Help double-checking the answer to a problem.

Question

deoxna · Answer

I need help with question 3, specifically. I got the final answer as:
$$\frac{ dx }{ dt } = \frac{ 450 }{ \sqrt{1000} } \approx 14 $$

theviper · Answer

hi $$\Huge{\color{red}{\text{Welcome to OPENSTUDY}\quad\color{blue}{\ddot{\smile}}}}$$

deoxna · Answer

Would it help if I wrote out my process? I'm fairly new to this part of calculus so I'm not very sure I'm doing it right. I just fell like I made it unnecessarily complicated....

tkhunny · Answer

Always.

ash2326 · Answer

Let after some time t, the horizontal distance is x, so the distance of airport from aircraft D is
$$ D=\sqrt {x^2+10^2}$$
10000 meters=10 km
$$\frac{dD}{dt}=\frac{1}{2\sqrt{x^2+10^2}}\times 2x \frac{dx}{dt}$$
at t= 2 minutes 
$$x=\frac 2 {60} \times 900=30 km$$

so x=30 km
$$\frac{dD}{dt}=\frac{1}{2\sqrt{{30}^2+10^2}}\times 2\times 30 \frac{dx}{dt}$$

$$\frac{dx}{dt}=900 kmph$$
$$\frac{dD}{dt}=\frac{1}{2\sqrt{{30}^2+10^2}}\times 2\times 30 \times 900=\frac{27000}{\sqrt{1000}}$$

deoxna · Answer

Ok, I see what I did wrong. I converted the 900 km/h to 15 km/min and it worked all right except in the final answer. Your answer is in km/hr and mine in km/min, but even after converting mine to km/hr its not right. Shouldn't it work out regardless of units used (as long as they are maintained)

tkhunny · Answer

You do have a units problem, but ash2326 has a formulation problem.

@ash2326 $D(x) = \sqrt{(15\cdot x)^{2} + 10^2}$

It's okay at the end where the denominator uses 30 km, but the derivative needs some work.

deoxna · Answer

@tkhunny Wait, why would that be the function for D? X already is the horizontal distance between the airport and the plane's "shadow"...

ash2326 · Answer

@DeoxNA But the distance between plane and airport is D|dw:1361338712135:dw|

ash2326 · Answer

@tkunny In my solution x= distance
not distance per min. I have derived for the distance D, not distance/min

ash2326 · Answer

@tkhunny

deoxna · Answer

@ash2326 That's what I meant to say (though not very clearly) in my.... :P

tkhunny · Answer

Well, you can redefine D(x) to include the 15 (like I suggested) or you need to get it into the definition of dx/dt.  Take your pick.  You can't do neither.  You don't have the speed of the aircraft in the definition of your derivative.

deoxna · Answer

My last comment should end in "comment" (I just can't write today)...