# Thread: Related Rates Issue: Cannot locate the proper formula

1. ## Related Rates Issue: Cannot locate the proper formula

On a morning of a day when the sin will pass directly overhead, the shadow of an 80-ft building on a level ground is 60 ft long. At the moment in question, the angle $\displaystyle \theta$ the sun makes with the ground is increasing at the rate of $\displaystyle 0.27 \deg$/min. At what rate is the shadow decreasing? (Remember to use radians and express your answer in inches per minute, to the nearest degree.)

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

It's basically a triangle. y is 80, x is unknown, as is r (or z, whichever way you want to look at it).

The givens are that $\displaystyle \frac{d\theta}{dt}=0.27\deg/min$

I have to find $\displaystyle \frac{dx}{dt}$ when x = 60. It is going to be negative because the shadow length is decreasing.

My problem is that I cannot find the proper formula to go along with this equation. I've tried to use the area of a triangle, but since I don't know $\displaystyle \frac{dA}{dt}$, that gets me nowhere. I think that using trigonometric ratios is the way to go, but I keep getting mixed numbers, ranging from decimals down into the thousandths, all the way up to the thousands. The answer (according to the book) is 7.1 in/min.

Solving the problem is not necessary. If you can help me find the right formula, I can take it from there.

I thank you for your time if you chose to post here.

2. Originally Posted by atac777
On a morning of a day when the sin will pass directly overhead, the shadow of an 80-ft building on a level ground is 60 ft long. At the moment in question, the angle $\displaystyle \theta$ the sun makes with the ground is increasing at the rate of $\displaystyle 0.27 \deg$/min. At what rate is the shadow decreasing? (Remember to use radians and express your answer in inches per minute, to the nearest degree.)

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

It's basically a triangle. y is 80, x is unknown, as is r (or z, whichever way you want to look at it).

The givens are that $\displaystyle \frac{d\theta}{dt}=0.27\deg/min$

I have to find $\displaystyle \frac{dx}{dt}$ when x = 60. It is going to be negative because the shadow length is decreasing.

My problem is that I cannot find the proper formula to go along with this equation. I've tried to use the area of a triangle, but since I don't know $\displaystyle \frac{dA}{dt}$, that gets me nowhere. I think that using trigonometric ratios is the way to go, but I keep getting mixed numbers, ranging from decimals down into the thousandths, all the way up to the thousands. The answer (according to the book) is 7.1 in/min.

Solving the problem is not necessary. If you can help me find the right formula, I can take it from there.

I thank you for your time if you chose to post here.
The formula you need, where x is the lenth of the shadow, is:

$\displaystyle \tan\theta = \frac{80}{x}$

$\displaystyle \theta = \arctan\frac{80}{x}$

Differentiate that and you will have an equation in three variables: the rate of angular change, the length of the shadow, and the rate of change of the length of the shadow.

As a general rule with these types of problems, if you are looking at a two linear rates of change, use the Pythagorean theorum; if the problem involves an angular rate of change, use sin ratios.

3. Originally Posted by sinewave85
The formula you need, where x is the lenth of the shadow, is:

$\displaystyle \tan\theta = \frac{80}{x}$

$\displaystyle \theta = \arctan\frac{80}{x}$

Differentiate that and you will have an equation in three variables: the rate of angular change, the length of the shadow, and the rate of change of the length of the shadow.

As a general rule with these types of problems, if you are looking at a two linear rates of change, use the Pythagorean theorum; if the problem involves an angular rate of change, use sin ratios.
I can't differentiate inverse trigonometric functions yet; we haven't covered them in class (that's on test number 4, which will be late next month)

Here's what I'm doing

$\displaystyle \frac{d}{dt}\tan \theta = 80x^-1\frac{d}{dt}$

$\displaystyle \frac{d\theta}{dt}\sec^2\theta = -80x^-2\frac{dx}{dt}$

I then drop the $\displaystyle x^-2$ to the denominator, divide both sides by $\displaystyle \frac{-80}{x^2}$, plug in the numbers, and solve for $\displaystyle \frac{dx}{dt}$

However, when I do that, i end up with -.003375 as my answer. I can't seem to figure out what I'm doing wrong.

