Inverse Functions (1 Viewer)

[FDownes]

So... I've been having a fair bit of trouble with this particular exercise I'm working on at the moment, so be prepared for a rush of questions.

You ready? Okay, here we go;

1. a) Find the area enclosed between the curve y = 1/√(1 - x²), the x-axis and the lines x = 0 and x = 1/2.
(This part is easy, the answer's pi/6 units²)

1. b) This area is rotated about the y-axis. Find the exact volume of the solid formed.

2. Find the area enclosed between the curve y = cos^-1x, the y-axis and the lines y = 0 and y = pi/4.

3. Find ∫x²/√(1 - x⁶) dx by using the substitution u = x³.

4. a) Show that (2x² + 5)/(1 + x²)(4 + x²) = 1/(1 + x²)(4 + x²).
(Ignore this question, I've figured this one out).

4. b) Hence evaluate ₁∫² (2x² + 5)/(1 + x²)(4 + x²) dx correct to 2 decimal places.

5. a) Differentiate x cos^-1x - √(1 - x²).

5. b) Find the area bounded by the curve y = cos^-1x, the x-axis and the lines x = 0 and x = 1/2.

6. Use d/dx [x sin^-1x + √(1 - x²) to help find the area enclosed between the curve y = sin^-1x, the x-axis and the line x = 0 and x = 1.

Phew... That's all of them.

 

[tommykins]

I will not give you solutions (as they are far too long) but I will push you in the right path.

1. a) Find the area enclosed between the curve y = 1/√(1 - x²), the x-axis and the lines x = 0 and x = 1/2.
(This part is easy, the answer's pi/6 units²)

1. b) This area is rotated about the y-axis. Find the exact volume of the solid formed.

Make x the subject (which leads to x² anyways) and integrate that with V = pi int x² dy and also, sub x = 0 and x = 1/2 to obtain your y domains.

2. Find the area enclosed between the curve y = cos^-1x, the y-axis and the lines y = 0 and y = pi/4.

y = cos^-1x but since we're trying to find the area on the y axis, make x the subject. So simply integrate x = cos y dy from pi/4 -> 0

3. Find ∫x²/√(1 - x⁶) dx by using the substitution u = x³.

Using x³ = u, dx = du/3x² . Since the x² is at the top, these "cancel out" and so it just becomes 1/3 ∫1/√(1 - u²).

4. b) Hence evaluate ₁∫² (2x² + 5)/(1 + x²)(4 + x²) dx correct to 2 decimal places.

The first part was meant to make 2 fractions, integrate those 2 fractions instead of 1. (Hence the "hence" word and the point of showing that this one fraction can be split up into the other two)

5. a) Differentiate x cos^-1x - √(1 - x²).

let u = x, u' = 1, v = cos^-1 x and v' = -1/√(1-x²)

y ' = cos^-1 x -x/√(1-x²) + x/√(1-x²)
= cos^-1 x

5. b) Find the area bounded by the curve y = cos^-1x, the x-axis and the lines x = 0 and x = 1/2.

let f'(x) = cos^-1x
f(x) = x cos^-1x - √(1 - x²). (from above).
Area = f(1/2) - f(0)

x cos^-1x - √(1 - x²).

6. Use d/dx [x sin^-1x + √(1 - x²) to help find the area enclosed between the curve y = sin^-1x, the x-axis and the line x = 0 and x = 1.

Same protocol as above.

Jeez all these SUP scripts are annoying

 

[lyounamu]

Yeah. I did that question but I cannot post that up. It's just TOO LONG! So listen to tommykins. He's got everything under control.

 

[hereSIR]

y = cos<SUP>-1</SUP>(sin x)

dy/dx = -1/(root of (1-sin^2 x)) . d/dx (sinx)
= -1/(root of cos^2 x) . -cos x
= cos x / absolute value of (cos x)
= 1 when cos x > 0 (and hence pi >_ x > pi/2 or -pi >_ x > -pi/2)
OR -1 when cos x < 0 (and hence -pi/2 <x < pi/2)

This might sound stupid, but can someone explain why in the above solution:

the (root of cos^2 x) = Absolute value of (cos x) and not just cosx

Why the absolute value?

 

[lacklustre]

I might hijack this thread for a bit.

Can anyone help me find the exact value of:

cos^-1(cos 7pi/6)

I'm a little confused.

I solved the inside of the brackets (cos 210) which equals -cos30. But this is not in the domain of 0<(or = to)x<(or = to)pi. How can i fix this?

 

[lyounamu]

This might sound stupid, but can someone explain why in the above solution:

the (root of cos^2 x) = Absolute value of (cos x) and not just cosx

Why the absolute value?

square root of (cos^2 (x)) = absolute value of (cos x)

 

[lyounamu]

I might hijack this thread for a bit.

Can anyone help me find the exact value of:

cos^-1(cos 7pi/6)

I'm a little confused.

