Primitive functions.. (2 Viewers)

[Makro]

1. Find the primitives of:

a) y'= (root)x

b) y' = x^-3

c) y' = 1/x⁸

For a & b, I just got the sign wrong. With C I'm pretty clueless. Thanks in advance.

 

[lolokay]

y' = x^a
y = x^a+1/(a+1) +C

just do this for them all

 

[Makro]

I have. I ended up getting:

1. y = x^3/2/3/2 + C
y = 2x^3/2/3 + C
y = 2(root)x³/3 + C

2. y = x^-2/-2 + C

 

[Aerath]

 

[Makro]

Thank you Aerath

Little bit of explanation on the answer for the first two I've posted?

 

[Finx]

For y' = root(x)

 

[Makro]

If and x = 7, when t = 0, find x when t = 4.

Not sure how to do this, I failed. I think I'm missing a step or doing a step wrong in the process. Also the numerous x and y values have me confused.

Also keep failing:

Given and and y = 3 when x = 1, find the equation of y in terms of x.

I don't understand this at all, I get to y' and then I don't know how to find C.

Please, please help me. Been doing this exercise for over an hour, it's frustrating me.

 

[Makro]

For y' = root(x)

That's what I got, but the answers are insisting the answer is a negative fraction. Same as for the #2 in the original post.

 

[lolokay]

if the answers say that 1. gives a negative result, then the answers are wrong

for 2. you got a negative (unless you're saying the answers give a positive?)

 

[Finx]

I don't understand this at all, I get to y' and then I don't know how to find C.

Please, please help me. Been doing this exercise for over an hour, it's frustrating me.

It tells you that y' = 0, (thats what the dy/dx thing means) so C must be 0.

 

[lolokay]

why do the questions you have use partial derivatives? since those aren't dealt with in HSC maths I'll just use the usual notation:



when t = 0, x = 7
7 = 1/3(0-3)³ + C
7 = -9 + C
C = 16

when t = 4,
x = 1/3(4-3)³ + 16
= 1/3 + 48/3
= 49/3
or 16 1/3


are you given anything else in the other question?

 

[Makro]

if the answers say that 1. gives a negative result, then the answers are wrong

for 2. you got a negative (unless you're saying the answers give a positive?)

Right, I was reading my own answer wrong, thanks.

It tells you that y' = 0, (thats what the dy/dx thing means) so C must be 0.

I don't quite understand, would you be able to do the working out? Thank you.

 

[Makro]

why do the questions you have use partial derivatives? since those aren't dealt with in HSC maths I'll just use the usual notation:



when t = 0, x = 7
7 = 1/3(0-3)³ + C
7 = -9 + C
C = 16

when t = 7,
x = 1/3(7-3)³ + 16
= 64/3 + 48/3
= 112/3


are you given anything else in the other question?

My answers give me, x = 16/1/3.

No, they're both the questions word for word.

We've only just started, this is still in the chapter of geometrical applications of calculus (last exercise) and then we start the chapter of integration. So I don't know any of the notation except for the first rule that was posted.

 

[Finx]

Wait, I think I was wrong. Let me try it out.

 

[lolokay]

sorry, i subbed the wrong value of t into the expression


I think y = 4x² - 8x + 7 would be the solution to the second one. i don't complete get the question though (the way it's worded maybe?)

 

[Makro]

sorry, i subbed the wrong value of t into the expression


I think y = 4x² - 8x + 7 would be the solution to the second one. i don't complete get the question though (the way it's worded maybe?)

Thats right. Can you show me?

 

[lolokay]

maybe if I rephrase the question it would help?

y" = 8

when x=1, y' = 0

when x=1, y=3

fine the equation of y. (it's just integrating and evaluating constants)

 

[Makro]

maybe if I rephrase the question it would help?

y" = 8

when x=1, y' = 0

when x=1, y=3

fine the equation of y. (it's just integrating and evaluating constants)

y'' = 8
y' = 8x + C

at x = 1, y' = 0
0 = 8 + C
C = -8

y' = 8x - 8
y = 4x² - 8x + D

at x = 1, y = 3

3 = 4(1)² - 8 (1) + D
7 = D

y = 4x² - 8x + 7
____________________________
Thanks a ton. I thought I couldn't do it, and typed up what I could, then I ended up getting the question

____________________________

f''(x) = 5x⁴, f'(0)=3 & f(-1) = 1, find f(2)

I ended up getting 14/1/2, while the ans was 20.5

 

[Finx]

f''(x) = 5x⁴, f'(0)=3 & f(-1) = 1, find f(2)

f''(x) = 5x^4

f'(x) = x^5 + C
f'(0) = (0)^5 + C = 3
.'. C = 3

f(x) = (x^6)/6 + 3x + D
f(-1) = [(-1)^6/6] + 3(-1) + D = 1
1/6 - 3 + D = 1
D = 3/5/6

.'. equation = (x^6)/6 + 3x + 3/5/6

f(2) = [(2^6)/6] + 3(2) + 3/5/6
= 20/1/2

 

[Makro]

. Must be something I'm missing. I'll be asking a teacher tomorrow, there's about 5 or 6 out of 15 that I can't do, so yeah, hmm.

Thanks for all the time and effort for anyone who posted. Much appreciated.