Related Rates and Exponential Growth

[ardentmed]

Hey guys,

I've two more word problem questions this time.

Question:

So for the first one, I know that
y=T-Ts where Ts = 20.

Thus, if T(0) = 90, then T'(70) = -1

T'(t) = k (T-Ts)
k= -1/50 (via substitution)

Now, we must find y.

y'(t) = ky and y(t) = T(t) - Ts
y(0) = 90- 20
y(0) = 70


Now, using the exponential growth formula, we get:

ln(50)/70 = -1/50 * t

Thus,
t=16.82 minutes.


As for the second question, we know that we can compute x and y by multiplying the speeds by 4 since t = 4 at 4PM.

s = √( 180^2 + 100^2) which is 205.912603.

Thus, now we can solve for ds/dt via the Pythagorean formula.

ds/dt = 25.739 km/hr.

Thanks in advance.

 

[ardentmed]

Can anyone help me out with this question?

Thanks in advance.

 

[ardentmed]

Is the derivative of the expression equal to:

k(T-Ts)?

Am I on the right track? I know that Y=T-Ts, but I'm lost as to how that becomes the first formula.

Does anyone have any advice on how to do this?

Thanks in advance.

 

[Dethrone]

Since no one is answering this question, I will try and tackle number two. I may be wrong, though.

We have $D = x^2 + y^2$
differentiating w.r.t time, we get:

$$\frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{D}$$

So we want x and y at 4P.M, which is easy to calculate:
Ship A is 500km away at noon, so 4 hours later while traveling at 45km/h, we get 500 - 180 = 320 km away = $x$. Next, Ship B sails at 25km/h north for 4 hours, which gives us 100 km = $y$. This forms a right angled triangle, where x = 320 and y = 180, this gives us $20\sqrt{281}$ as $D$.

Now, we still need $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at 4 pm, which is given in the question, but we have to be careful. $\frac{dx}{dt}$ is -45, because it is decreasing the size of the triangle, or the distance between the ships by traveling East. On the other hand, $\frac{dy}{dt}$ is +25 because it is increasing the size of the triangle, or the distance between the ships. Putting this all together, we get:

$$\frac{dD}{dt}=\frac{320(-45)+100(25)}{20\sqrt{281}} = -35.49 km/h$$
This makes sense with relation to this question, because Ship A is traveling towards Ship B faster than it is moving away from it. Thus, the distance between the ships are actually decreasing at the time 4pm. Of course, this holds true only at the exactly 4pm.

As I said before, I'm not completely sure if this is correct; it would be nice if another member could confirm.

 

[MarkFL]

1.) Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

$$\frac{dT}{dt}=-k(T-M)$$ where $$T(0)=T_0,\,0<k\in\mathbb{R}$$ and $$T>M$$.

The ODE is separable and may be written:

$$\frac{1}{T-M}\,dT=-k\,dt$$

Integrating, using the boundaries, and dummy variables of integration, we find:

$$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv$$

$$\ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt$$

$$t=\frac{1}{k}\ln\left(\frac{T_0-M}{T(t)-M} \right)$$

Now, let's let $$T_1=T(t_1)=70$$ (we will assume temperatures are in degrees Celsius). Now, if we know the rate at which the object is cooling at this time, we may call this rate $R_1$ and state from Newton's law:

$$-k(T_1-M)=-R_1\implies \frac{1}{k}=\frac{T_1-M}{R_1}$$

And so we may now state:

$$t_1=\frac{T_1-M}{R_1}\ln\left(\frac{T_0-M}{T_1-M} \right)$$

Now, we have a formula into which we may plug the given data:

$$T_1=70,\,M=20,\,R_1=1,\,T_0=90$$

and we obtain:

$$t_1=\frac{70-20}{1}\ln\left(\frac{90-20}{70-20} \right)=50\ln\left(\frac{7}{5}\right)\approx16.82$$

So, I agree with your result. (Yes)

 

[ardentmed]

Since no one is answering this question, I will try and tackle number two. I may be wrong, though.

We have $D = x^2 + y^2$
differentiating w.r.t time, we get:

$$\frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{D}$$

So we want x and y at 4P.M, which is easy to calculate:
Ship A is 500km away at noon, so 4 hours later while traveling at 45km/h, we get 500 - 180 = 320 km away = $x$. Next, Ship B sails at 25km/h north for 4 hours, which gives us 100 km = $y$. This forms a right angled triangle, where x = 320 and y = 180, this gives us $20\sqrt{281}$ as $D$.

Now, we still need $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at 4 pm, which is given in the question, but we have to be careful. $\frac{dx}{dt}$ is -45, because it is decreasing the size of the triangle, or the distance between the ships by traveling East. On the other hand, $\frac{dy}{dt}$ is +25 because it is increasing the size of the triangle, or the distance between the ships. Putting this all together, we get:

$$\frac{dD}{dt}=\frac{320(-45)+100(25)}{20\sqrt{281}} = -35.49 km/h$$
This makes sense with relation to this question, because Ship A is traveling towards Ship B faster than it is moving away from it. Thus, the distance between the ships are actually decreasing at the time 4pm. Of course, this holds true only at the exactly 4pm.

As I said before, I'm not completely sure if this is correct; it would be nice if another member could confirm.

Thanks for the insightful response. Just out of curiosity, where did the y=180 come from?

Also, isn't the answer positive since the question is asking for speed as a scalar quantity, hence "how fast?" I may be mistaken here.

Thanks a ton for the help.

 

[Dethrone]

Thanks for the insightful response. Just out of curiosity, where did the y=180 come from?

I've explained this above, and I never used that number directly.

Also, isn't the answer positive since the question is asking for speed as a scalar quantity, hence "how fast?" I may be mistaken here.

Thanks a ton for the help.

I explained it to. What they want is velocity, and you can represent direction in 2D by a positive or negative quantity. For example, acceleration is negative, and that's possible. The question also never asked for "speed", you were given "velocity": they gave you a magnitude (45km) and a direction (east). The negative answer that I arrived at makes sense. If you visualize the situation, the ship (ship A) moving right is moving much faster than the ship (ship B) moving upwards. The distance between them is therefore decreasing, or in other words, negative. Furthermore, at 4pm, ship A has not passed directly below ship B yet, and until that point, the relative distance between them will always be decreasing. If you need further clarification, let me know.

P.S: Do you not have an answer key for these questions?