2 cars moving from the same point, related rates.
2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?
Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$
Plugging in the values after 2 hours I got
$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$
I am not sure where to go from because I can't find the value of z.
calculus
asked Dec 9 at 0:38
Eric Brown
716
add a comment |
2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?
Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$
Plugging in the values after 2 hours I got
$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$
I am not sure where to go from because I can't find the value of z.
calculus
asked Dec 9 at 0:38
Eric Brown
716
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37
add a comment |
2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?
Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$
Plugging in the values after 2 hours I got
$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$
I am not sure where to go from because I can't find the value of z.
calculus
asked Dec 9 at 0:38
Eric Brown
716
2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?
Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$
Plugging in the values after 2 hours I got
$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$
I am not sure where to go from because I can't find the value of z.
calculus
calculus
asked Dec 9 at 0:38
Eric Brown
716
asked Dec 9 at 0:38
Eric Brown
716
asked Dec 9 at 0:38
Eric Brown
716
asked Dec 9 at 0:38
Eric Brown
716
asked Dec 9 at 0:38
Eric Brown
716
716
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37
add a comment |
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37
add a comment |
2 Answers
2
active
oldest
votes
Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$
We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$
By using Pythagorean theorem we have
$x^2+y^2=z^2$
Now implicitly differentiate with respect to $t$ to get
$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$
Now plug in the numbers. After $2$ hours
$$x=60(2)=120$$
and $$y=25(2)=50$$
$$z^2=(120)^2+(50)^2$$
Can you take it from here?
answered Dec 9 at 0:45
Key Flex
7,47941232
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
add a comment |
Hint.
The distance between the two cars is given by
$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$
so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$
NOTE
Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$
answered Dec 9 at 0:45
Cesareo
8,1833516
add a comment |
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2 Answers
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2 Answers
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active
oldest
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votes
Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$
We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$
By using Pythagorean theorem we have
$x^2+y^2=z^2$
Now implicitly differentiate with respect to $t$ to get
$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$
Now plug in the numbers. After $2$ hours
$$x=60(2)=120$$
and $$y=25(2)=50$$
$$z^2=(120)^2+(50)^2$$
Can you take it from here?
answered Dec 9 at 0:45
Key Flex
7,47941232
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
add a comment |
Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$
We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$
By using Pythagorean theorem we have
$x^2+y^2=z^2$
Now implicitly differentiate with respect to $t$ to get
$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$
Now plug in the numbers. After $2$ hours
$$x=60(2)=120$$
and $$y=25(2)=50$$
$$z^2=(120)^2+(50)^2$$
Can you take it from here?
answered Dec 9 at 0:45
Key Flex
7,47941232
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
add a comment |
Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$
We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$
By using Pythagorean theorem we have
$x^2+y^2=z^2$
Now implicitly differentiate with respect to $t$ to get
$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$
Now plug in the numbers. After $2$ hours
$$x=60(2)=120$$
and $$y=25(2)=50$$
$$z^2=(120)^2+(50)^2$$
Can you take it from here?
answered Dec 9 at 0:45
Key Flex
7,47941232
Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$
We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$
By using Pythagorean theorem we have
$x^2+y^2=z^2$
Now implicitly differentiate with respect to $t$ to get
$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$
Now plug in the numbers. After $2$ hours
$$x=60(2)=120$$
and $$y=25(2)=50$$
$$z^2=(120)^2+(50)^2$$
Can you take it from here?
answered Dec 9 at 0:45
Key Flex
7,47941232
edited Dec 9 at 19:47
answered Dec 9 at 0:45
Key Flex
7,47941232
answered Dec 9 at 0:45
Key Flex
7,47941232
answered Dec 9 at 0:45
Key Flex
7,47941232
7,47941232
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
add a comment |
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
– Eric Brown
Dec 9 at 0:53
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
– N. F. Taussig
Dec 9 at 16:54
add a comment |
Hint.
The distance between the two cars is given by
$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$
so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$
NOTE
Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$
answered Dec 9 at 0:45
Cesareo
8,1833516
add a comment |
Hint.
The distance between the two cars is given by
$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$
so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$
NOTE
Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$
answered Dec 9 at 0:45
Cesareo
8,1833516
add a comment |
Hint.
The distance between the two cars is given by
$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$
so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$
NOTE
Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$
answered Dec 9 at 0:45
Cesareo
8,1833516
Hint.
The distance between the two cars is given by
$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$
so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$
NOTE
Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$
answered Dec 9 at 0:45
Cesareo
8,1833516
edited Dec 10 at 12:18
answered Dec 9 at 0:45
Cesareo
8,1833516
answered Dec 9 at 0:45
Cesareo
8,1833516
answered Dec 9 at 0:45
Cesareo
8,1833516
8,1833516
add a comment |
add a comment |
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In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
– David K
Dec 10 at 13:37