Related Rates: An insect starts from the origin and crawls

[xtrmk]

Will someone be kind enough to help me? Thank you

At time t = 0 an insect starts from the origin crawling along a straight line at a rate of 3 ft/min. Two minutes later a second insect starts from the origin but in a direction perpendicular to that of the first, and at a speed of 5ft/min. How fast is the distance between them changing when the first insect has traveled 12 ft?

t = 0
dx/dt = 3ft/min

t=2
dx/dt = 5ft/min

I dont think i am setting it up or starting it right. any help is appreciated

 

[tkhunny]

The rates are insufficient. You must also know where each bug is. Let's assume Bug1 is on the x-axis and Bug2 is on the y-axis.

Displacement of Bug1 = 3*t, where t is time in minutes.

At any time, Bug1 is at (Bug1(t),0)

Displacement of Bug2 = 5*(t-2), where t is time in minutes from the beginning of the experiment.

At any time AFTER t = 2 min, Bug2 is at (0,Bug2(t))

A little Pythagorean Theorem should get you right up next to a complete solution.

 

[xtrmk]

still unclear. could you or someone else explain it in a different way

 

[soroban]

Re: Related Rates

Hello, xtrmk!

When someone or something has a headstart,
. . I have my own approach . . .

At time t = 0 an insect starts from the origin crawling along a straight line
at a rate of 3 ft/min. .Two minutes later a second insect starts from the origin
but in a direction perpendicular to that of the first, and at a speed of 5ft/min.
How fast is the distance between them changing when the first insect has traveled 12 ft?

    C *
      |   *
      |       *     x
   5t |           *
      |               *
      |                   *
      * - - - - * - - - - - - *
      O    6    A      3t     B

Bug #1 starts at \(\displaystyle O\) and walks for 2 minutes at 3 ft/min.
. . It walk 6 feet to point \(\displaystyle A.\)
In the next \(\displaystyle t\) seconds, it walk \(\displaystyle 3t\) feet to point \(\displaystyle B.\)

During the same \(\displaystyle t\) seconds, bug #2 walks \(\displaystyle 5t\) feet from \(\displaystyle O\) to \(\displaystyle C.\)

Pythagorus says: \(\displaystyle \:x^2\:=\5t)^2\,+\,(3t\,+\,6)^2 \:=\:34t^2\,+\,36t\,+\,36\)

Differentiate with respect to time: \(\displaystyle \:2x\cdot\frac{dx}{dt}\:=\:68t\,+\,36\;\;\Rightarrow\;\;\L\frac{dx}{dt}\:=\:\frac{1}{x}(34t\,+\,18)\;\) [1]


But #1 has travelled 12 feet: \(\displaystyle \:6\,+\,3t\:=\:12\;\;\Rightarrow\;\;t\,=\,2\) seconds.
And bug #2 has moved \(\displaystyle 5\cdot2\:=\:10\) feet.

The right triangle has legs \(\displaystyle 12\) and \(\displaystyle 10\).
The hypotenuse is: \(\displaystyle \:x\:=\:\sqrt{12^2\,+\,10^2} \:=\:\sqrt{244}\:=\:2\sqrt{61}\) feet.

Substitute into [1]: \(\displaystyle \L\:\frac{dx}{dt}\:=\:\frac{1}{2\sqrt{61}}[34(2)\,+\,18] \:=\:\frac{43}{\sqrt{61}}\:\approx\:5.5\) ft/sec.

 

[galactus]

I done this a wee bit differently than Soroban.

He beat me to the post, but what the heck.

\(\displaystyle \L\\D^{2}=(3t)^{2}+(5(t-2))^{2}\)

\(\displaystyle \L\\D^{2}=34t^{2}-100t+100\)

Differentiate:

\(\displaystyle \L\\2D\frac{dD}{dt}=(68t-100)\)

As Soroban showed, When bug 1 has traveled 12 feet, bug 2 has traveled

10 feet. Using ol' Pythagoras we get \(\displaystyle D=2\sqrt{61}\)
..........................................\(\displaystyle {\uparrow}\)
.......................\(\displaystyle \text{Soroban, haha}\)

\(\displaystyle \L\\4\sqrt{61}\frac{dD}{dt}=(68(4)-100)\)

\(\displaystyle \L\\\frac{dD}{dt}=\frac{172}{4\sqrt{61}}\approx{5.5055} \;\ ft/min\)