Locus of Curves: Solving 4.1.2 with 4.1.10 & 4.1.17

[bugatti79]

Hi Folks,

I am struggling to see how eqn 4.1.17 is determined using eqn 4.1.10 for problem 4.1.2. It is not clear to me what $$\frac{d\phi}{dt}$$ is.

I have inserted the eqns I think might be useful into the one jpeg. Any ideas? Sorry I can't make the picture any clearer given the size limit.

Thanks

 

[jvicens]

Hi there,

Thank you for your question. I can understand your confusion with determining eqn 4.1.17 using eqn 4.1.10 for problem 4.1.2. Let me try to explain it to you.

First, let's review eqn 4.1.10, which is the equation for the rate of change of a variable, in this case, let's call it x. This equation states that the rate of change of x is equal to the derivative of x with respect to time, or \frac{dx}{dt}.

Now, let's look at the problem 4.1.2. From the given information, we can see that x is related to another variable, let's call it \phi, through eqn 4.1.12. This means that x is a function of \phi, or x = f(\phi).

To determine eqn 4.1.17, we need to use the chain rule, which states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In this case, the composite function is x = f(\phi), and the outer function is \phi, while the inner function is t. So, using the chain rule, we can write:

\frac{dx}{dt} = \frac{dx}{d\phi} \frac{d\phi}{dt}

Substituting eqn 4.1.10 for \frac{dx}{dt} and eqn 4.1.12 for x, we get:

\frac{d\phi}{dt} = \frac{d}{dt} (f(\phi)) = \frac{dx}{d\phi} \frac{d\phi}{dt}

Rearranging this equation, we get:

\frac{d\phi}{dt} = \frac{1}{\frac{dx}{d\phi}} \frac{dx}{dt}

Now, let's look at eqn 4.1.17. This equation is the same as eqn 4.1.10, but with x replaced by \phi. This means that it represents the rate of change of \phi with respect to time, or \frac{d\phi}{dt}. So, by substituting our previous equation for \frac{d\phi}{dt} into eqn