Question

10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...

12) Find the differential dy of 3x^(2/3)
MY ANSWER: 9/(2cube root of x)
(Sorry, I don't know how to write a cube root thing)


20) Find the differential dy of the given function.
y=sec^2x/(x^2+1)
MY ANSWER: [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2

30) The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube.
MY ANSWER: 20.25

Please tell me if I got any of them wrong, and showing me what I'm supposed to be doing would be helpful! I had to teach this section to myself, so I'm struggling more than usual.

Answer 1

Never mind number 20! Someone already helped me with it!

Answer 2

I already did 20, scroll down

Answer 3

10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
-------------------------
dy = 0 - 4 x^3 dx
x = 2 and dx = .01
dy = -4 (8)(.01) = -.32 correct
y at x = 4 is y = -254
y at x = 4.01 = -256.56
delta y = -256.56 + 254 = -2.56

12) Find the differential dy of 3x^(2/3)
MY ANSWER: 9/(2cube root of x)
(Sorry, I don't know how to write a cube root thing)
------------------------
y = 3 x^(2/3)
dy/dx = (2) x^-(1/3)
dy = 2/(x^1/3)

20) Find the differential dy of the given function.
y=sec^2x/(x^2+1)
MY ANSWER: [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2

30) The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube.
MY ANSWER: 20.25
-----------
v = x^3
dv = 3 x^2 dx
dv = 3 (225)(.03) = 20.25 right

Answer 4

10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01
MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
-------------------------
dy = 0 - 4 x^3 dx
x = 2 and dx = .01
dy = -4 (8)(.01) = -.32 correct
ok but then x = 2 not 4 !!!!!
y at x = 2 is y = -14
y at x = 2.01 = -14.3224
delta y = -.3224
that is better :)

Answer 5

Thank you soooo much!