4. Originally Posted by atac777
I can't differentiate inverse trigonometric functions yet; we haven't covered them in class (that's on test number 4, which will be late next month)

Here's what I'm doing

$\displaystyle \frac{d}{dt}\tan \theta = 80x^-1\frac{d}{dt}$

$\displaystyle \frac{d\theta}{dt}\sec^2\theta = -80x^-2\frac{dx}{dt}$

I then drop the $\displaystyle x^-2$ to the denominator, divide both sides by $\displaystyle \frac{-80}{x^2}$, plug in the numbers, and solve for [tex]\frac{dx}{dt}[\math]

However, when I do that, i end up with -.003375 as my answer. I can't seem to figure out what I'm doing wrong.
Hmm, that is a puzle. I wonder why you would be given this problem without knowing how to differentiate inverse fuctions. Give me a second to think through how to do this without having to differentiate an inverse function.

5. Originally Posted by sinewave85
Hmm, that is a puzle. I wonder why you would be given this problem without knowing how to differentiate inverse fuctions. Give me a second to think through how to do this without having to differentiate an inverse function.
If you can't figure it out, let me know and I'll contact my professor about the problem.

6. Originally Posted by atac777
If you can't figure it out, let me know and I'll contact my professor about the problem.
It should work. The thing is that without taking the arctan, you have an extra variable -- the angle. We can describe the angle by solving $\displaystyle \theta = \arctan\frac{80}{60}$ and then using that angle for $\displaystyle sec^{2}{\theta}$. I don't have a calculator with me right now, but I am assuming you do. Also, are you making sure to convert the degrees to rads before you calculate?

7. Originally Posted by sinewave85
Also, are you making sure to convert the degrees to rads before you calculate?
0.27 degrees = $\displaystyle \frac{3\pi}{2000}$ rad?

8. Originally Posted by atac777
0.27 degrees = $\displaystyle \frac{3\pi}{2000}$ rad?
Yes, that is right.

9. Yes, it works. I ran it through Google. The thing to keep in mind, and that I should have pointed out earlier, is that your calculator is going to give you an answer that is in feet per minute which you need to convert to inches per minute. The calculated answer is about 0.5890486225, which multiplied by 12 gives you 7.068583471, which rounds to your answer of 7.1.

10. Originally Posted by sinewave85
Yes, that is right.
I keep getting either -2.78 in/min, or -0.000058 in/min.

11. Originally Posted by atac777
I keep getting either -2.78 in/min, or -0.000058 in/min.
Ok, let me type it out with all of the variables filled in.

12. Originally Posted by atac777
I can't differentiate inverse trigonometric functions yet; we haven't covered them in class (that's on test number 4, which will be late next month)

Here's what I'm doing

$\displaystyle \frac{d}{dt}\tan \theta = 80x^-1\frac{d}{dt}$

$\displaystyle \frac{d\theta}{dt}\sec^2\theta = -80x^-2\frac{dx}{dt}$

I then drop the $\displaystyle x^-2$ to the denominator, divide both sides by $\displaystyle \frac{-80}{x^2}$, plug in the numbers, and solve for $\displaystyle \frac{dx}{dt}$

However, when I do that, i end up with -.003375 as my answer. I can't seem to figure out what I'm doing wrong.
$\displaystyle \frac{d\theta}{dt} = 0.004712389, x = 60, \theta = 0.927295218$

$\displaystyle \frac{dx}{dt} = \frac{d\theta}{dt}\sec^{2}{\theta}\left(\frac{-x^{2}}{80}\right)$

$\displaystyle \frac{dx}{dt} = (0.004712359)\sec^{2}{(0.927295218)}\left(\frac{-(60)^{2}}{80}\right)$

$\displaystyle \frac{dx}{dt} = -0.5890486225\mbox{ft/min}\rightarrow\approx-7.1\mbox{in./min}$

13. Originally Posted by sinewave85
$\displaystyle \frac{d\theta}{dt} = 0.004712389, x = 60, \theta = 0.927295218$

$\displaystyle \frac{dx}{dt} = \frac{d\theta}{dt}\sec^{2}{\theta}\left(\frac{-x^{2}}{80}\right)$

$\displaystyle \frac{dx}{dt} = (0.004712359)\sec^{2}{(0.927295218)}\left(\frac{-(60)^{2}}{80}\right)$

$\displaystyle \frac{dx}{dt} = -0.5890486225\mbox{ft/min}\rightarrow\approx-7.1\mbox{in./min}$
Got it! Thank you so much for your help

14. Originally Posted by atac777
Got it! Thank you so much for your help
You are welcome! It was a good refresher for me, as I am only a few units ahead of you in calculus.

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### on a morning of a day when the sun will pass directly overhead the shadow of an 80 ft building

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