Question

Suppose f(x,t)=x^4sin(2t).

(a) At any point (x,t), the differential is
df=

(b) At the point (−2,π/3), the differential is
df=

(c) At the point (−2,π/3) with dx=0.1 and dt=−0.2, the differential is
df=

Answer 1

\[df=(4x^3\sin ^22t+(2\cos 2t)x^4 )d(x,t)\]

Answer 2

b) \[4(-2)^3(\sin 2\pi/3)^2+2\cos (2\pi/3)(-2)^4\]

Answer 3

\[(4x^3\sin ^22t)dx+(2x^4\cos 2t)dt\]

Answer 4

Would you guys like my help?

Answer 5

yes

Answer 6

OK give me a few minutes to type up a short lesson for you. Alright?

Answer 7

If given a function z of two variables, such that\[z =f(x, y)\]then\[dz =\frac{∂z }{ ∂x }dx +\frac{ ∂z }{ ∂y }dy\]Do you guys know what that means?

Answer 8

partial direvative

Answer 9

OK then let's do this one step at a time. As I usually say to my students, when you're learning something that is relatively new, then take baby steps. Hence

Step 1:\[\frac{∂z }{∂x }=?\]Can you tell me please?

Answer 10

No @tukajo it is not! In order to take the partial derivative of a function with respect to x, the other variables must be treated as constants. Care to try again?

Answer 11

\[4x^3\]

Answer 12

\[4x^3\sin 2t\]

Answer 13

Yes @tukajo I mean you are right.

Answer 14

Now put it all together and write dz. Please do that and show me.

Answer 15

And @tukajo @Jonask is right. In explaining the general rule I used the variable y but in the question it is the variable t.

Answer 16

No! Why is it dx/dz and dy/dz? That's not what I wrote.

Answer 17

Look at my explanation again.

Answer 18

It should be\[dz =4x ^{3}\sin (2t)dx +2x ^{4}\cos (2t)dt\]

Answer 19

Absolutely! Can you please do that so I can check.

Answer 20

But you have to simplify of course. Exact values please!! Remember the special trig ratios.

Answer 21

No! the -64dy is incorrect.

Answer 22

\[\cos (\frac{ 2\pi }{ 3 })=?\]

Answer 23

Right and\[2(-2)^{4}=?\]

Answer 24

No, (-2)(-2)(-2)(-2) = ?

Answer 25

Alright, so now please present your final answers again.