Related Rates [2 problems]

[Guest]

One side of a rectangle increases at 2 cm/s, while the other side decreases at 3cm/s. How fast is the area of the rectangle changing when the first side equals 20cm and the second side equals 50cm?


so I'm given: ds/dt = 2cm/s ( rate of one side)
ds/dt= -3cm(rate of 2nd side)
side 1: 20cm
side 2: 50 cm
need to find dsr/dt (rate of change of side rate)
A=l*w

I dunno how you would do this.

2nd question: oil spilled from a ruptured tanker spreads in a circle whos area increases at a constant rate of 6km^2/h. How fast is the radius of the spill increasing when the area is 9pikm^2?

im given da/dt=6km^2/h (rate of area)
A= 9pikm^2
need to find dr/dt (rate of radius)

What formula do I use? cuz I tried doing this:
A= pir^2
da/dt= pi2r(dr/dt)
but then where do I put the area? so..Im prett sure I'm not doing this right

thanks for the help

 

[soroban]

Hello, bittersweet!

One side of a rectangle increases at 2 cm/s, while the other side decreases at 3cm/s.
How fast is the area of the rectangle changing when the first side equals 20cm and the second side equals 50cm?

so I'm given: ds/dt = 2cm/s ( rate of one side)
ds/dt= -3cm(rate of 2nd side) \(\displaystyle \;\) . . . both sides are called "s" . . . like a square?
side 1: 20cm
side 2: 50 cm
need to find dsr/dt (rate of change of side rate)

A=L*W
If you're going to use "L" and "W", why not start with them?

Let \(\displaystyle L\) = first side, and \(\displaystyle W\) = second side.

We have:\(\displaystyle \,A\:=\:LW\)

Differentiate with repect to time: \(\displaystyle \L\,\frac{dA}{dt}\:=\:L\left(\frac{dw}{dt}\right)\.+\,W\left(\frac{dL}{dt}\right)\)

We are given: \(\displaystyle \frac{dL}{dt}\,=\,+2,\;\frac{dW}{dt}\,=\,-3,\;L\,=\,20,\;W\,=\,40\)

Therefore: \(\displaystyle \L\,\frac{dA}{dt}\:=\20)(2)\,+\,(50)(-3) \:=\; -110\) \(\displaystyle \,\text{cm}^2\text{/sec}\)

2) Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 km²/hr.
How fast is the radius of the spill increasing when the area is \(\displaystyle 9\pi\) km²?

What formula do I use? cuz I tried doing this:
\(\displaystyle A\:=\:\pi r^2\;\;\Rightarrow\;\;\frac{dA}{dt}\:=\: 2\pi r\left(\frac{dr}{dt}\right)\)
but then where do I put the area?

You don't "put" the area anywhere . . .


Your equations are correct.

We have: \(\displaystyle \L\,\frac{dA}{dt}\:=\:2\pi r\left(\frac{dr}{dt}\right)\)

And we want: \(\displaystyle \L\,\frac{dr}{dt}\:=\:\frac{1}{2\pi r}\left(\frac{dA}{dt}\right)\)

We are given: \(\displaystyle \,\frac{dA}{dt}\,=\,6\) . . . but we need \(\displaystyle r\).
That's where we use the area . . .

Since \(\displaystyle A\,=\,\pi r^2\), we have: \(\displaystyle \,9\pi\:=\:\pi r^2\;\;\Rightarrow\;\;r\,=\,3\)


Now we can "put" the radius into our equation:

\(\displaystyle \L\;\;\frac{dr}{dt}\;=\;\frac{1}{2\pi(3)}(6)\;=\;\frac{1}{\pi}\)\(\displaystyle \,\text{km/hr.}\)