Implicit differentiation

[Shadow]

Hey all,

 

we're reviewing trigonometry in our math class, and worked the following problem:

 

A boat is moving at a steady speed towards the base of a cliff. At one point it is seen under an angle of 20° from the top of the cliff. Ten time units later it is seen under an angle of 30°. What is the velocity [math]v[/math] of the boat?

 

 

We can solve this using trigonometry; if we denote the horizontal black strip [math]x[/math], the horizontal red strip [math]y[/math] and the whole horizontal strip (that is both the red and the black one) [math]z = x + y[/math], we have

 

[math]z=30 \cdot \tan(70^{\circ})[/math]

[math]x = 30 \cdot \tan(60^{\circ}) = 30 \cdot \sqrt{3}[/math]

[math]y = z - x = 30 \cdot ( \tan(70^{\circ}) - \sqrt{3})[/math]

[math]v = \frac{y}{10} = 3 \cdot (\tan(70^{\circ}) - \sqrt{3})[/math]

 

Since I'd recently been looking at implicit differentiation and related rates problems I thought it'd be interesting to try and solve this problem using calculus, but the result I arrive at is completely different. Here is how I work the problem:

 

[math]\Phi(t)[/math] is the function giving the angle between the cliff and the line connecting the boat and the top of the cliff at a certain time.

[math]x(t)[/math] is the function giving the distance of the boat from the base of the cliff at a given time.

 

We also know that [math]\frac{d\Phi}{dt}= \frac{60 - 70}{10} = -1[/math]

 

[math]\tan(\Phi(t)) = \frac{x(t)}{30}[/math]

[math]\Phi(t) = \tan^{-1}(\frac{x(t)}{30})[/math]

[math] \Phi' = \frac{\frac{x'}{30}}{\frac{x^2}{900}+1} = \frac{30x'}{x^2+900}[/math].

 

We are interested in [math]x'[/math], we know that [math]\Phi' = -1[/math], so we still need to find [math]x[/math], which is where I think I made the error, since I don't know at what time I should calculate [math]x[/math]. I find [math]x[/math] by calculating [math]\tan(60^{\circ}) = \frac{x}{30}[/math], thus [math] x = 30\sqrt{3}[/math]. But, after plugging this value into the equation above, I get [math]x' = -120[/math], which is incorrect. So...where's the problem?

 

Thanks.

[Shadow]

No one?

[timo]

From quickly looking over it:

1) You forgot the units.

2) How do you know that [math]\frac{d\Phi}{dt} = -1[/math] (in whatever units)? Hint: it's wrong.

3) What is [math]x'[/math] supposed to be? A slash usually denotes the derivative with respect to a variable called "x". You probably meant the derivative with respect to time. In principle, that would often be written as [math]\dot x[/math]. But I think you should write out [math]\frac{dx}{dt}[/math] explicitly, for now.

Edited February 11, 2011 by timo

[Shadow]

Hey timo, thanks for the response. I deliberately left the units out, just because I was lazy; I'm aware that units are very important in physics problems, but I didn't see any major role they would play in the mathematical perspective, so I just assumed the unit of distance to be "unit of distance" (ie. the cliff is 30 distance units high), the unit of time to be "unit of time" (ie. the whole scenario plays out in 10 time units) and the unit of velocity to be "unit of distance over unit of time" (ie. the velocity of the boat was circa three units of distance over units of time), since this was a mathematical problem, not a physics one. But for completeness, I believe the height was given in meters and the time in seconds, which would make the final velocity in meters per second.

 

I thought [math]\frac{d\Phi}{dt} = -1 deg/s[/math] because the angle changes by ten degrees in ten seconds, although I'm still not comfortable enough with calculus to be sure that it can be calculated in this way; could you outline the difference between [math]\frac{d\Phi}{dt}[/math] and what I did, ie. [math]\frac{\Phi_1 - \Phi_0}{t_1 - t_0}[/math]?

 

A slash usually denotes the derivative with respect to a variable called "x".

