Related Rates problem

[jtw2e2]

A lighthouse sits 1 mile from a straight coast line. The lamp makes 4 revolutions per minute. How quickly is the light sweeping along the coastline when the leading edge is at a point 3 miles from the point directly across from the lighthouse?

So: d?/dt = 8?/min
And we want dx/dt when x = 3

Here is my preliminary sketch.

Then, tan? = x/1 and x = tan?
I differentiated with respect to time and found dx/dt = sec[sup:aqclsz7j]2[/sup:aqclsz7j]? (d?/dt) which is the same as 1/cos[sup:aqclsz7j]2[/sup:aqclsz7j]? * (d?/dt).

Now, this is where I'm stuck -- finding ?. I tried using arctan and came up with dx/dt = 251.3274...

When I tried the Pythagorean theorem to find the hypotenuse, I got ?10, which means that cos? = 1/?10. So for the denominator cos[sup:aqclsz7j]2[/sup:aqclsz7j]?, we have (1/?10)[sup:aqclsz7j]2[/sup:aqclsz7j].
Continuing the final formula would be 1/(1/?10)[sup:aqclsz7j]2[/sup:aqclsz7j] * (8?/min) = 251.3274....

Am I missing something here? That number just seems way too high.

Thanks for any help you could offer. HW due in the morning!

 

[soroban]

Hello, jtw2e2!

A lighthouse sits 1 mile from a straight coast line.
The lamp makes 4 revolutions per minute.
How quickly is the light sweeping along the coastline when the leading edge
is at a point 3 miles from the point directly across from the lighthouse?

So: d?/dt = 8?/min
And we want dx/dt when x = 3

Then, tan? = x/1 and x = tan?
I differentiated with respect to time and found dx/dt = sec[sup:33980xnm]2[/sup:33980xnm]? (d?/dt)

All this is correct . . . but why fool around with trig identities?

\(\displaystyle \text{When }x = 3\text{, we have a right triangle with: }\;\begin{Bmatrix}opp &=& 3 \\ adj &=& 1 \\ hyp &=& \sqrt{10}\end{Bmatrix}\)
. . \(\displaystyle \text{Then: }\:\sec\theta \:=\:\frac{\sqrt{10}}{1} \:=\:\sqrt{10}\)

\(\displaystyle \text{Then we have: }\:\frac{dx}{dt} \:=\:\left(\sqrt{10}\right)^2\cdot 8\pi \:=\:80\pi\text{ miles/minute}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{Since }\frac{dx}{dt} \:=\:\sec^2\!\theta\,\frac{d\theta}{dt}\,\text{ and }\,\frac{d\theta}{dt}\,\text{ is a constant }k,\)

. . \(\displaystyle \text{we have: }\:\frac{dx}{dt} \:=\:k\sec^2\theta\)


\(\displaystyle \text{Note that: }\:\lim_{\theta\to\frac{\pi}{2}} \sec\theta \:=\:\infty\)

. . \(\displaystyle \text{Hence: }\;\lim_{\theta\to\frac{\pi}{2}} \frac{dx}{dt} \;=\;\lim_{\theta\to\frac{\pi}{2}} k\sec^2\theta \:=\:\infty\)

\(\displaystyle \text{That is, as the angle approaches }90^o\text{, the light-beam is approaching }in\!finite\text{ speed.}\)


\(\displaystyle \text{It is not surprising that, somewhere along the coastline, the speed is about 15,000 mph.}\)

 

[jtw2e2]

All this is correct . . . but why fool around with trig identities?

Mainly because I'm a moron and didn't know there was any other way to do it. :lol: I can repeat processes but after 30 Related Rates problems I still lack the ability to "know" what to do next.

Thanks for showing an easier way and explaining about the limit as theta --> infinity.

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ \frac{d\theta}{dt} \ = \ \frac{8\pi miles}{min}, \ find \ \frac{dx}{dt} \ when \ x \ = \ 3.\)

\(\displaystyle x \ = \ tan(\theta), \ hence \ \frac{dx}{dt} \ = \ sec^{2}(\theta)\frac{d\theta}{dt} \ (Implicit \ differentation \ with \ respect \ to \ t)\)

\(\displaystyle Now, \ when \ x \ = \ 3, \ \theta \ = \ arctan(3), \ ergo \ we \ have \ \frac{dx}{dt} \ = \ sec^{2}[arctan(3)]\frac{d\theta}{dt},\)

\(\displaystyle Ergo, \ \frac{dx}{dt} \ = \ (10)\bigg(\frac{8 \pi miles}{min}\bigg) \ = \ \bigg(\frac{80 \pi miles}{min}\bigg).\)