Two Questions? (1 Viewer)

[RHINO7]

Could someone please help me with these two questions--

i) If the volume of a cube is increasing at the rate of 23mm^3s^-1, find the increase in its surface area when its side is 140mm.

ii) The area of an equilateral triangle is increasing at the rate of 2cm^2s^-1.

Show that the area of the triangle is given by A=square root of 3x^2/4, where x is the side of the triangle.

 

[ssglain]

i) Let the sides of the cube have length x mm. Then volume = V = x³, surface area = A = 6x².

With this type of question, it helps to first establish which rate you are asked to find. In this case, dA/dt. This can be broken down into two related rates which can be derived from the functions for volume and surface area:
dA/dt = (dA/dx)*(dx/dt)

Now, dA/dx = 12x, dV/dx = 3x²

Also, dx/dt = (dx/dV)*(dV/dt)
Notice that dx/dV can be obtained by reciprocating dV/dx, and dV/dt is given as 23.

I will leave the rest for you.

ii) Let the altitude of the equilateral triangle = h cm.
By Pythagoras' theorem, h = √(x²+x²/4) = (√3/2)x
.: Area = (1/2)hx = (√3/4)x cm²

 

[jannny]

That question is so hard. I don't get how ssglain did it =/

for question
i) why do u need to break it down into two related rates?

 

[Forbidden.]

That question is so hard. I don't get how ssglain did it =/

for question
i) why do u need to break it down into two related rates?

All I can say is there are two variables and the function rule can help solve it by splitting it.

 

[RHINO7]

I still don't understand question i).

 

[ssglain]

That question is so hard. I don't get how ssglain did it =/

for question
i) why do u need to break it down into two related rates?

I understand why you find this one hard. This is very much the hardest type of rates question that you will get in a 2U paper - in fact, I'm not entirely sure whether they are allowed to ask it.

What f3nr15 said is pretty much it, but I will elaborate a bit further:

In order to obtain dA/dt, there must be a direct relationship between the two variables A and t, e.g. A = f(t). Unfortunately, there is no such relationship so you must consider those that have already been established, and the only relevant one is A = 6x². This means you can represent dA/dt using the variables A, t, and x.

The next step is where some students get very confused. The easiest way (although mathematically incorrect) is to consider dA/dt as a fraction, then clearly dA/dt = (dA/dx)*(dx/dt). So now you have represented dA/dt as two rates dA/dx and dx/dt which can be found using what you know.

dA/dx can be obtained easily by differentiating A = 6x²

dx/dt on the other hand is a bit trickier, but you do have two other conditions to help: V = x³ and dV/dt = 23.
Similarly as above dx/dt = (dx/dV)*(dV/dt)
Now, differentiating V = x³ gives dV/dx = 3x². Reciprocating this gives dx/dV (again, think of this as a fraction) and you alreayd have dV/dt.
[Notice that in the first situation concerning dA/dt you have two variable and need to introduce a third, in this situation concerning dx/dt you have three variables (V, x , t) and need to eliminate one of them (V)]

Therefore,
dA/dt = (dA/dx)*(dx/dt)
= (12x)*(23/3x²)
= 92/x
When x = 140, dA/dt = 0.657 mm²/s

Almost always if you cannot derive a rate by one simple differentiaton, you will need to introduce a third variable and split the rate. Don't hesitate to ask if you still don't understand, although I'm not sure how long it would take me to reply because I have trials to study for - so anyone who understands this please jump in and take over.

 

[RHINO7]

Okay, thanks. I understand now.

 

[bos1234]

Just using the same principal as saqlain but neater
Vol. of cube increasing at

so

must be found

=

=

.

Let the sides = x.

Formula for surface area is =

where

Formula for volume is

multiplying,

=

x

=0.69