TrueTears question thread

[TrueTears]

Can anyone help with Q 17 and Q 19? Many thanks!

[TrueTears]

Given f:R ->R where f(x) = sin(2x-3). Show that a restriction of f, namly F, defined on [1,2] has an inverse . Find the rule of
and stating its domain and range

[kamil9876]

Question 19: In the second phase, u hav to treat the 20cm string in two seperate parts, one is the arc AP and the other is the remaining straight line segment PB'.

so we know

Familiarise urself with the arclength formula
You can use it to work out AP:


 
PB' can be worked out by applying elementary trig to the right angled triangle OPB' :



Now sub these two expressions of PB' and AP into equation 1.

[kamil9876]

for question 17:

U must find an expression for the area of the blue area, and equate it the the area of the white section of the circle:

Let this area be and the radius of the circle be

Blue area:

Draw the line . Notice that the angle AOX is . Also, notice that the area is now split in half i.e: the area of triangle AOX is






So

The white area:

the area of the whole circle is . To get the area of the white part you need to multiply the whole area by the fraction . Hence:




  cancels out as well as a .

[kamil9876]

And now for the question that isn't associated with a number:

A bit dissapointed with the lack of in inside the trig function domain.

Restricting the domain to [1,2] means that the thing inside the bracket will go from -1 to 1. SO the function exhibits the values: sin(-1), sin(-0.9), sin(-0..... sin(1) etc. (there are infinite possibilities since there are inifite real numbers between -1 and 1)

In other words: you could say that the range of f(x) for this domain is exactly the same as the range of sin(x) for domain [-1,1]. Such a range is monotonic (contains no turning points) since the turning points occur at and and there are no other turning points between these two values. But the domain [-1,1] is in between these two values hence f(x) has no turning points, in fact it is always increasing. This means that for each x-value there is only one value of f(x) and conversely for each value of f(x) there is only one x-value. Hence the inverse function does exist.

Oh and for the other part of the question: the domain is trivially [1,2] while the range is [sin(-1),sin(1)] (as shown before) of f(x). SO for the inverse function, the domain and range will swap since the x and y values are being swapped (by definition of inverse).

[TrueTears]

Thanks kamil!!

yeah also dw about Q 19, i've figured it out.

also these 2


1. Prove

2.

[kamil9876]

Question1:

Multiply the numerater and denominator of RHS by








Multiply numerator and denominator by





this may help.

[kamil9876]

and question 2 looks awfully familliar to the tan(x+y) expansion.







Multiply numerator and denominator by

[TrueTears]

nice thanks kamil i got it

[TrueTears]

Just another question:

a) Find a suitable parametric equation for the ellipse

Make sure that the entire ellipse is given by your parametric equation

So i did this part. But can anyone just check if it is right because there are no answers to these questions



so  and where

b) i. The point , , is a point on the ellipse . S is the point .

The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP. Calculate the coordinates of Q ( in terms of )

ii. As varies, the point P moves around the ellipse. As P moves, so does Q. Find the cartesian equation of the locus of Q as varies. State the type of curve traced out by Q and give the coordinates of the centre of this curve.

If anyone could help with b) i and ii that would be great!

[kurrymuncher]

yep, a) is right

[TrueTears]

thanks ! i kinda have an idea for b) but just can't get it haha

[kamil9876]

If the line is extended by 3, then the rise is multiplied by 3, and hence so is the the run in order to keep the gradient the same. Think of similair triangles or parralel vectors. i.e: vector SQ=3SP and so the i component of SQ is 3 times as big as the i of SP and same for j.

Hence the rise for SP is while the rise for SQ is and so

The run of SP is but the run for SQ is and so

[kamil9876]

for part 2, relate the two values for x and y that we got earlier (note that in my previous post, x and y are the x and y values of point Q). Now the way u relate them is by using the pythagorean identity the same way that u do for parametric equations of elipses, where in this case the parameter is and the parametric equations are the two equation for x and y.

[TrueTears]

If the line is extended by 3, then the rise is multiplied by 3, and hence so is the the run in order to keep the gradient the same. Think of similair triangles or parralel vectors. i.e: vector SQ=3SP and so the i component of SQ is 3 times as big as the i of SP and same for j.

Hence the rise for SP is while the rise for SQ is and so

The run of SP is but the run for SQ is and so

thanks kamil i get most of that but just near the end how do you get the run for SQ? where did the come from?