MHB Kaska's question at Yahoo Answers involving related rates

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Related rates
AI Thread Summary
A lighthouse is located 170 feet from a shoreline, with a spotlight revolving at 12 revolutions per minute. The problem involves calculating the rate at which the spotlight moves along the shoreline when it is 11 feet from the closest point to the lighthouse. The correct formula derived for the speed of the spot is dx/dt = (ω/y)(y² + x²), where ω is the angular velocity and y is the fixed distance from the lighthouse. Substituting the values yields a result that aligns with the initial answer, confirming the calculation's accuracy without needing trigonometric functions. Understanding the relationship between the variables is crucial for solving related rates problems effectively.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Calculus related rates question, Help please!?

A lighthouse is fixed 170 feet from a straight shoreline. A spotlight revolves at a rate of 12 revolutions per minute, (24 rad/min ), shining a spot along the shoreline as it spins. At what rate is the spot moving when it is along the shoreline 11 feet from the shoreline point closest to the lighthouse?

I got the answer: 4080π(sec(0.0646)^2) but it seems to be wrong, I don't understand what I am doing wrong, any help?

Here is a link to the question:

Calculus related rates question, Help please!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Kaska,

Let's generalize a bit and derive a formula we can then plug our data into.

The first thing I would do is draw a diagram:

View attachment 618

As we can see, we may state:

$\displaystyle \tan(\theta)=\frac{x}{y}$

Now, let's differentiate with respect to time $t$, bearing in mind that while $x$ is a function of $t$, $y$ is a constant.

$\displaystyle \sec^2(\theta)\cdot\frac{d\theta}{dt}=\frac{1}{y} \cdot\frac{dx}{dt}$

Since we are being asked to find the speed of the spot, whose position is $x$, we want to solve for $\dfrac{dx}{dt}$:

$\displaystyle \frac{dx}{dt}=y\sec^2(\theta) \cdot\frac{d\theta}{dt}$

Let's let the angular velocity be given by:

$\omega=\dfrac{d\theta}{dt}$

and from the diagram and the Pythagorean theorem, we find:

$\displaystyle \sec^2(\theta)=\frac{x^2+y^2}{y^2}$

Hence, we have:

$\displaystyle \frac{dx}{dt}=\frac{\omega}{y}(y^2+x^2)$

Now we may plug in the given data:

$\displaystyle \omega=24\pi\frac{1}{\text{min}},\,y=170\text{ ft},\,x=11\text{ ft}$

$\displaystyle \frac{dx}{dt}=\frac{24\pi}{170}(170^2+11^2)\,\frac{\text{ft}}{\text{min}}=\frac{348252\pi}{85}\, \frac{\text{ft}}{\text{min}}$

This is equivalent to the answer you obtained (accounting for rounding), however we have done away with the need to use a trig. function.
 

Attachments

  • kaska.jpg
    kaska.jpg
    2.4 KB · Views: 140
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top