Related Rates Problem

[CanadaAotS]

Right now in calculus we're doing related rates, specifically to do with triangles and the useage of Pythagorean Theorem.

 

The problem goes like this:

 

A police helicopter is flying north at 60km/h at a constant altitude of 1 km. On the highway below, a car is travelling east at 45 km/h. When the chopper passes over the highway, the car is 2 km west of the point directly below it. At this moment, how fast is the distance between the car and the chopper changing? Is this distance increasing or decreasing? Explain.

 

Now the answer they give in the back is 40.25 km/h and they say it's decreasing.

 

I figured this out so far with c being the distance the car travels down the highway, h being the distance the helicopter travels along its northerly path and d is the distance between the two and t as time:

 

[math]\frac{dc}{dt} = 45 \mbox{ km/h}[/math]

 

[math]\frac{dh}{dt} = 60 \mbox{ km/h}[/math]

 

What I need to find out is [math]\frac{dd}{dt}[/math].

 

To get a better feel of what I was doing I drew a diagram in my note book (which I recreated in paint brush and is attached to this thread).

 

I was thinking that [math]d = \sqrt{c^2 + a^2}[/math] (a is the altitude which is the constant of 1 km) but that only works when the helicopter is hovering over the point where the cars path and the helicopters path intersect.

 

If anyone could help me out with this it'd be appreciated. BTW it apparently took my Calc teacher 12 hours to get this lol.

[ecoli]

first figure out which sides of the triangle are changing and which parts are constant. I think that hint may help you out... if not, come back.

 

btw... surpized it took so long for you teacher to figure it out. I haven't tried the problem yet, but it doesn't seem that different then any other related rates problem I've done.

[woelen]

Right now in calculus we're doing related rates' date=' specifically to do with triangles and the useage of Pythagorean Theorem.

 

The problem goes like this:

 

A police helicopter is flying north at 60km/h at a constant altitude of 1 km. On the highway below, a car is travelling east at 45 km/h. When the chopper passes over the highway, the car is 2 km west of the point directly below it. At this moment, how fast is the distance between the car and the chopper changing? Is this distance increasing or decreasing? Explain.

 

Now the answer they give in the back is 40.25 km/h and they say it's decreasing.

 

I figured this out so far with c being the distance the car travels down the highway, h being the distance the helicopter travels along its northerly path and d is the distance between the two and t as time:

 

[math']\frac{dc}{dt} = 45 \mbox{ km/h}[/math]

 

[math]\frac{dh}{dt} = 60 \mbox{ km/h}[/math]

 

What I need to find out is [math]\frac{dd}{dt}[/math].

 

To get a better feel of what I was doing I drew a diagram in my note book (which I recreated in paint brush and is attached to this thread).

 

I was thinking that [math]d = \sqrt{c^2 + a^2}[/math] (a is the altitude which is the constant of 1 km) but that only works when the helicopter is hovering over the point where the cars path and the helicopters path intersect.

 

If anyone could help me out with this it'd be appreciated. BTW it apparently took my Calc teacher 12 hours to get this lol.

I agree with the answer given here.I did the math and the answer is correct. Took me 5 minutes .

 

The distance can be written as d = sqrt(c*c+h*h+a*a), here a is the constant altitude of 1 km.

 

Now dd/dt can be written as (2*c*dc/dt + 2*h*dh/dt + 2*a*da/dt) / 2d.

 

Plug in the values (we work completely in hours and km, then no factors pop up in the derivatives):

 

c equals -2 (the car is at the left and approaching the helicopter)

dc/dt equals 45.

 

h equals 0 (the helicopter is precisely above the highway). So the value of dh/dt does not matter here (it does matter at any other point on its trajectory however).

 

a equals 1 and is constant, so da/dt equals 0.

 

Now the result is simple:

 

d = sqrt(c*c + h*h + a*a) = sqrt((-2)*(-2) + 0*0 + 1*1) = sqrt(4+1) = sqrt(5).

 

dd/dt = (2*(-2)*45 + 0 + 0) / (2*sqrt(5))

 

If you approach this numerically, then you see that this is near -40.25 km/h.

[CanadaAotS]

ok ok, now if I knew that [math]d = \sqrt{c^2 + h^2 + a^2}[/math] I would've been able to do the problem... why does d = that? like I said in the original post, I knew that when the helicopter hovered at 0 that [math]d = \sqrt{c^2 + a^2}[/math], I did not know that that was because h^2 is included and since its 0 it's absent...

[woelen]

ok ok, now if I knew that [math]d = \sqrt{c^2 + h^2 + a^2}[/math] I would've been able to do the problem... why does d = that? like I said in the original post, I knew that when the helicopter hovered at 0 that [math]d = \sqrt{c^2 + a^2}[/math'], I did not know that that was because h^2 is included and since its 0 it's absent...

Apply Pythagoras' rule twice.

 

If you have three sides A (left-right), B(forward backward) and C(up-down), all perpendicular to each other, then you can first apply pythagoras at the gound: sqrt(A*A+B*B). This line still is perpendicular to the line going upwards. Now apply pythagoras again:

 

sqrt(sqrt(A*A+B*B) * sqrt(A*A+B*B) + C*C) = sqrt(A*A+B*B+C*C)

 

This is the cumbersome way of explaining this:-) . In general, the length of a vector x in N dimensions equals [math]\sqrt(x^Tx)[/math], but in order to understand that you need some basic linear algebra. It is written as [math]||x||_2[/math], the 2-norm of a vector.