Related Rate - Rate of The Change in Distance of Two Ships

[Astrum]

Homework Statement

Ship A is 100 km west of ship B. Ship A is moving at 35 km/h (west), ship B is moving at 25 km/h (north). Find the rate of change of the distance between the two ships after 4 hours.

Homework Equations

The Attempt at a Solution

the distance between the two ships is shown by

$a^{2}+b^{2}=c^{2}$

Am I right to assume:

$[\frac{ds_{a}}{dt}]^{2}+[\frac{ds_{b}}{dt}]^{2}=[\frac{ds_{c}}{dt}]^{2}$

So, I need to find the rate of change of C? After 4 hours, the ships are 260 km apart.

I need a helping hint.

 

[Dick]

Homework Statement

Ship A is 100 km west of ship B. Ship A is moving at 35 km/h (west), ship B is moving at 25 km/h (north). Find the rate of change of the distance between the two ships after 4 hours.

Homework Equations

The Attempt at a Solution

the distance between the two ships is shown by

$a^{2}+b^{2}=c^{2}$

Am I right to assume:

$[\frac{ds_{a}}{dt}]^{2}+[\frac{ds_{b}}{dt}]^{2}=[\frac{ds_{c}}{dt}]^{2}$

So, I need to find the rate of change of C? After 4 hours, the ships are 260 km apart.

I need a helping hint.

Can you write the distance apart as a function of the time t? Your derivative formula isn't correct.

 

[Astrum]

Can you write the distance apart as a function of the time t? Your derivative formula isn't correct.

You mean like d=vt? I'm highly confused. I'm not sure how to relate the total distance C with d=vt.

 

[Dick]

You mean like d=vt? I'm highly confused. I'm not sure how to relate the total distance C with d=vt.

Pick coordinates. Suppose A is at (0,0) at time t=0 and B is at (100,0). Where are they at another time? You figured out the distance at t=4. Just do it for any t.

 

[Astrum]

at time t, they're at $d_{a}=35t$

$d_{b}=25t$.

 

[Dick]

at time t, they're at $d_{a}=35t$

$d_{b}=25t$.

You can't describe locations in the plane with a single number. Use coordinates.

 

[Astrum]

You can't describe locations in the plane with a single number. Use coordinates.

Ok, I'm not following. Let me try another example.

A pile of gravel is being poured from a belt at a rate of 20 m cubed per min. The base will always be half the height. How fast the the pile changing when h=10m?

So, you take the equation for the volume of a cone. take the derivative with respect to h. And we also have the change in volume with respect to time. How do we relate all this to time?

We have $\frac{dV}{dh}=\frac{1}{4}\pi h^{2}$

And: $\frac{dV}{dt}=20m^{3}/s$

Now what? We have these two derivatives. This is the part I don't understand.

 

[Dick]

Ok, I'm not following. Let me try another example.

A pile of gravel is being poured from a belt at a rate of 20 m cubed per min. The base will always be half the height. How fast the the pile changing when h=10m?

So, you take the equation for the volume of a cone. take the derivative with respect to h. And we also have the change in volume with respect to time. How do we relate all this to time?

We have $\frac{dV}{dh}=\frac{1}{4}\pi h^{2}$

And: $\frac{dV}{dt}=20m^{3}/s$

Now what? We have these two derivatives. This is the part I don't understand.

Ok. How about using $\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}$? That's the chain rule. You can use the same approach for the first one if you clearly define what a, b and c are.

 

[symbolipoint]

Assume at the start that ship B is at a reference point.
At any time t, let a be distance of ship A from the reference point, and let b equal the distance of ship B from the reference point.

$\[<br /> a = 35t + 100,\quad b = 25t<br /> \]$

Now, let h be the distance between the two ships at any time t, so that, by pythagorean theorem,
$$\[<br /> \begin{array}{l}<br /> h^2 = a^2 + b^2 \\ <br /> h^2 = \left( {35t + 100} \right)^2 + \left( {25t} \right)^2 \\ <br /> \end{array}<br /> \]$$
and using suitable steps,

$$\[<br /> h = 5\sqrt {74t^2 + 280t + 400} <br /> \]$$


You should be able to understand, and then continue and finish the exercise from this.

 

[Astrum]

Ok. How about using $\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}$? That's the chain rule. You can use the same approach for the first one if you clearly define what a, b and c are.

$\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}$ = $\frac{1}{4}\pi h^{2}\frac{dh}{dt}$ = 20

At this point we plug in 10 for h? so that means $\frac{dh}{dt}$=.255?

 

[Dick]

$\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}$ = $\frac{1}{4}\pi h^{2}\frac{dh}{dt}$ = 20

At this point we plug in 10 for h? so that means $\frac{dh}{dt}$=.255?

Sure.

 

[Astrum]

Sure.

So that begs the question, WHY is this correct? I can't find anything to explain it clearly. I don't know why you just plug in dh/dt

 

[symbolipoint]

Astrum, what are you asking in post #7? How fast is height, h changing when it is 10 meters?

Let me try.

b is base, r is radius, given base is half of height h. Radius is half of b.
$$\[<br /> \begin{array}{l}<br /> {\rm }b = {\textstyle{1 \over 2}}h \\ <br /> {\rm }r = {\textstyle{1 \over 2}}b \\ <br /> \end{array}<br /> \]<br />$$

$$\[<br /> V = \frac{\pi }{{3 \cdot 8}}h^3 <br /> \]$$
V and h are functions of time. V is a function of h, but both of them are functions of time. That is why you use Chain Rule when you find derivative of the right-hand side.

note: may need to check my V result carefully in case arithmetic mistakes; but I believe they're good.
EDIT: Now the formatting in the post is not working. I'll try in the next post.

 

[symbolipoint]

base b is half of height h,
radius is half of b,

$$\[<br /> \begin{array}{l}<br /> b = \frac{1}{2}h \\ <br /> r = \frac{1}{2}b \\ <br /> \end{array}<br /> \]$$

$$\[<br /> V = \frac{\pi }{{3 \cdot 8}}h^3 <br /> \]$$

 

[Astrum]

Yes. I'm just confused as to how you can get dh/dt in there. My confusion comes form not understand implicit differentiation 100%.

 

[symbolipoint]

Yes. I'm just confused as to how you can get dh/dt in there. My confusion comes form not understand implicit differentiation 100%.

Implicit differentiation is unnecessary. You can start (or continue) with the formula for volume of the cone, and this is a function of height, h. You need a bit of substitution as I showed to put the expression in the form as I have.

THEN,

1. Differentiate V with regard to t,
2. Use Chain Rule to differentiate h with regard to t, because h is a function of t.

 

[Dick]

Yes. I'm just confused as to how you can get dh/dt in there. My confusion comes form not understand implicit differentiation 100%.

Like I was saying in post 8, implicit differentiation is really just using the chain rule. Do you understand the chain rule?

 

[symbolipoint]

Implicit differentiation is unnecessary. You can start (or continue) with the formula for volume of the cone, and this is a function of height, h. You need a bit of substitution as I showed to put the expression in the form as I have.

THEN,

1. Differentiate V with regard to t,
2. Use Chain Rule to differentiate h with regard to t, because h is a function of t.

I am possibly unclear about what is and is not implicit differentiation. I have not reviewed it specifically, so I have learned (or relearned) how to find derivative without deciding if I am doing so "implicitly" or not.

Simply, V and h both are function of t; so one differentiates both sides of the formula with respect to t.