Here for help:

for what?

Let me enter the question first

A right triangle has sides 6, 8 and 10. The side with 6 is decreasing 1/min and the side with 8 is increasing 2/min. What is the rate of change of its area?

The answer gives +4, is that correct?

Is this a related rates question?

Yes

and do you want me to help solve, or just me to find the answer?

To check whether the answer is 4

I will appreciate that.

I'll just stream-of-consciousness solve it here

(leaving to make a picture real quick)

Let the time be t, then the area is (6-t)(4+t) right?

1/2 * (6-t)(8+2t)

which yes, is (6-t)(4+t)

guess you don't need the picture then

And finding dA/dt gives 2-2t

but I don't think that will be very useful

Oops I didn't mention that I am requested to find the rate at the second minute

oh yeah that's needed information

so let's declare x = 6, dx/dt = -1, y = 8, dy/dt = 2

A = 1/2 * xy

dA/dt = 1/2 * (x * dy/dt + y * dx/dt)

oh wait x = 4, y = 12

I was suspecting the answer, as the first minute the triangle has value 25 and the second minute has area 24, it should be decreasing as well

As the answer given is positive.

the rate at which the area is decreasing could be decreasing

oh wait thinking of second derivative

dA/dt = 1/2 * (4 * 2 + 12 * -1) = 1/2 * (8 - 12) = 1/2 * -4 = -2

What I found is -2/min (first derivative)

I also found that, so let's just say the answer given is wrong

Obviously the second derivative is -2

I'm not sure how that's obvious, also UNITS

The first derivative is 2-2t

Nah. Units are the same.

wouldn't second derivative be /min^2?

That is called acceleration, right?

Btw thanks a lot.

second derivative of area would be the rate of change of the rate of change of area, so acceleration of area

also, I tutor for fun often, but thanks for thanking!

Thanks for thanking for thanking too

:D

Also I would like to ask if how would you apply Cramer's rule on a 3x3 equation? I have seen many vids using varieties of techniques

Never learned it or forgot it, sorry

might have to call the mods - Deus and Gareth both have math backgrounds

testing :D

hellloooooo

Hi

I'm deeply offended

Sorry. :P

Cramer's rule says that to solve Ax=b (A an nxn matrix, x,b n,1 column vectors) you compute the determinants of A (call that D) and of the matrices you get by replacing one column of A with b (call that D_i when you replace the i'th column). Then x_i = D_i/D.

So for n=3 you need to be able to compute determinants of 3x3 matrices.

I don't know what definition of determinant you've been given, but for 2x2 and 3x3 matrices it turns out that a nice easy-to-remember procedure works (but BE CAREFUL it doesn't work for n>3). Look at all the diagonals of the matrix, including ones that "wrap around". For each of them, multiply together all the elements along that diagonal. Then add up the ones for diagonals running NW-to-SE and subtract the ones for diagonals running SW-to-NE.

So if your matrix is (a b c; d e f; g h i) then you have the obvious NW-to-SE diagonal aei. But you also have (starting at b) bfg, which wraps around from the second to the third row, and (starting at c) cdh, which wraps around from the first to the second row. Those all have positive sign.

In the other direction we have ceg which doesn't have to wrap, but also bdi and afh.

So the determinant is (aei+bfg+cdh) - (ceg+bdi+afh).

So to apply Cramer's rule, you just need to do that four times, once with your original matrix A and once for each of the three modified ones.

Again, that method of calculating the determinant works for 2x2 and 3x3 matrices but NOT for larger ones. (Nor for smaller ones: it would make the determinant of any 1x1 matrix zero.) The thing that's true in general is that to calculate the determinant of a matrix you look at all ways of picking one element from each row, and for each of those you multiply all the elements together and multiply by +1 or -1 depending on "the sign of the permutation" (details available if wanted ...

... but never mind for now) and all all those up, and for 2x2 and 3x3 matrices the only ways to pick one element from each row are ones that just go along a diagonal in one of the two "45-degree" directions, but that's very untrue for larger matrices.

*all all -> add all

well, it doesn't work the same way for 2×2 -- you need to not wrap around for 2×2

@00xxqhxx00, see above Cramer's rule stuff

oh, yeah, of course you're right