Let's practise the basics

Calculus Level 5

Find the minimum of

[ ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t + 3 cos ⁡ 2 t ( sin ⁡ t + 2 ) 2 ] 2 . \left[ \frac{(\sin{t}+2)^2}{\cos^2t}+\frac{3\cos^2t}{(\sin{t}+2)^2} \right]^2.

From a calculus exam I recently gave.


The answer is 16.

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Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Otto Bretscher
Nov 9, 2015

In a first step, let's find the minimum p p of the first summand, ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t \frac{(\sin{t}+2)^2}{\cos^2t} . Taking the derivative, 2 ( sin ⁡ t + 2 ) ( 2 sin ⁡ t + 1 ) cos ⁡ 3 t \frac{2(\sin{t}+2)(2\sin{t}+1)}{\cos^3t} , we see that the minimum is attained when sin ⁡ t = − 1 / 2 \sin{t}=-1/2 , so that p = ( 3 / 2 ) 2 3 / 4 = 3 p=\frac{(3/2)^2}{3/4}=3 .

Making the substitution u = ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t u=\frac{(\sin{t}+2)^2}{\cos^2t} , we can write the given function f f as u + 3 u u+\frac{3}{u} ; this function is increasing for u > 3 u>\sqrt{3} . Since the minimal value of u u is 3, the minimal value of u + 3 u u+\frac{3}{u} is m = 3 + 3 3 = 4 m=3+\frac{3}{3}=4 , so that m 2 = 16 m^2=\boxed{16} .

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Flipping the order of your paragraphs would help to motivate the approach taken. Otherwise, "find the minimum of the first summand" is bad form for problems where we want to find the minimum of the expression.

  1. Using u = ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t u = \frac{(\sin{t}+2)^2}{\cos^2t} , we want to find the minimum of F ( u ) = ( u + 3 u ) 2 F(u) = ( u + \frac{3}{u} ) ^2 .
  2. The range of u u is [ 3 , ∞ ) [ 3, \infty ) .
  3. The shape of F ( u ) F(u) looks like ....
  4. Hence, the minimum occurs at ...

Nice! I chose the substitution x = sin ⁡ ( t ) + 2 cos ⁡ ( t ) x = \dfrac{\sin(t) + 2}{\cos(t)} , so that

f ( x ( t ) ) = x 2 + 3 x 2 ⟹ d f d t = d f d x d x d t = 2 x ( 1 − 3 x 4 ) ( 1 + 2 sin ⁡ ( t ) cos ⁡ 2 ( t ) ) . f(x(t)) = x^{2} + \dfrac{3}{x^{2}} \Longrightarrow \dfrac{df}{dt} = \dfrac{df}{dx} \dfrac{dx}{dt} = 2x\left(1 - \dfrac{3}{x^{4}}\right) \left(\dfrac{1 + 2\sin(t)}{\cos^{2}(t)}\right).

Now x ≠ 0 x \ne 0 since sin ⁡ ( t ) + 2 > 0 \sin(t) + 2 \gt 0 for all real t , t, so the critical points occur when

  • 1 + 2 sin ⁡ ( t ) = 0 ⟹ sin ⁡ ( t ) = − 1 2 , 1 + 2\sin(t) = 0 \Longrightarrow \sin(t) = -\dfrac{1}{2}, corresponding to a value for f f of 4 4 , and

  • x 2 = 3 ⟹ ( sin ⁡ ( t ) + 2 ) 2 = 3 cos ⁡ 2 ( t ) ⟹ sin ⁡ 2 ( t ) + 4 sin ⁡ ( t ) + 4 = 3 ( 1 − sin ⁡ 2 ( t ) ) x^{2} = \sqrt{3} \Longrightarrow (\sin(t) + 2)^{2} = \sqrt{3}\cos^{2}(t) \Longrightarrow \sin^{2}(t) + 4\sin(t) + 4 = \sqrt{3}(1 - \sin^{2}(t))

⟹ ( 1 + 3 ) sin ⁡ 2 ( t ) + 4 sin ⁡ ( t ) + ( 4 − 3 ) = 0 , \Longrightarrow (1 + \sqrt{3})\sin^{2}(t) + 4\sin(t) + (4 - \sqrt{3}) = 0,

which has no real roots as the discriminant b 2 − 4 a c = 12 − 12 3 < 0. b^{2} - 4ac = 12 - 12\sqrt{3} \lt 0.

As f ( t ) > 0 f(t) \gt 0 for all real t t and f ( t ) → ∞ f(t) \rightarrow \infty as cos ⁡ ( t ) → 0 \cos(t) \rightarrow 0 any critical points we found above will correspond to (local) minima, we can conclude that the global minimum of f f is indeed 4. 4.

Brian Charlesworth - 5 years, 7 months ago

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Thank you for taking an interest in my silly little problem, Brian! I don't understand the step where you go from x 2 = 3 x^2=\sqrt{3} to sin ⁡ t + 2 = 3 cos ⁡ t \sin{t}+2=\sqrt{3}\cos{t} .

Otto Bretscher - 5 years, 7 months ago

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Whoops, sorry! It should have been ( sin ⁡ ( t ) + 2 ) 2 = 3 cos ⁡ 2 ( t ) , (\sin(t) + 2)^{2} = \sqrt{3}\cos^{2}(t), which yields no real solutions anyway.

I wonder how many of your students just forged ahead with taking the derivative without any helpful substitution? That direct approach would have ended up with a messy equation to solve.

Brian Charlesworth - 5 years, 7 months ago

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@Brian Charlesworth – Let me admit that I did suggest a substitution to my students... we are leaning the chain rule in my Calc I class.

Otto Bretscher - 5 years, 7 months ago

Liked you solution a lot and got my answer right but ended up writing 4,5,6 (did not saw m 2 m^2 :-) ), please try this

Department 8 - 5 years, 7 months ago

I applied A.M.-G.M. inequality and got the answer as 12.Can you please tell me where I am making the mistake?

PRITHWI SINHA - 5 years, 7 months ago

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The lower bound of 12 \sqrt{12} is not attained since you cannot make the two terms equal . In general, AM-GM does not give you a minimum but merely a lower bound... you need to check whether the value is attained.

Otto Bretscher - 5 years, 7 months ago

Instead of AM-GM inequality try using weighted means inequality.

Tanishq Varshney - 5 years, 7 months ago

The range of ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t \frac{(\sin t + 2)^2}{\cos^2 t} is [ 3 , ∞ ) [3, \infty) , so the value of 12, which is achieved when ( sin ⁡ t + 2 ) 2 cos ⁡ 2 t = 3 \frac{(\sin t + 2)^2}{\cos^2 t} = \sqrt{3} , is impossible.

Jake Lai - 5 years, 7 months ago

How do you say that minimum value of u + 3 u u+\frac{3}{u} happens at minimum value of u?

Harish Sasikumar - 5 years, 7 months ago

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Because u + 3 u u+\frac{3}{u} is increasing for u ≥ 3 u\geq \sqrt{3}

Otto Bretscher - 5 years, 7 months ago

Fun weekend problem, Otto! Thanks for publishing.

tom engelsman - 5 years, 7 months ago

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