Find the Rate of Change of Theta as a function of time

[Bluewolf1986]

Here is my question from my Calculus 1: Differential Calculus Course I need help on:

Suppose that a camera is fixed at (0,0) in the coordinate plane (measured in feet). An actor starts at (10,10) and moves down (negative y direction) at 1 foot per second, and moves right (positive x direction) at 1 foot per second, so that his position at time t is (10+t,10−t).

Let θ be the angle between the positive x direction and the line of sight from the camera to the actor as a function of t. Find the rate of change θ as a function of time t.

(Type ∗ for multiplication; / for division; ∧ for exponentiation. The functions sqrt(x), ln(x), sin(x), etc. are known. Type e and pi for the mathematical constants e and π.)

Here's the steps I have taken so far attempting to use implicit differentiating that turned out to be wrong:

1) dθ/dt= Sec^2(-1(10)-10+t)+(1)(10)-(10-t)/(10+(10+t)^2

2) Cos^2(-10+t)-(10-t)/(10+(10+t)^2

3) Cos^2(-20)/(20+t)^2

4) Cos^2(-20/(400+40t+t^2)

5) Cos^2( -1/20+40t+t^2)

Please help with an initial step or formula to solve this problem. Thank you!

 

[JeffM]

Forget about differentiating until you have defined the function that is to be differentiated.

What is it?

[MATH]f( \theta ) = WHAT?[/MATH]

 

[Bluewolf1986]

The function of theta is not defined, the instructors expect the students to figure that out by the information in the problem. That is exactly why I am having trouble with this problem, because it is expected to be able to contrive a function by somehow "thinking through it". Thank you for your response though, I appreciate your attempt to help me with this conundrum.

 

[JeffM]

I was not being flippant or dismissive. When you actually try to apply math to practical problems, no one provides you with equations or functions. That is where the creativity comes in for applied math. So the approach of your course is excellent.

My point is that differentiable calculus concerns an operation on a differentiable function. You specified a derivative without first specifying the function. I am simply asking what function you believe is the relevant one and why. Without that, we do not know where you went awry (if you did).

It looks as though you defined

[MATH]tan( \theta ) = \dfrac{10 + t}{10 - t}.[/MATH]
Is that what you did?

If so, does differentiating that tangent function give you [MATH]\dfrac{d \theta}{dt}[/MATH]?

To get [MATH]\dfrac{d \theta}{dt}[/MATH], the normal way to proceed is:

[MATH]\theta = f(t) \implies \dfrac{d \theta}{dt} = f'(t).[/MATH]

 

[lev888]

I would start with drawing everything (coordinate plane, the actor at some point (10+t,10−t), etc.
Figure out how angle depends on t, write down the expression - that's your function.

 

[Steven G]

Forget about differentiating until you have defined the function that is to be differentiated.

What is it?

[MATH]f( \theta ) = WHAT?[/MATH]

I never heard of a better response than what you just stated.

 

[Bluewolf1986]

I was not being flippant or dismissive. When you actually try to apply math to practical problems, no one provides you with equations or functions. That is where the creativity comes in for applied math. So the approach of your course is excellent.

My point is that differentiable calculus concerns an operation on a differentiable function. You specified a derivative without first specifying the function. I am simply asking what function you believe is the relevant one and why. Without that, we do not know where you went awry (if you did).

It looks as though you defined

[MATH]tan( \theta ) = \dfrac{10 + t}{10 - t}.[/MATH]
Is that what you did?

If so, does differentiating that tangent function give you [MATH]\dfrac{d \theta}{dt}[/MATH]?

To get [MATH]\dfrac{d \theta}{dt}[/MATH], the normal way to proceed is:

[MATH]\theta = f(t) \implies \dfrac{d \theta}{dt} = f'(t).[/MATH]

It's fine, I know what you meant when you stated that. To answer your question, I did define tan(\theta)=10+t/10-t. I forgot to include that step. I will attempt to find another function that will find the difference in theta over the difference in time when differentiated. I absolutely agree that the approach of the course I'm taking is excellent and that applied math takes a lot of creativity. Some of these problems are easier for me to solve than others, such as this one. However, I always enjoy the learning process either way. Thank you again for your attempt to help me solve this problem, it was very helpful and I genuinely appreciate it.

 

[JeffM]

It's fine, I know what you meant when you stated that. To answer your question, I did define tan(\theta)=10+t/10-t. I forgot to include that step. I will attempt to find another function that will find the difference in theta over the difference in time when differentiated. I absolutely agree that the approach of the course I'm taking is excellent and that applied math takes a lot of creativity. Some of these problems are easier for me to solve than others, such as this one. However, I always enjoy the learning process either way. Thank you again for your attempt to help me solve this problem, it was very helpful and I genuinely appreciate it.

Actually, I feel a bit apologetic about my initial response.

The question seems to be asking you to find

[MATH]\dfrac{d \theta }{dt} = f'(t).[/MATH]
So the straight forward way to proceed is to ask yourself what is

[MATH]\theta = f(t).[/MATH]
But, without explicitly saying so, you proceeded from

[MATH]g( \theta) = f(t), \text { where } g( \theta ) = tan ( \theta).[/MATH]
I was trying to induce the observation that

[MATH]\theta = g^{-1}(g ( \theta )) = g^{-1}(f(t)).[/MATH]
I was too curt.