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Calculus Level 5

The time of a complete oscillation of a simple pendulum of length l l is given by T = 2 π ⋅ l g T = 2\pi \cdot \sqrt{\frac{l}{g}} , where g g is a constant. Using differentials , by what per cent should the length be changed in order to correct a loss of 2 minutes per day?

Let the percent change in length be represented as A B % \frac{A}{B} \% , where A A and B B are coprime positive integers, find A + B A+B .


The answer is 23.

Mohit Gupta by Mohit Gupta , 20, India

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1 solution

T = 2 π l g ln ⁡ ( T ) = ln ⁡ ( 2 π ) + ln ⁡ ( l ) 2 − ln ⁡ ( g ) 2 D i f f e r e n t i a t e b o t h s i d e s ( 100 ) d T T = ( 100 ) d l 2 L % c h a n g e i n l e n g t h = ( 100 ) d l L % c h a n g e i n l e n g t h = 200 d T T = 200 2 ( 24 ) . ( 60 ) = 5 18 T=2\pi \sqrt { \frac { l }{ g } } \\ \ln { (T } )=\ln { (2\pi ) } +\frac { \ln { (l) } }{ 2 } -\frac { \ln { (g) } }{ 2 } \\ Differentiate\quad both\quad sides\\ (100)\frac { dT }{ T } =(100)\frac { dl }{ 2L } \\ \%\quad change\quad in\quad length\quad =(100)\frac { dl }{ L } \\ \%\quad change\quad in\quad length\quad =\quad 200\frac { dT }{ T } \\ =200\frac { 2 }{ (24).(60) } =\frac { 5 }{ 18 }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

1 day=1440 min=T

I know how everyone solved this, T = 1440 , T 1 = 1442 ⇒ d l l × 100 = d T T × 200 = 2 1440 × 200 = 5 18 T=1440,{ T }_{ 1 }=1442\Rightarrow \cfrac { dl }{ l } \times 100=\cfrac { dT }{ T } \times 200=\cfrac { 2 }{ 1440 } \times 200=\cfrac { 5 }{ 18 }

the point is that d l 2 l = d T T \cfrac { dl }{ 2l } =\cfrac { dT }{ T } can be used only when d t → 0 , d T → 0 dt\rightarrow 0,dT\rightarrow 0

Now, T 1 T = l + Δ l l ⇒ 1 + Δ l l = ( T 1 T ) 2 = ( 1442 1440 ) 2 = ( 721 720 ) 2 ⇒ Δ l l = ( 721 720 ) 2 − 1 = 1441 518400 ⇒ Δ l l × 100 = 1441 5184 ≃ 5 18 \cfrac { { T }_{ 1 } }{ T } =\sqrt { \cfrac { l+\Delta l }{ l } } \Rightarrow 1+\cfrac { \Delta l }{ l } ={ \left( \cfrac { { T }_{ 1 } }{ T } \right) }^{ 2 }={ \left( \cfrac { 1442 }{ 1440 } \right) }^{ 2 }={ \left( \cfrac { 721 }{ 720 } \right) }^{ 2 }\\ \\ \Rightarrow \cfrac { \Delta l }{ l } ={ \left( \cfrac { 721 }{ 720 } \right) }^{ 2 }-1=\cfrac { 1441 }{ 518400 } \\ \\ \Rightarrow \cfrac { \Delta l }{ l } \times 100=\cfrac { 1441 }{ 5184 } \simeq \cfrac { 5 }{ 18 }

so A B = 1441 5184 , A B ≠ 5 18 A n s = A + B = 1441 + 5184 = 6625 \cfrac { A }{ B } =\cfrac { 1441 }{ 5184 } ,\cfrac { A }{ B } \neq \cfrac { 5 }{ 18 } \\ \\ Ans=A+B=1441+5184=6625

Ayush Verma - 5 years, 2 months ago

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