1 . ax^2/b and I've used V(x) to emphasise that it is a function of 'x'.

hope this helps.

Satiric · Apr 13, 2009

Drongoski said:
Imagine the full isosceles triangle base 2a; go x metres from bottom up the perpendicular bisector of the triangle. draw a line parallel to the base (the top of the trough). Then the smaller lower triangle is similar to the original cross=sectional isosceles triangle. Therefore, since similar triangles are proportional, x/(its base) = b/2a so thst base of smaller triangle is 2ax/b and the area of triangle = half x base x height = (1/2) . (2ax/b) .x = ax^2/b and for its volume multiply by length = 1 metre so that V(x) = 1 . ax^2/b and I've used V(x) to emphasise that it is a function of 'x'.

hope this helps.

yep. got it
thanks a lot.

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