Answer:
After 0.24 hours Rahul and Kumi are closest to each other.
Step-by-step explanation:
It is given that Rahul is 4 kilometers due south of the post office and traveling toward it at 10 kilometers per hour. It means length of leg1 after t hours is
[tex]leg_1=4-10t[/tex]
At noon, Kumi is 2 kilometers due east of the post office and jogging easy away from it at 5 kilometers per hour. It means length of leg2 after t hours is
[tex]leg_1=2+5t[/tex]
Using Pythagoras theorem,
[tex]Hypotenuse^2=leg_1^2+leg_2^2[/tex]
[tex]S^2=(4-10t)^2+(2+5t)^2[/tex]
Differentiate with respect to t.
[tex]2S\frac{dS}{dt}=2(4-10t)(-10)+2(2+5t)(5)[/tex]
[tex]2S\frac{dS}{dt}=2(-40+100t+10+25t)[/tex]
Cancel out common factors.
[tex]S\frac{dS}{dt}=125t-30[/tex]
Divide both sides by S.
[tex]\frac{dS}{dt}=\frac{125t-30}{S}[/tex]
[tex]\frac{dS}{dt}=\frac{125t-30}{\sqrt{(4-10t)^2+(2+5t)^2}}[/tex] .... (1)
Equate [tex]\frac{dS}{dt}=0[/tex], to find critical values.
[tex]\frac{125t-30}{\sqrt{(4-10t)^2+(2+5t)^2}}=0[/tex]
[tex]125t-30=0[/tex]
[tex]125t=30[/tex]
[tex]t=\frac{30}{125}=0.24[/tex]
The critical value of the distance function is 0.24.
Differentiate (1) with respect to t.
[tex]\frac{dS}{dt}=\frac{125t-30}{\sqrt{(4-10t)^2+(2+5t)^2}}[/tex]
[tex]\frac{d^2S}{dt^2}=\frac{64 \sqrt{5}}{(25 t^2 - 12 t + 4)^{3/2}}[/tex]
Substitute t=0.24,
[tex]\frac{d^2S}{dt^2}=\frac{64 \sqrt{5}}{(25 (0.24)^2 - 12 (0.24) + 4)^{3/2}}\approx 34.94>0[/tex]
Since [tex]\frac{d^2S}{dt^2}>0[/tex], therefore the distance of Rahul and kumi is minimum at t=0.24.
Therefore, after 0.24 hours Rahul and Kumi are closest to each other.
