Related Rate Problem

[cyugsi2]

Hello board I'm having difficulty with one of the homework problem I have been assigned. the problem is

at what rate is the diagonal of a cube increasing if its edges are increasing at a rate of 2cm/s

I am horrible at word problems but what I do know is dx/dt will be 2cm/s and the thing that would probably lead me to finding out what rate the diagonal of a cube is increasing is the pythagorean theorem. what i'm having trouble how to use the pythagorean theorem to form an equation that connects the diagonal of the original cube and the diagonal of the cube its edges increased . any help would be appreciated

 

[tkhunny]

I am horrible at word problems

Really, just never say this or think this again. It is a little like learning a new language. Many have survived it.

at what rate is the diagonal of a cube increasing if its edges are increasing at a rate of 2cm/s

You need some sort of relationship between the edge and the diagonal.

If the edge is 'r'.

The volume (V) is r^3

The surface area (SA) is 6*r^2

The diagonal (D) is \(\displaystyle \sqrt{r^{2}+r^{2}+r^{2}} = \sqrt{3\cdot r^{2}}\)

You must know these things. It is acceptable to expect you to know them for an exam (WITHOUT a cheat sheet!)

And there you have it \(\displaystyle D = \sqrt{3\cdot r^{2}}\) or, to simplify your life, \(\displaystyle D^{2} = 3\cdot r^{2}\)

Find dD/dr and you are almost done.

 

[cyugsi2]

I'm still struggling with this problem I thought I was trying to figure out dD/dt

 

[Deleted member 4993]

I'm still struggling with this problem I thought I was trying to figure out dD/dt

Yes you are - however you are given dr/dt.

dD/dt and dr/dt are related. What is the relationship??

 

[cyugsi2]

Ok so I'm trying to figure it out but i might be going completely in the wrong direction what I'm thinking is

the relation is the rate in they are asking of the problem. now to figure out that rate I would be using like triangles since the triangle that has the original diagonal and the diagonal of a cube expanded at 1 second will have the same angles and ratio of the length sizes of the triangles. That is how I find out what dD/Dr is right? Would it be correct to say that (d/d-4)=(sqrt(r^2+r^2)+sqrt(2))/(sqrt(r^2+r^2)) this looks way to complicated so I think I might have completely gone in the wrong direction...

 

[cyugsi2]

wait wait I thought about it again and I could write that as d/(d-4)=r/r-sqrt(2) does that make more sense?

 

[Deleted member 4993]

TK showed you the relation between 'D' (the cube diagonal) and r (the edges of cube)

\(\displaystyle D \ = \ r \cdot \sqrt{3}\)

If you differentiate both sides with respect to 't' you get

\(\displaystyle \frac{dD}{dt} \ = \ \frac{dr}{dt} \cdot \sqrt{3}\)

you know

\(\displaystyle \frac{dr}{dt} \ = \ 0.02 \frac{m}{s}\)

Now continue......

 

[tuanphan]

Thanks

man, you're so cool, thanks for the direction

TK showed you the relation between 'D' (the cube diagonal) and r (the edges of cube)

\(\displaystyle D \ = \ r \cdot \sqrt{3}\)

If you differentiate both sides with respect to 't' you get

\(\displaystyle \frac{dD}{dt} \ = \ \frac{dr}{dt} \cdot \sqrt{3}\)

you know

\(\displaystyle \frac{dr}{dt} \ = \ 0.02 \frac{m}{s}\)

Now continue......

 

[rjump]

TK showed you the relation between 'D' (the cube diagonal) and r (the edges of cube)

\(\displaystyle D \ = \ r \cdot \sqrt{3}\)

If you differentiate both sides with respect to 't' you get

\(\displaystyle \frac{dD}{dt} \ = \ \frac{dr}{dt} \cdot \sqrt{3}\)

you know

\(\displaystyle \frac{dr}{dt} \ = \ 0.02 \frac{m}{s}\)

Now continue......

Why does sqrt{3} stay the same after you differentiate?

 

[daon2]

Why does sqrt{3} stay the same after you differentiate?

Why does the 2 "stay the same" in the following statement? \(\displaystyle d/dx(2x) = 2\)?

I guarantee there is a (probably bold, or colored) box with a list of rules for basic differentiation in your textbook, and likely in the back of your book too. You'll find your answer there.