MedVision ad

Related Rates (1 Viewer)

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Two Straight roads meet an an angle of 60 degrees. A starts from the intersection and travel along one road at 40km/h. One hour later B starts from the intersection and travels along the other road at 50km/h. At what rate is the distance between them changing three hours after A starts.
 

Shikobe

New Member
Joined
Feb 10, 2009
Messages
15
Gender
Male
HSC
2010
This question is basically the same type as the question 17(Fitz book).
So, instead of using Pythagoras theorem too find the distance, you use the cosine rule and thats basically it.
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
I did. But i didnt get it. Its actuall q18 (fitz book)

Youll get something like x^2 +y^2 - xy = distance

But how do you differentiate that?
 

lychnobity

Active Member
Joined
Mar 9, 2008
Messages
1,292
Gender
Undisclosed
HSC
2009
I did. But i didnt get it. Its actuall q18 (fitz book)

Youll get something like x^2 +y^2 - xy = distance

But how do you differentiate that?
Implicitly.

Treat y as a function, then differentiate wrt x

differentiate this: x^2 +y^2 + (- xy) = distance

and it becomes this: 2x + 2y(y') + (-y - y'x) = 0

2x + 2y'y - y - y'x = 0

2x - y + y'(2y - x) = 0

y' = (y - 2x)/(2y - x)
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Implicitly.

Treat y as a function, then differentiate wrt x

differentiate this: x^2 +y^2 + (- xy) = distance

and it becomes this: 2x + 2y(y') + (-y - y'x) = 0

2x + 2y'y - y - y'x = 0

2x - y + y'(2y - x) = 0

y' = (y - 2x)/(2y - x)
But implicit isnt part of the 3unit course. Although i did try and the answer wasnt correct :(
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top