Related rates problem - kinda stuck on this one

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Homework Statement


Suppose you take a car trip, traveling east along a very long highway, starting at time t 0. Let x t be the number of gallons of gasoline used during the first t hours, and let st be the distance traveled in that time. Because you’re using very cheap gasoline that’s not good for the engine, your fuel efficiency (measured in miles/gallon) decreases as fuel is consumed. The graph of fuel efficiency as a function of gasoline used is shown in the graph below.
upload_2018-8-1_12-48-0.png

You record the total fuel used at various times during the trip and make the graph below, showing amount of fuel used as a function of time.
upload_2018-8-1_12-48-23.png

Using only these two graphs, estimate your instantaneous velocity 15 hours after the start of the trip. Explain how you arrived at that value

Homework Equations


Given in the graphs.
A hint was given not to try and find formulas for the graphs
x t = number of gallons used for the first t hours
s(t) = distance traveled in that time

The Attempt at a Solution


So, for 15 hrs, according to the second graph. we can tell that the overall gallons of fuel used at t = 15 is x =9( I think). From that we can tell that the interval of fuel efficiency that we will use is (30, 25) as the efficiency is 25 at x = 9. However, if x t is the number of gallons used, then our solution gives us 15x, which contrasts with the second graph we're given. Honestly, I'm just not sure how to proceed with the question. All the related rates problems I've done so far have provided functions instead of graphs, and here I can't think of a way to find the relationship between miles per gallon and the instantaneous velocity.
 

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  • #2
Orodruin
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How would you typically find the velocity if you have dustance as a function of time?
 
  • #3
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How would you typically find the velocity if you have dustance as a function of time?
I'm pretty sure you take the derivative of the d(t) function
 
  • #4
Orodruin
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I'm pretty sure you take the derivative of the d(t) function
Right. Or s(t) to follow the notation in your problem. Do you have the distance as a function of time? If not, can you construct such a function using the functions you do have?
 
  • #5
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Right. Or s(t) to follow the notation in your problem. Do you have the distance as a function of time? If not, can you construct such a function using the functions you do have?

s(t) isn't given, but I can construct a ds/dx as a function of time instead. ds/dx in this case is miles/gallon. What I'm confused with, however, is trying to create the s(t function from the ds/dx function. I tried multiplying the ds/dx by t function by the gallons given at that time, but it didn't seem to work
 
  • #6
Orodruin
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You do not need to find s(t), you need to express v = ds/dt. What are the functions you are working with and how can you use them to express ds/dt?

You need not be using ds/dx as a function of t. It is the derivative of s with respect to x. What you need is s as a function of t.
 
  • #7
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You do not need to find s(t), you need to express v = ds/dt. What are the functions you are working with and how can you use them to express ds/dt?

You need not be using ds/dx as a function of t. It is the derivative of s with respect to x. What you need is s as a function of t.
I think I have ds/dt already, by combining the ds/dx function and the x/t function, I graphed the ds/dt function. By checking the value at t = 15, the velocity appears to be 25 mph. Is this correct?
 
  • #8
Orodruin
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I think I have ds/dt already, by combining the ds/dx function and the x/t function, I graphed the ds/dt function. By checking the value at t = 15, the velocity appears to be 25 mph. Is this correct?
This really does not describe what you did. How did you "combine" the functions?
 
  • #9
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This really does not describe what you did. How did you "combine" the functions?
So, if you look at the both the functions, x is a common variable between them, representing the number of gallons. Using this, you can find the miles per gallon at a specific time. For example, if you look at the second graph, the x(gallon) value at t = 15 hours is approximately 9. You can take this x value and see what the miles per gallon value is on the ds/dx function when x = 9. So, this way, you can find the miles per gallon value, in this case 25, at a specific time. I think this is how you combine the functions.
 
  • #10
Orodruin
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So, if you look at the both the functions, x is a common variable between them, representing the number of gallons. Using this, you can find the miles per gallon at a specific time. For example, if you look at the second graph, the x(gallon) value at t = 15 hours is approximately 9. You can take this x value and see what the miles per gallon value is on the ds/dx function when x = 9. So, this way, you can find the miles per gallon value, in this case 25, at a specific time. I think this is how you combine the functions.
No this is incorrect. What you are computing is the value of ds/dx when t = 15, not what ds/dt is. The units do not even make sense as what you are getting has dimension of length/fuel volume (you cannot just assume that the units will change accordingly). Consider the two functions s(x) and x(t). How would you determine the distance travelled as a function of t, i.e., s(t)?
 
  • #11
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No this is incorrect. What you are computing is the value of ds/dx when t = 15, not what ds/dt is. The units do not even make sense as what you are getting has dimension of length/fuel volume (you cannot just assume that the units will change accordingly). Consider the two functions s(x) and x(t). How would you determine the distance travelled as a function of t, i.e., s(t)?
You plug in t for x in s(x), is what I'm assuming
 
  • #12
Delta2
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You plug in t for x in s(x), is what I'm assuming
Not exactly, you plug in the expression for x(t) instead of just x. That is s(x(t)).

But you said you are not allowed to find the functions formulas from the graphs.

