With Respect to Related Rates

[kimmy_koo51]

I just want to double check to make sure I have this question correct.

Oil leaking from a damaged oil tanker forms as a circular slick on the ocean surface. The radius of this circle is increasing at 1.5 m/min. How fast is the area of the slick increasing when the radius is 100 m.

dr/dt = 1.5 m/min
r= 100 m

A = pie r^2
dA/dt = pie(2r)(dr/dt)
dA/dt = pie (2)(100)(1.5)
da/dt = 300 pie

Therefore, the area is increasing at 300pie m/min when the radius is 100m.

Something seems off to me. Can someone double check my work and everything?

Thank you in advance.

 

[galactus]

Enough with the 'pie' already....DUH!. :roll:

It's a Greek letter, not your dessert. PI.

Yes, your answer is correct. Good work.

 

[kimmy_koo51]

Excellent, thank you.

 

[skeeter]

what's "off" are your units for dA/dt, the rate of change of area ...

dA/dt = 300pi m²/min

also ... "pie" is something you eat; "pi" is the lower case Greek letter that represents the constant ratio of the circumference of a circle to its diameter.

 

[kimmy_koo51]

What about this question:

A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 km north of the intersection, and the car is 0.8 km to the east, the police determine with radar that the distance between them and the car is increasing at 20km/h. If the cruiser is moving 60km/h at the instant of measurement, what is the speed of the car.

My work:

Let y be the police cruiser
Let x be the car being chased
Let r be the distance between them.
y= 0.6 km
dy/dt = 60 km/h
x=0.8 km
dx/dt = ?
r = ?
dr/dt = 20 km/r

Solve for r

r^2 = y^2 + x^2
r^2 = 0.6^2 + 0.8^2
r = 1

Differentiate

2r dr/dt = 2y dy/dt + 2x dx/dt
Sub everything in and get a value of 70 km/h for dx/dt

 

[skeeter]

your answer is correct, but I would like to know what exactly did you substitute in for dy/dt to determine dx/dt?

 

[kimmy_koo51]

I substituted in 60 for dy/dt

 

[skeeter]

I substituted in 60 for dy/dt

then dx/dt wouldn't work out to be 70 km/hr, would it?

r dr/dt = y dy/dt + x dx/dt

(1)(20) = (.6)(60) + (.8)(dx/dt)

20 = 36 + .8(dx/dt)

16 = .8(dx/dt)

20 km/hr = dx/dt

where is the mistake?

 

[kimmy_koo51]

My mistake or a mistake in the problem?

 

[skeeter]

your mistake.

 

[kimmy_koo51]

That wouldn't make sense, having a high speed car chase, and the one being chased driving at 20 km/h, now would it?

 

[kimmy_koo51]

I have no idea what I did wrong.

 

[skeeter]

why do you think I asked you about the value you used for dy/dt?

what is actually happening to the distance "y" during the chase?

 

[kimmy_koo51]

Distance y is decreasing, isn't it?

 

[skeeter]

if a value is decreasing, what can you say about its rate of change (its derivative)?

 

[kimmy_koo51]

It is also decreasing! Ohh! That makes the 60 km/h a negative?

 

[skeeter]

correct on the negative ... note that the derivative itself is not decreasing, it's constant.

Glad you figured that out yourself ... now you'll remember it.

 

[kimmy_koo51]

Thanks a lot!