Related Rates: What is the speed of the plane at that point?

[Guest]

I think my problem with this is that im not drawing the picture correctly-- this problem is just not making sense to me, although it is probably not that hard and im just doing something silly.. The book does say that it is helpful to use law of cosines though.

"An airplane which is climbing at an angle of 45 degrees passes directly over a ground radar station at an altitude of 1 mi. Later, a reading shows that the distance from the radar station to the plane is 5 mi. and is increasing at 7 mi./min. What is the speed, in mi/hr, of the plane then?"

Any suggestions would be greatly appreciated. Thank you.

 

[stapel]

I think my problem with this is that im not drawing the picture correctly....

So how did you draw the picture? (Please reply with the image, a link to the image, or a detailed description of the image.) Thank you.

Eliz.

 

[Guest]

how can i draw it on here?

 

[stapel]

how can i draw it on here?

You can't. You need to post it (from some remote hosting somewhere), post a link to it (on said remote hosting), or provide a detailed description.

Thank you.

Eliz.

 

[galactus]

Is this your image?.

You could try the law of cosines. You can use your data to find a and then da/dt(the planes speed).

Solve \(\displaystyle 5^{2}=a^{2}+b^{2}-2abcos(\frac{3{\pi}}{4})\) for a.

Then differentiate the law of cosine formula:

\(\displaystyle c^{2}=a^{2}+b^{2}-2abcos(\frac{3{\pi}}{4})\)

\(\displaystyle c^{2}=a^{2}+b^{2}-2ab(\frac{-1}{\sqrt{2}})\)

\(\displaystyle c^{2}=a^{2}+b^{2}+\sqrt{2}ab\)

Be careful. differentiate with respect to time, enter in your data and solve for da/dt. It's easier than it looks. db/dt=0

 

[Guest]

Hey, thanks for the help, but my answer is still not coming out correctly.

I solved for A in the equation and got A to be = rad26 - rad 2

then after differentiating i got da/dt = (2c dc/dt)/(2a+rad2b). I plugged in my answers and i got 7.97 but my instructor told us that we should be getting 7.07 (5 rad 2), thanks again...

 

[stapel]

It would really help if you showed all of your steps, not just the final answers. For instance, how did you get "A = 24 radians"? Which angle is A?

Thank you.

Eliz.

 

[Guest]

I used Galactus's equation to solve for A and to find da/dt.

 

[Guest]

oh, i see why you're probably confused- when i said rad26 i meant radical. Sorry about that.

 

[Guest]

24=a^2+2radical2*a
so a = radical 26 -radical 2


2c dc/dt = 2a da/dt + 2b db/dt + radical 2(da/dt*b + db/dt *a)
da/dt = (2c dc/dt)/ (2a + radical2 *b)

does anyone know what im doing incorrectly?

 

[galactus]

Here we go swimmer:

You have: \(\displaystyle \L\\c^{2}=a^{2}+b^{2}+\sqrt{2}ab\)

Differentiate:

\(\displaystyle \L\\2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}+\sqrt{2}(a\frac{db}{dt}+b\frac{da}{dt})\)

\(\displaystyle \L\\2(5)(7)=2(3\sqrt{2})\frac{da}{dt}+2(1)(0)+\sqrt{2}((3\sqrt{2})(0)+(1)\frac{da}{dt})\)

\(\displaystyle \L\\70=7\sqrt{2}\frac{da}{dt}\)

\(\displaystyle \L\\\frac{da}{dt}=5\sqrt{2}\;\ mi/min.\)

As for the rad thing. I would suggest using 'sqrt'. Better yet, take time to learn some LaTex.

Have fun