MATHS study thread. (4 Viewers)

[Riviet]

oh, ok, thanx.
Find the derivative of x In x.

d/dx(x ln x)=x(1/x)+lnx(1)
=1+lnx

Next Question (last question in my 2u assessment today ):
Find the equation of the parabola with vertex (3,13) and a y-intercept of 4.

 

[pLuvia]

d/dx(x ln x)=x(1/x)+lnx(1)
=1+lnx

Next Question (last question in my 2u assessment today ):
Find the equation of the parabola with vertex (3,13) and a y-intercept of 4.

y = ax²+bx+c

c = 4
(3,13) satisfy the equation

13 = 9a+3b+4
9 = 9a+3b
3 = 3a+b ..........................1

-b/2a = 3
-b = 6a
b = -6a ............................2

.:sub 2 into 1

3 = 3a+(-6a)
3 = -3a
a = -1
b = 6

.: equation of parabola is y = -x²+6x+4

You already do natural log in 2u? or is it for 3u?

 

[kido_1]

this is how I think..

d/dx(x ln x)=x(1/x)+lnx(1)
=1+lnx

Next Question (last question in my 2u assessment today ):
Find the equation of the parabola with vertex (3,13) and a y-intercept of 4.

Knowing that the vertex is (3,13), it means the axis of symettry must be 3. (The x-side of the point is the axes of symmetry!!)

Using the formula
y=ax^2
and the points (3,13) and (0,4)
3-0=3
and 13-4=9
.;x=3
y=9
9=a(3)^2
9=9a
.: a=1

-b/2a =3 (axes of symettry)
since a=1
-b/2=3
-b=6
.:b=-6

c=4 (given(y-intercept))

.: the equation is
y= x^2-6x+4

hey reminding me how do u display x squared on the screen (I always use x^2)!!!

 

[pLuvia]

x[sup ]2[/sup ]

without the spaces

or to do this

x₂ use x[sub ]2[/sub ] also with out the spaces

edit:

how can the curve be y = x²-6x+4

when the vertex is way above the x axis? lol

 

[pLuvia]

try this

(a) sketch the curve y = 1sinx for 0< x <2pi
(b) The region bounded by the curve y = 1+sinx, the x axis and the ordinates x = 0 and x = 3pi/2 is rotated about the x axis, find the volume of the solid generated

 

[Riviet]

Kadlil's answer is correct since the parabola's vertex is above the y-intercept, therefore it must be concave down.

 

[kido_1]

am I bad...
stupid mistake, a=-1 , b=6 and c=4
the equation is y=-x^2+6x+4

do u know y
in the distanc u go -3 across and -9 down
that means X=-3 and Y=-9
-9= (-3)^2 a
-9= 9a
.; a=-1

-b/2a =3
-b/-2 =3
-b=-6
.; b=6

sorry for the mistake...

same answer as kadlil, but partially different way

 

[kido_1]

x²
i think I get it now, haha
wait im doing that integration problem...

 

[pLuvia]

Did you self tutor to learn integration?

 

[kido_1]

i am self-tutoring myself till the 27th of this month, then I start maths, english, chem, bio and physics coaching....

 

[Riviet]

Wow, which coaching school?

 

[YBK]

The line y= 71x-315 is a tangent to the curve y=14x^2+17x+13.
Find the point of contact.

Can't you just solve simulataneously???

71x-315 = 14x^2+17x+13
...........

 

[pLuvia]

Can't you just solve simulataneously???

71x-315 = 14x^2+17x+13
...........

possibly

oh well, I got the answer

 

[Riviet]

Someone gonna do kadlil's last question? I cbb either.

 

[YBK]

possibly

oh well, I got the answer

lol, those numbers are huge

 

[pLuvia]

lol, those numbers are huge

Yeh, the question seems made up like on the spot

 

[Riviet]

From the curve y = 1+sinx for 0< x <2pi, the region bounded by the curve y = 1+sinx, the x axis and the ordinates x = 0 and x = 3pi/2 is rotated about the x axis, the volume of the solid generated is:
V=(0->3pi/2)∫pi[1+2sinx+sin²x]dx
=(0->3pi/2)∫pi[3/2+2sinx+1/2(cos2x)]dx
=[3/2x-2cosx+1/4(sin2x)](0->3pi/2)
=9pi/4+2
=17pi/4 units³

Next Question (in my 3u assessment ):
Given f(x)=(x²-4x+3)/(x-2)³
i) find the x and y intercepts (2 marks)
ii) show that f'(x)=(-x²+4x-1)/(x-2)⁴ (3 marks)
iii) hence use f'(x) to find any stationary points and determine their nature (4 marks)
iv) state the equations of any horizontal or vertical asymptotes. (2 marks)
v) hence sketch f(x) showing all critical points, intercepts and asymptotes (5 marks)

 

[insert-username]

Given f(x)=(x²-4x+3)/(x-2)³
i) find the x and y intercepts (2 marks)

y-intercept: -3/2

x-intercepts: -1, -3


ii) show that f'(x)=(-x2+4x-1)/(x-2)4 (3 marks)

f(x)=(x²-4x+3)/(x-2)³

f'(x) = (vu' - uv')/v²

= [(x-2)³(2x - 4) - (x²-4x+3).3(x-2)²]/(x-2)⁶

= (x-2)²(2x² - 4x - 4x + 8 - 3x² + 12x -9)/(x-2)⁶

= (-x² + 4x -1)/(x-2)⁴


iii) hence use f'(x) to find any stationary points and determine their nature (4 marks)

S.P. at (2 ± √3). Can't be bothered find f''(x)


iv) state the equations of any horizontal or vertical asymptotes. (2 marks)

Vertical asymptote at x=2. That's all I can be bothered to do...


I_F

 

[kido_1]

Y-intercept= -1.5
X-intercepts= -1 and 3

ii) f(x)= (vu'-uv') / v²
= (x-2)² (2x²-8x-1+12x) / (x-2) ⁶
= (-x²+4x-1) / (x-2)⁴

iii) Stationary point is (2+/- root 3)
Nature-Real

iv) Vertical Asymptote x=2
Horizontal Asymptote y=...

 

[Riviet]

Can't be bothered find f''(x)

Haha same here, didn't have time to find the pof's, we only had 45 mins and this question was 1 of 2... the other was a hard parametrics question.