Calculus, derivatives in action

[wuffle]

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( don't know 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 shouldn't even matter at all, should I use implicit differentiation? even if I do, what for?

2nd problem, don't know how to start, can't even draw a diagram , t.t help!

 

[HallsofIvy]

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?

 

[Mark44]

1st problem, all I know is that perpendicular lines have similar slopes( don't know 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.

 

[wuffle]

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! that's what we got before!

doing second one

 

[wuffle]

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 can't 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 don't know how to relate this to derivatives.