I solved the inside of the brackets (cos 210) which equals -cos30. But this is not in the domain of 0<(or = to)x<(or = to)pi. How can i fix this?

cos^-1(cos7pi/6) = cos^-1(cos-pi/6)
= cos ^-1(-(square root of 3)/2)
= 5pi/6

 

[hereSIR]

square root of (cos^2 (x)) = absolute value of (cos x)

Sorry, but can you explain why it is?

 

[lyounamu]

Sorry, but can you explain why it is?

Sorry. If something is placed in the root, it is always positive (there is not square root of negative number), right? That's why it is in the absolute value.

And there are always two possible answers, not just one.

 

[hereSIR]

Sorry. If something is placed in the root, it is always positive (there is not square root of negative number), right? That's why it is in the absolute value.

But doesn't the squre inside the squareroot eliminate any possible negative values derived from cos, hence shouldn't squareroot of (square of cosx) be just cosx!

I know the solution above by you is correct, but just finding it difficult to understand the concept behind it.

example:
squareroot of (square x) = x not Absolute x, am I right?

hence it should follow

squareroot of (square cox x) = cos x not Absolute cos x.

If my example above is correct, the answer should only be one in the original eqn.

Thanks in advance

 

[lyounamu]

But doesn't the squre inside the squareroot eliminate any possible negative values derived from cos, hence shouldn't squareroot of (square of cosx) be just cosx!

I know the solution above by you is correct, but just finding it difficult to understand the concept behind it.

example:
squareroot of (square x) = x not Absolute x, am I right?

hence it should follow

squareroot of (square cox x) = cos x not Absolute cos x.

If my example above is correct, the answer should only be one in the original eqn.

Thanks in advance

Your answer only gives one answer out of two possible answers. If you write that in exam, you will only get the half mark.

I will give you an example. Follow it and ask if you don't understand.

Your explanation = square root of (square x) = x

My explanation = square root of (square x) = absolute value of x
However, absolute value of x = x if x > 0 or -x if x<0
You will see that x = -x if x<0 is also a possible answer.

Example 1
cos (x) = -cos (-x)
Example 2
x = -(-x)

 

[hereSIR]

Thanks, I will work on it!

 

[foram]

But doesn't the squre inside the squareroot eliminate any possible negative values derived from cos, hence shouldn't squareroot of (square of cosx) be just cosx!

I know the solution above by you is correct, but just finding it difficult to understand the concept behind it.

example:
squareroot of (square x) = x not Absolute x, am I right?

hence it should follow

squareroot of (square cox x) = cos x not Absolute cos x.

If my example above is correct, the answer should only be one in the original eqn.

Thanks in advance

When you take the square root in this case, you're only taking the positive root. Thats why theres no + or - in front of it. Thats why it's the same as an absolute value.

 

[lacklustre]

Another one for the smart kids:

cos(tan^-1(-2))

i drew up the triangle, but how do you know which side of the triangle to make negative?

 

[lyounamu]

Another one for the smart kids:

cos(tan^-1(-2))

i drew up the triangle, but how do you know which side of the triangle to make negative?

There is a domain given for x, which is -pi/2 < x < pi/2

So, draw a triangle in the 4th quadrant that satisfies the condition.

Let x = tan^-1(-2)
Therefore tan x = -2

And cos(tan^-1(-2)) = cos x
= 1/(square root of 5)
= 0.447213595...
= 0.44 (correct to two decimal places).
From the triangle,

 

[lacklustre]

There is a domain given for x, which is -pi/2 < x < pi/2

So, draw a triangle in the 4th quadrant that satisfies the condition.

hmm i am still little confused.

I drew a triangle with sides 1 and 2 and hypotenuse sqrt5. One of the sides has to be negative. How do you work out which side?

I'm prob not getting something (i am a bit slow unfortunately

 

[lyounamu]

hmm i am still little confused.

I drew a triangle with sides 1 and 2 and hypotenuse sqrt5. One of the sides has to be negative. How do you work out which side?

I'm prob not getting something (i am a bit slow unfortunately

Ok, when you draw the diagram on the plane. The triangle is situated in the 4th quadrant. Therefore, its opposite side must be negative because it runs downwards. Hypotenuse is always positive as it is a length. The adjacent side is positive because it is moving from the origin to the positive value.

Look at my edited post. I have got the answer up there.

So, since tan x = -2, cos x = 1/(square root of 5) and sin x = -2/(square root of 5)

 

[lacklustre]

Ok, when you draw the diagram on the plane. The triangle is situated in the 4th quadrant. Therefore, its opposite side must be negative because it runs downwards. Hypotenuse is always positive as it is a length. The adjacent side is positive because it is moving from the origin to the positive value.

Look at my edited post. I have got the answer up there.

So, since tan x = -2, cos x = 1/(square root of 5) and sin x = -2/(square root of 5)

Oh thanks. I did not know that this applied.

 

[lacklustre]

Edit: nevermind i get it now.

 

[lacklustre]

Sigh, another question:

sin^-1(2sqrt2)/3 - tan^-12

I get sin^-1((2sqrt2)-2)/3sqrt5

The answer is tan^-1(18-10sqrt2)/31

So can there be two answers to this question?
i.e. you can use sin(alpha - beta) or you could use tan(alpha - beta) right?