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Calculus, derivatives in action

  • Thread starter wuffle
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  • #1
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Homework Statement



I made a thread last week where I had a reasonable idea what was going on in 2 problems last week
this week I have 2 problems where i have NO idea where to start.

Show that the ellipse x2 +2 y2 =2 and the hyperbola 2 x2 -
2 y2 =1 intersect at right angles.

What is the first time after 3 o'clock that the hands of the clock are together?

Homework Equations





The Attempt at a Solution



1st problem, all I know is that perpendicular lines have similar slopes( dunno how to explain , for example 1 line has slope of 4 , perpendicular has to have a slope of -1/4).

but why does that matter? these are not lines so slopes shouldnt even matter at all, should I use implicit differentiation? even if I do, what for?

2nd problem, dunno how to start, cant even draw a diagram , t.t help!
 

Answers and Replies

  • #2
HallsofIvy
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Well, you posted this in the "Calculus and Beyond" section so you must know of "tangent" lines and derivatives. Two curves are perpendicular at a point of intersection if and only the tangent lines at those points are perpendicular.

For (2), have you drawn a picture? You know the hour hand must lie between "3" and "4" on the clock face. For the minute hand to be in the same position, the time must be between 3:15 and 3:20. Can you narrow that down?
 
  • #3
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wuffle said:
1st problem, all I know is that perpendicular lines have similar slopes( dunno how to explain , for example 1 line has slope of 4 , perpendicular has to have a slope of -1/4).
Your example is correct, but the lead-in text isn't. Perpendicular lines have slopes that are negative reciprocals of each other.
 
  • #4
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Well, you posted this in the "Calculus and Beyond" section so you must know of "tangent" lines and derivatives. Two curves are perpendicular at a point of intersection if and only the tangent lines at those points are perpendicular.

This helped a LOT, here's what I did:

Differentiate implicitly two functions, here's what I got

dy1/dx= -x/2y

dy2/dx=x/y

So since slopes must be perpendicular

-x/2y * x/y=-1
-x2/2y2=-1
This is where I got stuck for a long time, here's what I did next

x2 +2 y2=2
2x2-2y2=1 multiply by 2

4x2 -2y2=2 =>

x2 +2 y2=4x2 -2y2
3x2=6y2
x2=2y2
2=x2/y2
-1=-x2/2y2


but hey! thats what we got before!

doing second one
 
  • #5
26
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Well, you posted this in the "Calculus and Beyond" section so you must know of "tangent" lines and derivatives. Two curves are perpendicular at a point of intersection if and only the tangent lines at those points are perpendicular.

For (2), have you drawn a picture? You know the hour hand must lie between "3" and "4" on the clock face. For the minute hand to be in the same position, the time must be between 3:15 and 3:20. Can you narrow that down?
well yea obviously, since the hour arrow cant be more than half(3.175) because the minute one has to be in same range, so its somewhere 3.16-3.17 ish, but how exactly does that relate to calculus?rate of change of arrows?

obviously the minute arrow changes faster, but how much faster is question for me, i dunno how to relate this to derivatives.
 

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