Related Rates Question

[Tacobell]

Hello, this is my first post. I have a question of related rates that was giving me trouble, could someone help me, it is...

 

There is an isosceles triangle with the sides increasing at a rate of 2 ft/min, the base is 19 feet, what is the rate of change of area when the height is 8 ft?

 

Someone said to me take the derivative with respect to a variable other than t, this did not make sense to me because you want rate of growth per unit of time.

 

 

Tacobell

[Amr Morsi]

area=0.5*base*height=0.5*base*sqrt(S^2-(9.5)^2), where S is the length of the side.

 

Then, d(area)/dt=d(area)/dS*dS/dt= 0.5*19*S/sqrt(S^2-(9.5)^2) * 2 ft^2/min

 

At height=8 ft, S=sqrt(8^2+(9.5)^2) ft.

 

Substitute with S in the rate of change of area , and you will get the result.

[Bignose]

Amr,

 

in the future, please note that this forum tries not to just give answers to posters, especially when the problem seems very home-work like.

 

We try to help the questioner help themselves, by giving them ideas to look at, different points of view, asking them what knowns and unknowns they have and what equations relate them all, etc. We also will look over work that they post, and point out mistakes if we see them.

 

But, we usually try to not spoon feed answers. In short, no one gets much of anything out of that. Look around the Homework Help section to get an idea of what I am writing about here. The "not doing homework" rule applies to all sections, not just the homework help section.

 

Thanks

[triclino]

Hello, this is my first post. I have a question of related rates that was giving me trouble, could someone help me, it is...

 

There is an isosceles triangle with the sides increasing at a rate of 2 ft/min, the base is 19 feet, what is the rate of change of area when the height is 8 ft?

 

Someone said to me take the derivative with respect to a variable other than t, this did not make sense to me because you want rate of growth per unit of time.

 

 

Tacobell

 

Tacobell

 

The only thing that does change with time is the angle, θ between the side of the triangle and its height.

 

So if you can express the area of the triangle in terms of this angle and one of its sides ,then very easily you will find the:

 

dA/dt of the triangle,because you know the d(one side of the triangle)/dt = 2ft/min

[Amr Morsi]

O.K. Bignose. Sorry for that.

[Tacobell]

Thanks, for the help. This question is actually from a test I took last week. I just was wondering about the answer. I got the derivative of the formula for the area of a triangle then got the derivative of the Pythagorean theorem and solved both for height rate and set equal to each other (eliminating dh/dt) and then solved for the rate I wanted (dA/dt rate of area change). Is there anything wrong with this?

 

Tacobell

[Amr Morsi]

It's all right, Tacobell.

[triclino]

Here is an alternative proof:

 

Let A be the area of triangle

 

x ,the height

 

z the side of the triangle

 

y the base of the triangle

 

And , θ the angle between the height ,x and the side z

 

 

Now all the above except , θ ,change with time . We also know dz/dt .

 

So if you can express A in term of ,θ and z ,then the calculation of dA/dt will be an easy matter.

 

So,the area of the triangle is:

 

A = (y.x)/2.........................................................................................1

 

 

But :

 

sinθ = [math]\frac{\frac{y}{2}}{z} = \frac{y}{2z}\Longrightarrow[/math] 2zsinθ = y.........................................................................................2

 

 

cosθ = [math]\frac{x}{z}\Longrightarrow[/math] zcosθ = x......................................................................................................3

 

 

And substituting (2) and (3) into (1) we have:

 

 

A= sinθcosθ[math] z^2[/math]

 

Now we can differentiate w.r.t time ,t.

 

Thus :

 

[math]\frac{dA}{dt}[/math] = sinθcosθ[math]\frac{d(z^2)}{dt}[/math]=2zsinθcosθ[math]\frac{dz}{dt}[/math],but since dz/dt = 2ft/min and zcosθ =x we have that:

 

[math]\frac{dA}{dt}[/math]= 4xsinθ (ft\min). And since we want the rate when x=8ft ,then we have:

 

[math]\frac{dA}{dt}[/math] = 32sinθ ([math]\frac{ft^2}{min}[/math]).

 

Where θ =arctan[math]\frac{\frac{19}{2}}{8}[/math],if when x=8ft ,y=19ft