Upchuck Is Slipping!

Calculus Level 4

Ben's one of the awesome alien forms is Upchuck. He can eat up anything and transform those into energy to throw back.

This time Upchuck is at the South Pole. He's standing on a frozen lake, which necessarily indicates frictionless icy surface and he is at rest initially. Vilgax is shooting Upchuck with his destructive Cosmic Rifle. And excess to mention, Upchuck is eating the shots. Consequently, by Newton's third law, Upchuck is slipping backwards. At any time, t(in seconds) , you can express Upchuck's displacement(in meters) by this differential equation: d s = A t B + C t d t ds=\frac{At}{B+Ct}dt Where, A,B,C are constant. What's Upchuck's displacement(in meters, of course) at t = 100 s e c o n d s t=100 seconds ( Rounded up to the nearest integer )?

Given, A = ϕ ( G o l d e n R a t i o ) A=\phi (Golden Ratio) B = e B=e C = π C=\pi

Bonus: Have you produced the generalized equation for any values of the constant?


The answer is 270.

Muhammad Arifur Rahman by Muhammad Arifur Rahman , 23, Bangladesh

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1 solution

We just have to integrate the differential equation, right? d s = A t B + C t d t ds=\frac{At}{B+Ct}dt ∫ d s = ∫ A t B + C t d t \int ds=\int \frac{At}{B+Ct}dt = A ∫ t B + C t d t =\text{A}\int\frac{t}{B+Ct}dt = A ∫ ( 1 − B + C t − t B + C t ) d t =\text{A}\int(1-\frac{B+Ct-t}{B+Ct})dt = A ( ∫ d t − ∫ B + C t − t B + C t d t ) =\text{A}(\int dt-\int\frac{B+Ct-t}{B+Ct}dt) Now given, A = ϕ = 1 + 5 2 \text{A}=\phi=\frac{1+\sqrt{5}}{2} B = e = 2.71 … B=e=2.71\dots C = π = 3.14159 … C=\pi=3.14159\dots We use the integration technique.

Assume, u = B + C t … ( 1 ) u=B+Ct\dots (1) d u d t = d d t ( B + C t ) = C \frac{du}{dt}=\frac{d}{dt}(B+Ct)=C d t = d u C … ( 2 ) dt=\frac{du}{C}\dots (2)

And also, t = u − B C … ( 3 ) t=\frac{u-B}{C}\dots (3) Plugging 1,2,3 into the main, ∫ d s = A ( ∫ d t − ∫ u − u − B C u . d u C ) \int ds=\text{A}(\int dt-\int\frac{u-\frac{u-B}{C}}{u}.\frac{du}{C}) = A ( ∫ d t − ∫ u C − u + B C u C d u ) =\text{A}(\int dt-\int\frac{\frac{uC-u+B}{C}}{uC}du) = A ( ∫ d t − ∫ u C − u + B u C 2 d u ) =\text{A}(\int dt-\int\frac{uC-u+B}{uC^2}du) ∫ d s = A ( ∫ d t − 1 C ∫ d u + 1 C 2 ∫ d u − B C 2 ∫ 1 u d u ) \int ds=\text{A}(\int dt-\frac{1}{C}\int du+\frac{1}{C^2}\int du-\frac{B}{C^2}\int\frac{1}{u}du) s = A ( t − u C + u C 2 − B C 2 l n ( u ) ) s=\text{A}(t-\frac{u}{C}+\frac{u}{C^2}-\frac{B}{C^2}ln(u)) Let's define, t 1 = 0 s t_1=0s t 2 = 100 s t_2=100s u 1 = B + C t 1 = B u_1=B+Ct_1=B u 2 = B + C t 2 = B + 100 C u_2=B+Ct_2=B+100C Finally get Upchuck's displacement, s = A { ( 100 − 0 ) − B + 100 C − B C + B + 100 C − B C 2 − B C 2 l n B + 100 C B } s=\text{A}\{(100-0)-\frac{B+100C-B}{C}+\frac{B+100C-B}{C^2}-\frac{B}{C^2}ln\frac{B+100C}{B}\} From your calculator, s = 269.98 m s=269.98m Round up to the nearest integer to get it! D i s p l a c e m e n t , s = 270 m Displacement,\boxed{s=270m}

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