# Find out value of k such that .............?

http://www.walterzorn.com/grapher/grapher_e.htm

Use above grapher to graph following equations (no need to change any default settings on that page)

1/atan(x);

sqrt(k^2-x^2);

-sqrt(k^2-x^2)

where k is any constant value

Find value of "k" such that graph of 1/atan(x) "touches" (i.e. is tangent to) circle created by equation k^2-x^2

I am sure you will get amazed when you get the answer, can you explain me why do we get that value as an answer?

Bhaskar:

It is not sqrt(e)

(Hint: just check value of sqrt(e) find a constant near its value and you will get the answer)

Once you get the correct answer, please do not forget to explain it :)

### 7 Answers

- Cool DudeLv 41 decade agoBest Answer
It's not k = √(1+16/π^2)... I thought it was this for a while, and it does look pretty good when you graph it. The two curves do intersect at (1,4/π), but they're not tangent to each other here. To check, look at the slopes of the curves...

The derivative of 1/arctan(x) is -1/(1+x^2)(arctan(x))^2. Plug in x = 1, get -8/π^2.

The derivative of sqrt(k^2-x^2) is -x/sqrt(k^2-x^2). Plug in k = √(1+16/π^2) and x = 1, get -π/4.

In my book, -8/π^2 ≠ -π/4, but they are kind of close, so it's not a big surprise that the graph is convincing.

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My thoughts:

We must be able to simultaneously solve:

1/arctan(x) = sqrt(k^2-x^2)

and:

1/(1+x^2)(arctan(x))^2 = x/sqrt(k^2-x^2)

We can get rid of k here, and we're reduced to solving, say:

1/arctan(x) = (x+x^3)^(1/3)

This, again, shows that k = √(1+16/π^2) (which implies x = 1) is not a solution.

Once you get an solution for this (other than simply noting that it exists and is a little bigger than 1), you can get k, since we also get the equation:

sqrt(k^2-x^2) = (x+x^3)^(1/3)

Plug in x, solve for k, have a drink, it's your birthday.

- Scythian1950Lv 71 decade ago
Boy, this one took a lot of fiddling around. The exact answer for k is

√(1+16/π²) = 1.6189931866

Edit: Naw, this isn't it, it's close, but it's not the right answer. Well, back to the scribbling board.

Edit 2: I see that John B has caught my mistake. Yeah, I made a scribbling error somewhere.

Edit 3: The tangency occurs where k = 1.6189133 approximately, so I'll have to see if there's an exact value for this. JB was correct with his approximate answer.

Edit 4: Yeah, I don't know if there's any "elegant" solution to this. First, if x = a where the tangency occurs, then a must satisfy the equation:

a(a² + 1)(ArcTan(a))³ = 1

for which a = 1.008104 (this was why I thought it was 1). Then k can be found by the following equation:

k = √(a² + 1/(ArcTan(a))²)

I haven't been able to go from the first equation to k without resorting to approximations.

Edit 5: More accurate figures for a and k:

a = 1.0081028413648671

k = 1.6189133050111404

- 4 years ago
Use the discriminant: b^2 - 4ac. If you have an equation in the form of ax^2 + bx + c = 0, it will have two unique solutions if the discriminant is greater than zero, one solution if the discriminant is equal to zero, and no solutions if it's less than zero. So in the case of x^2 + kx + 79=0, we have a=1, b=k, and c=79. Plug this into the discriminant and you get k^2 - 4(1)(79), or k^2 - 316. So for the original equation to only have one root, this must equal zero. This is true when k = sqrt(316) or -sqrt(316). The reason the discriminant works becomes clear if you look at the quadratic formula. The discriminant is the part found underneath the radical. If you can take the square root, then the "plus or minus" in the quadratic formula gives you two answers. If the discriminant is zero then the square root of it is zero, and the "plus or minus" zero in the quadratic formula only gives you the same answer twice. If the discriminant is less than zero, then you have no solution, because you can't take the square root of a negative number.

- JBLv 71 decade ago
Hmm. I get k = 1.61891330501 approximately. I fail to see what should amaze me. I must be missing something. This does not equal phi =(1+ √5) / 2, although it is a bit close.

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Scythian is correct. Ignore all I wrote above. I can't believe I groped so close to it and didn't see it. I depended way too much on built in floating point solve routines and way too little on thinking about it.

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Oh my! I promise not to post on this thread again.

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- Mugen is StrongLv 71 decade ago
(√5 + 1)/2 = 1.618034 ?

edit

x = 1.008102841364867113

k = 1.618913305011140291