Minima of 9tan2x+4cot2x | 65 Characters

[Saitama]

Homework Statement

This is question is from my test paper.
Find the minimum value of 9tan²x+4cot²x.
I wasn't able to solve it during my examination.
But later i remembered that i can find its minima which unfortunately my teacher still hasn't taught. :(
I have a very little knowledge of it so my attempts maybe wrong.

Homework Equations

The Attempt at a Solution

I substituted y=tan²x

$$\frac{d}{dy}(f(y))=9-\frac{4}{y^2}$$
$$\frac{d^2}{d^2y}(f(y))=\frac{8}{y^3}$$

$$9-\frac{4}{y^2}=0$$
I get y=2/3

Substituting y=2/3 in the second derivative i get 27 but the answer is 12.

 

[NascentOxygen]

This sub-forum is for problems that can be solved without calculus. Does your teacher expect you to arrive at the answer without using calculus?

You seem to be saying the derivative of 9.tan²x = 9
and the derivative of 4/(tan²x) = 4/(x²)

Neither is correct. How did you calculate these?

 

[Saitama]

This sub-forum is for problems that can be solved without calculus. Does your teacher expect you to arrive at the answer without using calculus?

My teacher solved it without finding the minima but i still want to know where i am wrong.

Please moderators,, move it to the right sub-forum.

 

[Saitama]

You seem to be saying the derivative of 9.tan²x = 9
and the derivative of 4/(tan²x) = 4/(x²)

Neither is correct. How did you calculate these?

Didn't you see that i took tan²x=y?

 

[Saitama]

Bump :)

 

[Mark44]

I don't see any advantage in using the substitution you used. As far as I can tell, you need calculus techniques to find the minimum of the given function.

You have y = 9tan²(x) + 4cot²(x)

Take the derivative directly and set it to zero.

You'll need to use a few basic identities, but otherwise the problem is pretty straightforward.

 

[Saitama]

I don't see any advantage in using the substitution you used. As far as I can tell, you need calculus techniques to find the minimum of the given function.

You have y = 9tan²(x) + 4cot²(x)

Take the derivative directly and set it to zero.

You'll need to use a few basic identities, but otherwise the problem is pretty straightforward.

I already tried that before but i was stuck after finding out the derivative:-
$$\frac{dy}{dx}=18tan(x)sec^2(x)-8cot(x)cosec^2(x)$$

If i substitute the derivative to zero, then what i get is this:-
$$18sec^4(x)-8cosec^4(x)$$

I don't understand what should i do?

Could you please tell me where i am wrong in my method?

 

[Mark44]

Homework Statement

This is question is from my test paper.
Find the minimum value of 9tan²x+4cot²x.
I wasn't able to solve it during my examination.
But later i remembered that i can find its minima which unfortunately my teacher still hasn't taught. :(
I have a very little knowledge of it so my attempts maybe wrong.

Homework Equations

The Attempt at a Solution

I substituted y=tan²x

$$\frac{d}{dy}(f(y))=9-\frac{4}{y^2}$$
$$\frac{d^2}{d^2y}(f(y))=\frac{8}{y^3}$$

$$9-\frac{4}{y^2}=0$$
I get y=2/3

Substituting y=2/3 in the second derivative i get 27 but the answer is 12.

I already tried that before but i was stuck after finding out the derivative:-
$$\frac{dy}{dx}=18tan(x)sec^2(x)-8cot(x)cosec^2(x)$$

The equation above is correct, but the one below is not.

If i substitute the derivative to zero, then what i get is this:-
$$18sec^4(x)-8cosec^4(x)$$

You are confused. tan(x) $\neq$ sec²(x) and cot(x) $\neq$ csc²(x). Those are derivative formulas (one of them has the wrong sign), but that's not what you're doing in this step.

Rewrite all of the functions in terms of sines and cosines.

Here's a start:
18tan(x)/cos²(x) - 8/(tan(x)sin²(x)) = 0

Now replace the tangent factors.

I don't understand what should i do?

Could you please tell me where i am wrong in my method?

 

[Saitama]

I did as you said.
What i now get is:-
$$\frac{18sin^4(x)-8cos^4(x)}{cos^3(x)sin^3(x)}=0$$

Is it correct?

 

[Mark44]

Yes. You can simplify this to 18sin⁴(x) = 4cos⁴(x), and you can turn this into an equation involving the tan function.

Notice that cos(x) cannot equal zero, nor can sin(x) = 0, but these are places where either tan(x) is undefined or cot(x) is undefined, so you're not going to get a minimum for your function at any of these points.

One other thing: there are lots of solutions to the equation you're trying to solve.

