Two people start biking from the same point. One heads east at 13 ​mph, the other south at 18 mph. What is the rate at which the distance between the two people is changing after 25 minutes and after 35 ​minutes? What is the rate at which the distance between the two people is changing after 25 ​minutes?

Answer:

[tex]22.21\ \text{mph}[/tex]

[tex]22.2\ \text{mph}[/tex]

Step-by-step explanation:

[tex]\dfrac{da}{dt}=13\ \text{mph}[/tex]

[tex]\dfrac{db}{dt}=18\ \text{mph}[/tex]

[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between the two people is changing

At 25 minutes

[tex]a=13\times \dfrac{25}{60}=5.417\ \text{miles}[/tex]

[tex]b=18\times \dfrac{25}{60}=7.5\ \text{miles}[/tex]

[tex]c=\sqrt{5.417^2+7.5^2}\\\Rightarrow c=9.25\ \text{miles}[/tex]

Distance between the two people is

[tex]c^2=a^2+b^2[/tex]

Differentiating with respect to time we get

[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{5.417\times 13+7.5\times 18}{9.25}\\\Rightarrow \dfrac{dc}{dt}=22.21\ \text{mph}[/tex]

Distance between the two people is changing at a rate of [tex]22.21\ \text{mph}[/tex].

At 35 minutes

[tex]a=13\times \dfrac{35}{60}=7.583\ \text{miles}[/tex]

[tex]b=18\times \dfrac{35}{60}=10.5\ \text{miles}[/tex]

[tex]c=\sqrt{7.583^2+10.5^2}\\\Rightarrow c=12.95\ \text{miles}[/tex]

Distance between the two people is

[tex]c^2=a^2+b^2[/tex]

Differentiating with respect to time we get

[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{7.583\times 13+10.5\times 18}{12.95}\\\Rightarrow \dfrac{dc}{dt}=22.2\ \text{mph}[/tex]

Distance between the two people is changing at a rate of [tex]22.2\ \text{mph}[/tex].

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LINKS WILL BE REPORTED

Resolución de Problemas con Ecuaciones Lineales
o de Primer Grado.

Ejemplo

Utilizamos el lenguaje algebraico para hacer esa representación.

Sea x número de casos Infectados en Santiago
Sea 2x + 5,088 números de casos Infectados en Sto. Dgo.
Entonces x + 2x + 5,088 = 43,956
x + 2x = 43,956 - 5,088
3x = 38,868
x = 38,868
3
x = 12,912
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2x + 5,088 número de casos Infectados en Sto. Dgo. = 2 (12,912) + 5,088 = 31,000 43,956

NESECITO QUE ME AYUDEN A RESOLVER EL PROBLEMA



ACTIVIDAD 1 : Según el boletín #329, d/f 10/02/2021 del Ministerio de Salud Pública, el número total de casos infectados hasta esa fecha en el Distrito Nacional, era el doble más 12,585 casos más que los que había en la provincia de la Santiago. ¿Cuántos contagiados hay en cada una de las provincias si entre las dos suman 89,766 casos?
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