need help with some homework revision (1 Viewer)

[red802]

guys, i have trouble with a couple of questions and im not sure if i got them right

1) find the equation of the tangent ot the curve y=cos3x, at the point where x = pie/6, leave ur answer in exact form

2) use the sketch of y = 2sinx to find the number of solutions to the equation 2sinx = 1-(1/2)x for 0<x<2x

3)the sector in the diagram has an arc length of 12cm and an area of 9n/8 cm^2 n=pie

The diagram View attachment 13031

sry, also the length(l)=12cm of the circle

4) a piece of wire of length 10 metres is cut into two pieces and used to form 2 squares.

4i)If one piece of wire has length x metres, find the side length of each square
ii) Show that the combined area of the squares is given by
A=1/8 (x^2 - 10x +50)
iii) Find dA/dx and hence find the value of x that makes A a minimum
iv) Find the least possible value of the combined areas


Sry, theres a lot of questions, but if u have the time, please can u do it and show the working out, just to see if my answers are right and i did the right working out

 

[darkliight]

Ok, for question 1 ...

We have a curve y = cos(3x), its gradient at x is given by y'(x) = -3sin(3x). In particular, the gradient of y at x = pi/6 is , y'(pi/6) = -3sin(3pi/6) = -3sin(pi/2) = -3.

So we have the curves gradient at that point, now all we need is a point the tangent line goes through and we can use the formula y - y_1 = m(x - x_1).

We know that the tangent line shares a point with the curve at x = pi/6, so we can plug that into our equation for y and get y(pi/6) = cos(3pi/6) = cos(pi/2) = 0.

So our tangent line goes through the point (pi/6, 0) and has a gradient of -3. This gives us y - 0 = -3(x - pi/6), or y = -3x + pi/2.

 

[darkliight]

For question 2, the graph of 2sin(x) is the same as sin(x), except y now goes from -2 to 2 instead of -1 to 1.

Now draw the graph of the line 1-x/2 over the top of your 2sin(x) graph and see how many times they intesect on the domain 0<x<2pi(?)

 

[SoulSearcher]

4.i)If one piece of wire has length x metres, find the side length of each square
ii) Show that the combined area of the squares is given by
A=1/8 (x^2 - 10x +50)
iii) Find dA/dx and hence find the value of x that makes A a minimum
iv) Find the least possible value of the combined areas

i) Length of total wire is 10 metres. If the wire is cut into 2 pieces, and a length of the wire is x metres, then the other length is (10-x) metres. If both of the wires are made into a square, the side length of each square is x/4 and (10-x/4) metres.
ii) The area of each particular square is x²/16 and (10-x)²/16. Therefore the total combined area of the squares are
A = x²/16 + (10-x)²/16
= 1/16(x² + (10-x)²)
= 1/16(x² + 100 - 20x + x²)
= 1/16(2x² - 20x + 100)
= 1/8(x² - 10x + 50)
iii) dA/dx = 1/8(2x-10)
To find turning points, let dA/dx = 0
0 = 1/8(2x-10)
0 = 2x - 10
2x = 10
x = 5
d²A/dx² = 1/8 * 2 = 1/4, and since d²A/dx² > 0, dA/dx is a minimum for all values of x, therefore minimum value of A is when x = 5
iv) A = 1/8(x² -10x + 50)
when x = 5,
A = 1/8(25 - 10*5 + 50)
= 1/8(25 - 50 + 50)
= 25/8 metres²

 

[red802]

kool, thanks everyone, but is it still possible to help me with question 3

 

[SoulSearcher]

I'm not sure what you are asking in number three, but the radius should be 3pi/16 centimetres and the angle should be 64/pi radians.

 

[red802]

oh, yeah, im u have to find threatre and r, can u show the working out please

 

[Riviet]

3)the sector in the diagram has an arc length of 12cm and an area of 9n/8 cm^2 n=pie

The diagram View attachment 13031

sry, also the length(l)=12cm of the circle

l=r@
12=r@
@=12/r (1)
Area=1/2.r²@
9pi/8 = 1/2.r²@
Substitute (1),
9pi/8 = 1/2.r².12/r
6r=9pi/8
r=3pi/16

.'. @=l/r
=12/(3pi/16)
=64/pi

Yep, typo. [fixed now]

 

[SoulSearcher]

.'. @=l/r
=12/(3pi/16)
=64pi

Typo?

 

[red802]

thanks guys,