Sorry for asking this question but have you been taught the chain rule for derivatives? If yes, what equation can you make using the chain rule that relates the derivatives (with respect to x and/or t) of the functions s(x), x(t)?
 
  • #13
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Not exactly, you plug in the expression for x(t) instead of just x. That is s(x(t)).

But you said you are not allowed to find the functions formulas from the graphs.

Sorry for asking this question but have you been taught the chain rule for derivatives? If yes, what equation can you make using the chain rule that relates the derivatives (with respect to x and/or t) of the functions s(x), x(t)?
Yeah, I know the chain rule for derivatives. If you want to go by that and your previous relation, an equation could be s'(x(t)) * x'(t).
 
  • #14
Delta2
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Yeah, I know the chain rule for derivatives. If you want to go by that and your previous relation, an equation could be s'(x(t)) * x'(t).
An equation has a right and a left side connected by the =. You show us only the right side of the equation. Ok I can complete the left side for you it ll be
(s(x(t)))'=s'(x(t))*x'(t)
Now, (s(x(t)))' which i simply denote as s'(t)

(that might give some ambiguity in the symbolism I admit, the difference between s'(t) and s'(x(t)) is that in first the variable of differentiation is t, while in the second the variable of differentiation is x=x(t))

at t=15 is what we wish to find that is s'(15).

By that equation from chain rule you have to find x'(15) and s'(x(15)). Can you find these two values provided the two diagrams? I think you already found s'(x(15))=25 right? How can you find x'(15)?
 
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  • #15
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So, if you look at the both the functions, x is a common variable between them, representing the number of gallons. Using this, you can find the miles per gallon at a specific time. For example, if you look at the second graph, the x(gallon) value at t = 15 hours is approximately 9. You can take this x value and see what the miles per gallon value is on the ds/dx function when x = 9. So, this way, you can find the miles per gallon value, in this case 25, at a specific time. I think this is how you combine the functions.
Apparently this answer of 25 mph was right...who knows but thanks guys for the help!
 
  • #16
Orodruin
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Apparently this answer of 25 mph was right...who knows but thanks guys for the help!
Yes, but your reasoning is wrong.
 
  • #17
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I don't get 25 mph. I get 27 mph. From the 2nd graph, the equation for the gas usage vs time is ##x=0.04t^2##, so $$\frac{dx}{dt}=0.08t$$So, at 15 hours, $$\frac{dx}{dt}=1.2\ gallons/hr$$From the first graph, at t = 15 hr, $$\frac{ds}{dx}=22.5 mpg$$So, $$\frac{ds}{dt}=\frac{ds}{dx}\frac{dx}{dt}=(22.5)(1.2)=27\ mph$$
 
  • #18
Orodruin
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I don't get 25 mph. I get 27 mph. From the 2nd graph, the equation for the gas usage vs time is ##x=0.04t^2##, so $$\frac{dx}{dt}=0.08t$$So, at 15 hours, $$\frac{dx}{dt}=1.2\ gallons/hr$$From the first graph, at t = 15 hr, $$\frac{ds}{dx}=22.5 mpg$$So, $$\frac{ds}{dt}=\frac{ds}{dx}\frac{dx}{dt}=(22.5)(1.2)=27\ mph$$
The point was not to parametrise the given functions and use estimates. A reasonable estimate would give a fuel consumption of roughly 1 gallon/hour at t=15 h. It would also be quite strange to consume 0 gallons/hour at t=0 as even an engine revolving without producing propulsion uses fuel.

I also think that we should not be giving out full solutions since the OP has not yet solved the problem.
 
  • #19
Orodruin
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From the first graph, at t = 15 hr,
dsdx=22.5mpg​
Also, this cannot hold. Clearly x < 10 for t=15 h and ds/dx > 25 for x < 10.
 
  • #20
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Also, this cannot hold. Clearly x < 10 for t=15 h and ds/dx > 25 for x < 10.
Oops. I stand corrected. At 15 hr, x = 9 gal, and ds/dx=25.5 mpg. Thanks.
 
  • #21
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The point was not to parametrise the given functions
Where specifically does it say that the point was not to parameterize the given functions.? That is how I personally would have approached the problem. The first graph can obviously be parameterized by the straight line $$\frac{ds}{dx}=30-\frac{x}{2}$$The second graph is essentially a parabola that can accurately be parameterized as $$x=\frac{t^2}{25}$$
So, $$v=\frac{ds}{dt}=\frac{ds}{dx}\frac{dx}{dt}=\left(30-\frac{t^2}{50}\right)\left(\frac{2t}{25}\right)$$
 
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  • #22
Delta2
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Where specifically does it say that the point was not to parameterize the given functions.?
Check post #1, section 2. Homework Equations , it says a hint was given not to try to find formulas from the graphs.
 
  • #23
Orodruin
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Where specifically does it say that the point was not to parameterize the given functions.?

A hint was given not to try and find formulas for the graphs
Obviously you can solve it by parametrisation, but it does not seem that was the problem constructor’s idea, even if (s)he clearly used those functions to make the plots.

I still think we are solving the OP’s problem for him by having this conversation.
 

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