 

Really? Because I'm pretty sure I've seen plenty of examples being written something like [math]y = 3x+2[/math] and then [math]y' = 3[/math], which according to you should mean differentiation with respect to a variable called "y".

Anyhow, [math]x'[/math] was supposed to denote velocity.

[timo]

I thought [math]\frac{d\Phi}{dt} = -1 deg/s[/math] because the angle changes by ten degrees in ten seconds, although I'm still not comfortable enough with calculus to be sure that it can be calculated in this way; could you outline the difference between [math]\frac{d\Phi}{dt}[/math] and what I did, ie. [math]\frac{\Phi_1 - \Phi_0}{t_1 - t_0}[/math]?

That is actually not the problem. The problem is that you assume that if the angle changed by A degrees over ten seconds, then in the next ten seconds it will also change by A degrees. Or in other words: you assume that the rate by which the angle changes is constant over time. If that was the case, then what you did would have been ok. But it is not the case. The velocity is constant, but the rate by which the angle changes is not.

 

Because I'm pretty sure I've seen plenty of examples being written something like [math]y = 3x+2[/math] and then [math]y' = 3[/math], which according to you should mean differentiation with respect to a variable called "y".

No. What I meant is that the slash usually means a derivative with respect to x. That's what makes x' particularly strange. So according to me, [math] y' = \frac{dy}{dx} = \frac{d(3x+2)}{dx} = 3[/math]. Taking a derivative with respect to <something> means that <something> is the variable in the "denominator".

[Shadow]

Thanks timo, that makes sense. Okay, so now I have to express the angle in terms of time, right? The only way I can think of doing that would be [math]\Phi(t) =\tan^{-1}(\tan(70^{\circ}) - \frac{v \cdot t}{30})[/math] which means that [math] \frac{d \Phi}{dt} = -\frac{v}{30(\tan(70^{\circ}) - \frac{v \cdot t}{30})^2 + 1}[/math]. What next? Is [math] t=10[/math] or [math]t=0[/math]? And how do I find [math]\frac {d\Phi}{dt}[/math]?

Edited February 12, 2011 by Shadow

[Shadow]

Oops, made a mistake; [math]\frac{d \Phi}{dt} = -\frac{v}{30(\tan(70^{\circ}) - \frac{v \cdot t}{30})^2 + 30}[/math]

[timo]

It's not clear to me what you are trying to do. Obviously, if you know v, then you know [math]\Phi (t)[/math] and [math] \frac{d\Phi}{dt}(t)[/math] - by plugging in v. Perhaps this comment helps: The question is relatively easy to answer because you know that v, which is the same as dx/dt, is constant. Phi and its time-derivative [math]\frac{d\Phi}{dt}[/math] are not constant. In principle, you could write down something like [math]10^\circ = \int_{0}^{10} \! \frac{d\Phi}{dt} \, {\rm d}t[/math], plug in the term for [math]\frac{d\Phi}{dt}[/math] on the right-hand side, integrate, and solve for v. That would be the formal "for impressing first semester physics students" way to solve the problem.

Edited February 15, 2011 by timo

[Shadow]

Except that I don't know [math]v[/math]; that's what I'm supposed find out in the problem. I'm aware of the fact that I can calculate it easily using simple trigonometry, but the whole point was to try and arrive at the answer using calculus, if possible; are you saying it's not?

 

Also, I just used Mathematica to evaluate the integral and solve for v, since I haven't got as far as integration in my studies and was curious to find out if this method works, and the result differs from the one I get if I use geometry.

Edited February 15, 2011 by Shadow

[timo]

You already used trigonometry when constructing [math]\Phi(t)[/math], which is fine of course. Perhaps you should try to follow the formal Ansatz with the integral that I wrote down and see what happens.

[Shadow]

Like I said, integral's are beyond my current knowledge (not for long though!) so if they're necessary to solve this problem then I'll just leave it be come back to it after I've learned integration. But out of curiosity I did try to evaluate the integral and solve the equation using Mathematica, and the result I get for [math]v[/math] is not the correct answer.

Edited February 15, 2011 by Shadow