 

[Saitama]

Yes. You can simplify this to 18sin⁴(x) = 4cos⁴(x), and you can turn this into an equation involving the tan function.

Notice that cos(x) cannot equal zero, nor can sin(x) = 0, but these are places where either tan(x) is undefined or cot(x) is undefined, so you're not going to get a minimum for your function at any of these points.

One other thing: there are lots of solutions to the equation you're trying to solve.

Is it 18sin⁴(x)=8cos⁴(x)?
I get:-
$$tan^4(x)=\frac{8}{18}$$.

But then i would have to take the fourth root. Fourth root of 4 is $\sqrt{2}$ but then what is the fourth root of 3?

 

[Mark44]

Simplify to tan⁴(x) = 4/9, so tan²(x) = 2/3. (You can ignore the solution tan²(x) = -2/3.)

So tan(x) = ? and hence x = ?

 

[Saitama]

Simplify to tan⁴(x) = 4/9, so tan²(x) = 2/3. (You can ignore the solution tan²(x) = -2/3.)

So tan(x) = ? and hence x = ?

tan(x)=$\frac{\sqrt{2}}{\sqrt{3}}$ and x=tan^-1$\frac{\sqrt{2}}{\sqrt{3}}$

Right..?

 

[Mark44]

Partly. tan(x) = $\pm \sqrt{2/3}$

There is not just one solution, though - there are an infinite number of solutions.

 

[Ray Vickson]

Homework Statement

This is question is from my test paper.
Find the minimum value of 9tan²x+4cot²x.
I wasn't able to solve it during my examination.
But later i remembered that i can find its minima which unfortunately my teacher still hasn't taught. :(
I have a very little knowledge of it so my attempts maybe wrong.

Homework Equations

The Attempt at a Solution

I substituted y=tan²x

$$\frac{d}{dy}(f(y))=9-\frac{4}{y^2}$$
$$\frac{d^2}{d^2y}(f(y))=\frac{8}{y^3}$$

$$9-\frac{4}{y^2}=0$$
I get y=2/3

Substituting y=2/3 in the second derivative i get 27 but the answer is 12.

If you want to minimize f = 9y + 4/y without using calculus, here is how to do it: write f = w1*(9y/w1) + w2*((4/y)/w2), where w1, w2 > 0 and w1+w2=1. By the arithmetic-geometric inequality, w1*r1 + w2*r2 >= r1^w1 * r2^w2, with equality if and only if r1 = r2. So, with r1 = 9y/w1 and r2 = (4/y)/w2 we have f >= (9/w1)^w1*(4/w2)^w2*y^(w1-w2). Note that when w1 = w2 (so w1 = w2 = 1/2), we have f >= sqrt(18)*sqrt(8) = 12; that is, we *always* have f >= 12 for any y > 0. We get *equality* for r1 = r2, or 9y/w1 = (4/y)/w2, or 9y = 4/y (because w1 = w2 = 1/2). So, when 9y = 4/y (y = 2/3) we have f = 12, so that must be the minimum: f >= 12 for all y > 0 and we have achieved a value f = 12!

In general, to minimize f = c1*y + c2/y with c1, c2 > 0 we must take both terms of f equal; that is, c1*y = c2/y, so y = sqrt(c2/c1), and f_min = 2*c1*sqrt(c2/c1) = 2*sqrt(c1*c2).

Note: this type of thing is the basis of *Geometric Programming*, devised by Duffin, Peterson and Zener (yes, the Zener of diode fame).

RGV

 

[Saitama]

Partly. tan(x) = $\pm \sqrt{2/3}$

There is not just one solution, though - there are an infinite number of solutions.

tan(x)=$\pm \sqrt{2/3}$. If i substitute tan(x)=$\pm \sqrt{2/3}$ in the given equation i get the answer as 12. Is this the minimum value?

I don't get what do you mean by there are infinite solutions?

 

[Ray Vickson]

tan(x)=$\pm \sqrt{2/3}$. If i substitute tan(x)=$\pm \sqrt{2/3}$ in the given equation i get the answer as 12. Is this the minimum value?

I don't get what do you mean by there are infinite solutions?

How may values of x are there that satisfy the equation tan(x) = sqrt(2/3)? Note: there is only ONE minimum VALUE of f (namely, f_min = 12), but there may be more than one value of x that gives you that f-value.

RGV

 

[Saitama]

How may values of x are there that satisfy the equation tan(x) = sqrt(2/3)? Note: there is only ONE minimum VALUE of f (namely, f_min = 12), but there may be more than one value of x that gives you that f-value.

RGV

More values of x include the x+2kpi, where k is an integer.
Right..?

