Cupid's Arrow

[Mooch22]

Uh-oh! It's Valentine's Day and Cupid has you in his sights. he is located 10 feet about the soon-to-be-love-of-your-life (STBLOYL) and is aiming his arrow directly at you!! You are located 30 feet away from the STBLOYL.

1.) Cupid's arrows fly at a constant rate of 15 ft/sec. If you begin running to the STBLOYL at a constant rate of 5 ft/sec, how long will it take you to meet him/her assuming they don't move?

2.) Cupid begins to fly directly up at a constant rate of 12 feet/sec as soon as you begin running. After four seconds, how fast is the distance between you and Cupid changing?

3.) After how many seconds are you closest to Cupid?

HELP.... PLEASE!! I'M OK IN THE CUPID DEPARTMENT OF MY LIFE, JUST NOT CALCULUS! I'M DESPERATE HERE....!

 

[galactus]

2.) Cupid begins to fly directly up at a constant rate of 12 feet/sec as soon as you begin running. After four seconds, how fast is the distance between you and Cupid changing?

Here's an attempt at part 2. I was thinking, I had better amend my post.

Something doesn't seem kosher.

I think it is this way. Soroban?. skeeter?. Whatcha think?.

We need \(\displaystyle \frac{dD}{dt}\) at t=4, knowing \(\displaystyle \frac{dy}{dt}=12\ and

\frac{dx}{dt}=-5\).

Now, do Pythagoras this way, perhaps.

\(\displaystyle \L\\D^{2}=(30-x)^{2}+(y+10)^{2}\)

Now differentiate with respect to time:

\(\displaystyle \L\\D\frac{dD}{dt}=(30-x)\frac{dx}{dt}+(y+10)\frac{dy}{dt}\)

Now, because Cupid has risen 58 feet in 4 seconds(10+4(12)) and you

have travelled 20 feet in four seconds(30-20=10), \(\displaystyle \sqrt{10^{2}+58^

{2}}{\text{ implies}

D=2\sqrt{866}=58.86\)

Solve for \(\displaystyle \frac{dD}{dt}\)

\(\displaystyle \L\\2\sqrt{866}\frac{dD}{dt}=10(-5)+58(12)\)

\(\displaystyle \L\\\frac{646}{2\sqrt{866}}=\frac{323}{\sqrt{866}}=10.96\ ft/sec.\)

I am more confident with this solution then the last. Check it out.

As for #1, I don't see much to that except 30/5=6 seconds to reach your

STBLOYL. Unless I'm missing something.

Try 3 on your own. In the meantime I will give it a go. Post your attempt.
Okey-doke?.