l e t x r e p r e s e n t B i d e n ′ s d i s t a n c e f r o m t h e o r i g i n p o i n t let\ x\ represent\ Biden's\ distance\ from\ the\ origin\ point l e t x re p rese n t B i d e n ′ s d i s t an ce f ro m t h e or i g in p o in t
y r e p r e s e n t T r u m p ′ s d i s t a n c e f r o m t h e o r i g i n p o i n t \ \ \ \ \ \ y\ represent\ Trump's\ distance\ from\ the\ origin\ point y re p rese n t T r u m p ′ s d i s t an ce f ro m t h e or i g in p o in t
t r e p r e s e n t t h e t i m e a f t e r t h e y s t a r t e d w a l k i n g \ \ \ \ \ \ t\ represent\ the\ time\ after\ they\ started\ walking t re p rese n t t h e t im e a f t er t h ey s t a r t e d w a l kin g
z r e p r e s e n t t h e d i s t a n c e b e t w e e n t h e m \ \ \ \ \ z\ represent\ the\ distance\ between\ them z re p rese n t t h e d i s t an ce b e tw ee n t h e m
z = x 2 + y 2 z=\ \sqrt{x^2+y^2} z = x 2 + y 2
d x d t = 52 \frac{dx}{dt}=52 d t d x = 52
d y d t = 86 \frac{dy}{dt}=86 d t d y = 86
w e w i s h t o c a l c u l a t e d z d t we\ wish\ to\ calculate\frac{dz}{dt} w e w i s h t o c a l c u l a t e d t d z
∫ d x d t d t = 52 t + c \int{\frac{dx}{dt}dt}=52t+c ∫ d t d x d t = 52 t + c
∫ d y d t d t = 86 t + k \int{\frac{dy}{dt}dt}=86t+k ∫ d t d y d t = 86 t + k
at\ t=0,\
x = 0 , y = 0 x=0,\ y=0 x = 0 , y = 0
52 ( 0 ) + c = 0 → c = 0 → x = 52 t 52\left(0\right)+c=0\rightarrow c=0\rightarrow x=52t 52 ( 0 ) + c = 0 → c = 0 → x = 52 t
86 ( 0 ) + k = 0 → k = 0 → y = 86 t 86\left(0\right)+k=0\rightarrow k=0\rightarrow y=86t 86 ( 0 ) + k = 0 → k = 0 → y = 86 t
z = x 2 + y 2 = ( 52 t ) 2 + ( 86 t ) 2 = 10100 t z=\sqrt{x^2+y^2}=\ \sqrt{\left(52t\right)^2+\left(86t\right)^2}=\ \sqrt{10100}\ t z = x 2 + y 2 = ( 52 t ) 2 + ( 86 t ) 2 = 10100 t
d z d t = 10100 = 100.499 m / s \frac{dz}{dt}=\ \sqrt{10100}=100.499m/s d t d z = 10100 = 100.499 m / s
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