 

[Mark44]

More values of x include the x+2kpi, where k is an integer.
Right..?

Not quite. You're missing three-quarters of the values if you take multiples of 2$\pi$.

 

[Saitama]

Not quite. You're missing three-quarters of the values if you take multiples of 2$\pi$.

So what should be it?

 

[Mark44]

I used wolframalpha to graph the function. From that graph, the period appears to be π/2.

 

[NascentOxygen]

I don't get what do you mean by there are infinite solutions?

Well, one minimum value of y, but infinite values of x where y has that minimum.

Click on pan and slide the view towards y=12 here: http://fooplot.com/index.php?q0=9*tan%28x%29^2+4*cot%28x%29^2

The graph is periodic, and repeats at intervals of Pi. Don't let the double dip mislead you here. We see each cycle has two minima.

 

[stallionx]

Homework Statement

This is question is from my test paper.
Find the minimum value of 9tan²x+4cot²x.
I wasn't able to solve it during my examination.
But later i remembered that i can find its minima which unfortunately my teacher still hasn't taught. :(
I have a very little knowledge of it so my attempts maybe wrong.

Homework Equations

The Attempt at a Solution

I substituted y=tan²x

$$\frac{d}{dy}(f(y))=9-\frac{4}{y^2}$$
$$\frac{d^2}{d^2y}(f(y))=\frac{8}{y^3}$$

$$9-\frac{4}{y^2}=0$$
I get y=2/3

Substituting y=2/3 in the second derivative i get 27 but the answer is 12.

Sir,

Have you tried making substitution where u=tanx

and 4 u**2 + 9 u** (-2) = y

take differentials of 2 sides to get

8*u du -18 * u**(-3) du =dy

divide by dx both sides and set to 0 for a minimum

where the ultimate solution would be something like

x(min)= ATAN ( (2 / 9)**( 0.25) )

 

[Saitama]

Well, one minimum value of y, but infinite values of x where y has that minimum.

Click on pan and slide the view towards y=12 here: http://fooplot.com/index.php?q0=9*tan%28x%29^2+4*cot%28x%29^2

The graph is periodic, and repeats at intervals of Pi. Don't let the double dip mislead you here. We see each cycle has two minima.

Thanks for the help.

Sir,

Have you tried making substitution where u=tanx

and 4 u**2 + 9 u** (-2) = y

take differentials of 2 sides to get

8*u du -18 * u**(-3) du =dy

divide by dx both sides and set to 0 for a minimum

where the ultimate solution would be something like

x(min)= ATAN ( (2 / 9)**( 0.25) )

I substituted y=tan²x.

 

[Mark44]

Sir,

Have you tried making substitution where u=tanx

and 4 u**2 + 9 u** (-2) = y

take differentials of 2 sides to get

8*u du -18 * u**(-3) du =dy

divide by dx both sides and set to 0 for a minimum

where the ultimate solution would be something like

x(min)= ATAN ( (2 / 9)**( 0.25) )

For the second time, please read the entire thread before adding comments. The OP, Pranav-Arora, has already found the minimum points.

 

[rude man]

Nothing wrong with letting y = tan^2(x): f = 9y + 4/y
f' = 9 -4/y^2 = 0
y = 2/3
Check for minimum: f" = 8/y^3 > 0 so OK
So f = 9*2/3 + 4/(2/3) = 6 + 6 = 12

y = -2/3 is not a minimum since then y" < 0.

f = 12 is the only solution. There are an infinite number of values of x that satisfy the minimum solution, however, since x = arc tan√(2/3) = 0.68 +/- n*pi, n = +/- 1, 2, ...

 

[Mark44]

Nothing wrong with letting y = tan^2(x): f = 9y + 4/y
f' = 9 -4/y^2 = 0
y = 2/3
Check for minimum: f" = 8/y^3 > 0 so OK
So f = 9*2/3 + 4/(2/3) = 6 + 6 = 12

y = -2/3 is not a minimum since then y" < 0.

Since you are defining y = tan^2(x), it's not possible for y to equal -2/3.

f = 12 is the only solution. There are an infinite number of values of x that satisfy the minimum solution, however, since x = arc tan√(2/3) = 0.68 +/- n*pi, n = +/- 1, 2, ...

 

[rude man]

Right.

 

[NascentOxygen]

tan(x)=$\pm \sqrt{2/3}$. If i substitute tan(x)=$\pm \sqrt{2/3}$ in the given equation i get the answer as 12. Is this the minimum value?

I don't get what do you mean by there are infinite solutions?

There's a solution in all 4 quadrants between 0 and 2Pi. +sqrt(2/3) gives 2 solutions here, and -sqrt(2/3) provides the other two solutions.