Question

Find dw/dt without converting w to a function of t and using the appropriate chain rule W = xycosz; x = t; y = t^2 ; z= arccost

I don't think this is possible

Answer 1

convert W into a function of t. Any ideas?

Answer 2

dw/dt = dw/dx dx/dt * dw/dy dy/dt * dw/dz dz/dt ?

Answer 3

But there is an arccos 1/sqrt(). ohh nuuuu D:

Answer 4

my bad, I can't read for shiz.

Answer 5

Is this a multivariate calculus quesiton?

Answer 6

yes :(

Answer 7

bahaha, I'm not qualified to do this, but I'll give what I'd do:
dW/dt= dw/dx dx/dt+dw/dy dy/dt+dw/dz dz/dt

Answer 8

as for the arccosine you automatically assume that domain of z is within that of the arccosine , so the W(x,y,z) cancels out for jus the partial z

Answer 9

oh. Thanks for the advice

Answer 10

dz/dt is a formula that you should have memorized. d/dx arccos(x)=-1/sqrt(1-x^2)

Answer 11

That's the what the teacher said too

Answer 12

what's the main issue you are seeing?

Answer 13

partial derivitives scare me, I never do them right ;/

Answer 14

I get an angle that i put into the arccos and it's crazy bad. But the teacher said they would cancle too. So you guys must know what's up.

Answer 15

i've got the answer
It's hard :/

Answer 16

well, just follow the rule.
dW/dt= dw/dx dx/dt+dw/dy dy/dt+dw/dz dz/dt
assuming you already have the first two terms of that down, we'll just look at dw/dz dz/dt

dw/dz=-xy sin z; dz/dt=-1/sqrt(1-t^2)
dw/dz dz/dt=\(\Huge \frac{-xy\sin z}{\sqrt{1-t^2}}\)
we know that
x=t
y=t^2
z=arccos(t)
so
dw/dz dz/dt=\(\Huge \frac{-t^3\sin z}{\sqrt{1-t^2}}\)
at this point recall that sin^2 theta+cos^2 theta=1, but also that z must be between 0 and pi inclusive, so sin z must be between 0 and 1 (sin z is positive). Therefore we shall take the positive square root of sin z=sqrt(1-cos^2 t)

Answer 17

*cos^2 z

Answer 18

Thank you for your help though
if your curious it's w = t^4, dw/dt = 4t^3

Answer 19

oh, that's the same thing?

Answer 20

\(\Huge \frac{-t^3\sqrt{1-\cos^2(\arccos t)}}{\sqrt{1-t^2}}=-t^3\)

Answer 21

what's the same thing?

Answer 22

is that equivilent to the answer
;/ arccos is very hard for me to understand.

Answer 23

but it should be 4t^3

Answer 24

Well, I only evaluated dw/dz dz/dt

Answer 25

I believe I got the sign wrong, and should've gotten a t^3

Answer 26

dw/dz=-xy sin z; dz/dt=-1/sqrt(1-t^2)
yes, in this line they should cancel to make positive.

Answer 27

If you want, I am perfectly willing to start over and go through the entire problem with you the way it was supposed to be done without doing the substitutions for t

Answer 28

;D omg that would be great. But I must warn you i'm horrible at math

Answer 29

okay. So let's start over; I'll recite the chain rule.
\(\Huge \frac{dW}{dt}=\frac{\partial W}{\partial x} \frac{dx}{dt}+\frac{\partial W}{\partial y}\frac{dy}{dt}+\frac{\partial W}{\partial z}\frac{dz}{dt} \)

Answer 30

i reamber that because it always criss crosses with the x, then y, then the z
:( but so hard to use

Answer 31

and I think arcsin and arccos are always used to find theda

Answer 32

As we can see it's a three part problem, so let's just start with the x-portion, then the y-portion, then the z-portion.

Calculating dw/dx we get simply y cos z because y cos z is a constant, and x becomes 1
From the problem, y=t^2, and z=arccos t
Therefore dw/dx=(t^2)cos(arccos t)=t^3

Calculating dx/dt we get 1

therefore the first term
\(\Huge\frac{\partial W}{\partial x} \frac{dx}{dt}=t^3*1=t^3\)

Answer 33

ok

Answer 34

Now for the second part.
Calculating dw/dy we get simply x cos z because x cos z is a constant, and y becomes 1.
From the problem, x=t, and z=arccos t
Therefore dw/dy=(t)cos(arccos t)=t^2

Calculating dy/dt we get 2t.

Therefore the second term
\(\Huge \frac{dW}{dt}=\frac{\partial W}{\partial y}\frac{dy}{dt}=t^2*2t=2t^3\)

Answer 35

adding them together would give us 5t^3?

Answer 36

Now for the third part, the trickiest part.
Calculating dw/dz we get -xy sin(z) because xy is a constant, and cos(z) becomes - sin z.
From the problem, x=t, y=t^2,
therefore dw/dz=-t^3 sin(z)
Notice I have not gotten rid of the z. since it's not immediately apparent what to do with it, it's best to leave it alone for now.

Calculating dz/dt we get -1/sqrt(1-t^2)

The trick here is to recall the pythagorean identity from trigonometry, that is
\(\Large \sin^2\theta+\cos^2\theta=1\)
Because of this, \(\Large \sin\theta=\pm \sqrt{1-\cos^2\theta}\) (looks so similar to dz/dt)
But this is an identity, and thus applies to any variable; in this case z is of interest because we got dw/dz=-t^3 sin(z) earlier; \(\Large \sin z=\pm \sqrt{1-\cos^2 z}\)

The big question is, do we use the plus or the minus? The answer lies within analyzing the domain and range of the functions we were originally given.

Answer 37

why do we look in the domain?
because it has to be a real number?

Answer 38

recall that z=arccos(t), and that W=xy cos(z)
The range of the arccos function is between 0 and pi. Remember that we are interested in taking the sin of z. In other words, we want to look at all possible values of the sin of z to determine whether or not the sign is positive (if we know the sin of z is always positive, then we take the positive square root because the square root yields a positive number; if we know that the sin of z is negative, we take the negative square root for hte same reason). However, we know that the z=arccos(t), meaning that we are taking the sin of z from 0 to pi. If you look at a graph of sin z, you'll find that sin z from [0,pi] is really the first "bump" of the graph.
Thus because sin z is always positive, we know that our expression \(\pm\sqrt{1-\cos^2 z}\) is always positive, and because the square root of a number must be positive (in the real range), we must take the positive root because plus times plus makes plus.

Thus we know that \(\Huge \sin z=+\sqrt{1-\cos^2 z}\)