Related Rates | Page 1 | Mathematics | Guild Forums

Dr. Leonard McCoy

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Posted: Mon Oct 16, 2006 11:55 am

Ok... So we got to related rates in my calculus class. Most of them were a breeze, but I am having problems on two of them.

problem 1

Quote:

A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30cm wide at the bottom, 80cm wide at the top, and has the height 50cm. If the trough is being filled with water at the rate of 0.2m^3/min, how fast is the water level rising when the water is 30cm deep?

This being and odd, I know the answer is 10/3 cm/min.

problem 2

Quote:

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3.

Sorry about the pi/3 bit... But the answer to this one is 0.3 m^2/sec.

I would like to thank everyone that helps out in advance! It is weird this is the first time where I have the answer, but I can't figure out the work... Oh well.

Posted: Mon Oct 16, 2006 11:58 pm

Dr. Leonard McCoy

problem 1

Quote:

A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30cm wide at the bottom, 80cm wide at the top, and has the height 50cm. If the trough is being filled with water at the rate of 0.2m^3/min, how fast is the water level rising when the water is 30cm deep?

This being and odd, I know the answer is 10/3 cm/min.

If you cut away the trapezoid at some height h<50cm, the top base will be changed, but not the bottom nor the bottom angles. This allows you to calculate the volume V(h) of this restricted trapezoidal trough, which is just the volume of the water for some height. Then, using the chain rule dV/dt = [dV/dh][dh/dt], it is very simple to find dh/dt, as dV/dt is given as constant.

Dr. Leonard McCoy

problem 2

Quote:

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3.

Sorry about the pi/3 bit... But the answer to this one is 0.3 m^2/sec.

The area of a triangle is A = [ab/2] sin θ. Differentiate using the chain rule; youre already given dθ/dt as constant. Your answer is correct.

VorpalNeko
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Dr. Leonard McCoy

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Posted: Tue Oct 17, 2006 7:09 am

VorpalNeko

Dr. Leonard McCoy

problem 1

Quote:

A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30cm wide at the bottom, 80cm wide at the top, and has the height 50cm. If the trough is being filled with water at the rate of 0.2m^3/min, how fast is the water level rising when the water is 30cm deep?

This being and odd, I know the answer is 10/3 cm/min.

If you cut away the trapezoid at some height h<50cm, the top base will be changed, but not the bottom nor the bottom angles. This allows you to calculate the volume V(h) of this restricted trapezoidal trough, which is just the volume of the water for some height. Then, using the chain rule dV/dt = [dV/dh][dh/dt], it is very simple to find dh/dt, as dV/dt is given as constant.

The thing is, is that I have no idea how to get the volume of the trough. sweatdrop I am sure that by now I should, but I honestly have no idea... It has been about 5 years since geometry. That stuff is really hazy.
Everything is OK now. I figured it out! finally!!!

Posted: Tue Oct 17, 2006 7:17 am

VorpalNeko

Dr. Leonard McCoy

problem 2

Quote:

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3.

Sorry about the pi/3 bit... But the answer to this one is 0.3 m^2/sec.

The area of a triangle is A = [ab/2] sin θ. Differentiate using the chain rule; youre already given dθ/dt as constant. Your answer is correct.

Thank you so much for the help on this! It is greatly appreciated!

Dr. Leonard McCoy

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