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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?

Sagot :

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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