Question
477i 0Jt(2) J pu. (& "suogenba uuBQpnu-Stonej Suysn ofjeue %! (2)J J8q MouS( uaq T= z2 = (2) [ 13Treimw sic Irm? NINII - dnwv OZOELVOL "LOZO2) 31np237 2UO4123S II SDILVWJHIVW DNIUJJNIDN]) NINJC dnvv 0z08l a9jn12a- zuol12a8
477i 0Jt (2) J pu. (& "suogenba uuBQpnu-Stonej Suysn ofjeue %! (2)J J8q MouS( uaq T= z2 = (2) [ 13T reimw sic Irm? NINII - dnwv OZOELVOL "LOZO2) 31np237 2UO4123S II SDILVWJHIVW DNIUJJNIDN]) NINJC dnvv 0z08l a9jn12a- zuol12a8
Answers
Ilcrc, whire precipirare $\{\mathbb{R})$ obraincd on rrcarment wirh aqucous solurion of $\mathrm{BaCl}_{2}$ is of (a) $\mathrm{AgC}$. (b) $\mathrm{CaCl}_{2}$ (c) $\mathrm{BaSO}_{4}$ (d) $\mathrm{PbC}^{\prime} \mathrm{I}_{2}$
We're gonna find the 1st and 2nd derivative of our vector given. So in order to do that, we are going to take derivatives of each of our components. So we're going to have a two T in the I direction plus a three t squared in the J, and then just minus one in the K. So we can just write minus K when it is a one. Okay, now, for our second derivative, we are again going to take a derivative of our first relative this time. So we're going to have to in the I direction plus six T in the J direction and then R. K value actually zeros out. So we don't have to write it.
Okay. Hey, guys. In this problem, you're told that I
It's probably given the intuition I did 14 15 16 17 north To solve this, you're gonna need these little facts right here. That idea zero is one eye's ableto I I squared is equal to negative. What I cubed is equal to make by pattern repeats. Well, using this pattern, it is clear than on it. Of the 14 he is going to be the same is going to be the same. Power is going to be the same as I squared. I switch is negative once and literally I to the 15 is going to be I cubed, which is negative. I I to the 16th Who's gonna be equal to you? I to the fourth or high zero which is one and I to the 17 is equal to you. I the first, which is ah, which is I. And if you add all of these are you get the value zero and that is inter choice A in the problem Now the reason why I was able to get these steps is because you take the module is of the expert. You see that these exponents of five repeats of
Okay, This equation has three variables Z squared equals x squared plus y squared. So, we had to be given extra information. DX DT is two. Dy DT is three. Find DZ DT when X is five and y is 12. All right. So every one of these is changing with respect to T. So when you take the derivative of the square, you get to Z times the derivative Ozzy, which is DZ DT equals two X times the derivative of X. Dx DT plus two Y times the derivative of Y Dy DT. So we can divide the twos off of all of them. Okay? So find DZ DT when X is five, Ny is 12. So five is gonna go here. 12 is gonna go here but we don't have anything for the Z. So we have to stop and find that first. Use the original equation. So, Z squared equals five squared last 12 squared. That's 25 plus 1 44. That's 1 69. So, Z is plus minus The square root of 169, which is 13. That's where he's positive 13 1st. So 13 DZ DT equals five times two plus 12 times three. So DZ DT equals 10 plus 36. 46/13. Okay, And then when you plug in uh minus 13 for Z, you'll get dizzy. D T Equals -46/13