I’ve posted the problem at
http://members.home.net/mstudi/calc.gif
Please help!
I’ve gotten down to:
0 = (SD - 6)[(49 + SD^2)^0.5] + SD[(SD^2 - 12SD + 61)^0.5]
How would I solve this for SD?
I posted a link to the original problem in case someone can find an alternate method of solving for either CS or SD.
Studi
Are you locked in to solving this with calculus? Simple geometry works much better.
Extend AC 5 units past C to F. Extend BD 5 units past D to G. (I’m using F because the prime in A` is almost invisible.) The shortest distance from B to F is a straight line that crosses CD at H. BHD and BFG are similar triangles so HD/BD = FG/BG = .5, so HD = 3.5.
By symmetry (and if required the argument can be made more rigorous), the point S must coincide with H in order for x + y to be minimal.
It looks like you’ve done all the calculus already…this is more of an algebra question.
Anyway, there are (at least) two ways to solve an equation of the form sqrt(a)+sqrt(b)=0. The first is to square both sides, then isolate the sqrt(ab) term in the resulting expression and then square both sides again. The other way is to divide by sqrt(a), leaving you with an equation that has only one square root in it, then isolate that square root and then square both sides of the equation. Either way, you end up with an equation with no square roots left, which you can then solve however you like. Be sure to check whatever answers you get by plugging them back into the original equation, because squaring the equation may introduce additional solutions that you don’t want.
But Synergist is right anyway, the geometrical solution is much nicer.
Just one thing to note, in order to solve it algebriacally you inevitably have to incorporate the coefficients of the square roots into the square roots themselves. Normally, you would use the simple exponent rule: aSqrt(b) = Sqrt(a^2b), however, if a is negative then you don’t put the -1 into the equation and get -aSqrt(b) = -1*aSqrt(b) = -Sqrt(a^2b). In this case there can only be a solution if (d - 6) is negative because there are no real solutions to sqrt(a) + sqrt(b) = 0 other than a and b = 0. Realizing the restriction of d < 6 will lead you to the algebriac solution.
A better way, IMHO, to solve the equation 0 = sqrt(a) + sqrt(b) is to remember the formula for the difference of two perfect squares: x[sup]2[/sup] - y[sup]2[/sup] = (x + y)(x - y). Similarly, then, a - b = [sqrt(a) + sqrt(b)][sqrt(a) - sqrt(b)]. So, for your equation:
0 = (SD - 6)[(49 + SD^2)^0.5] + SD[(SD^2 - 12SD + 61)^0.5]
multiply both sides by
(SD - 6)[(49 + SD^2)^.05] - SD[(SD^2 - 12SD + 61)^0.5]
giving you
0 = (SD - 6)[sup]2[/sup](49 + SD[sup]2[/sup]) - SD[sup]2[/sup](SD[sup]2[/sup] - 12SD + 61)
Voila. No square roots to worry about. Just multiply it all out, and it simplifies itself quite nicely.
Yes, I have to use calculus.
For this relief, much thanks, all.
Studi
Wait, there’s still a problem.
If I substitute the answer (3.5) into the equation:
0 = (SD - 6)[(49 + SD^2)^0.5] + SD[(SD^2 - 12SD + 61)^0.5]
you get the left side, zero.
However, when I substitute 3.5 into the expanded thing, I don’t get zero.
I followed Punoqllad’s advice, and got this:
0 = (SD - 6)2(49 + SD2) - SD2(SD2 - 12SD + 61)
Let x = SD
I then multiplied it out to get:
0 = 2x^4 - 24x^3 + 146x^2 - 588x + 1764
How is this best solved? I’ve tried using the factor theorem (basically guess and check to find one value of x that works, and then dividing the above equation by the factor to simplify things)
Double-check your algebra. You should get a much nicer polynomial than the one you’ve got.
I just had a really nice and pretty solution, but decided not to post it, just in case we’re doing your homework for you.
If you clean up your work a little bit, it should come out much nicer. Suggestion:
Let s = segment CS, and (6 - s) = segment SD. That will clean up your equations quite a bit. Simplify every chance you get, and use fractions instead of negative powers.
At the end you should end up with a nice quadratic equation that factors into two terms. The positive one is CS, and subtracting that from 6 gives SD.
Damn, I really wanted to post that solution, too. It works out really nicely. haha
This is a great problem to solve with an analog computer. Here’s what you do:
[ul]
[li]Take a flat mirror. Lay it on a level surface.[/li][li]Mark two spots on the mirror six units apart (e.g. 6 cm). You can use the edge of a piece of tape as a mark.[/li][li]Take a flat-top pen (one that you can balance standing up). Measure off five units (e.g., 5 cm) from the end and mark it (the edge of a piece of tape works well). Balance the pen on one of the marks on the mirror.[/li][li]Take another flat-top pen. Mark it at seven units (7 cm) and place on the other mark on the mirror.[/li][/ul]
So now you’ve built the computer. To do the actual computation, we’ll make use of the fact that light takes the shortest path.
[ul]
[li]Using one eye, maneuver until the reflection of the mark on the far pen lines up visually with the mark on near pen. Put a mark on the glass where the reflection appears to be. For better accuracy, look at the reflection from the other direction. The reflection point should be the same from either direction.[/li][li]Finally, measure the distance from the reflection point’s mark to either of the pens. That’s your answer.[/li][/ul]
Let nature do the calculus for you!
No, you’re not doing my homework for me. My grade 13 final calculus exam is on Monday. I’m just reviewing.
I’ll try it again later. Right now, the name of the game is integration.
Studi
I’m getting a really pretty 2nd degree function, but it’s wrong. I’m getting:
35X^2 + 12X + 1763 = 0
where X is SD. But that can’t be right, because it doesn’t have any real solutions.
Are you sure it should be 12 in the original equation and not 11 (5+6)?
The 12 in the original equation is correct, it comes from (6-SD)[sup]2[/sup]+25, which is equal to x[sup]2[/sup].
Studi, if I may offer some exam advice: take the time during your exam to double-check all of your algebra. I get to mark exams with these kinds of problems on them from time to time, and if I had a nickel for every student who knew the calculus but lost way too many marks on the algebra…well, I’d have a lot of nickels.
Thanks for the advice Math Geek; you’re right, the actual calculus is very easy. I’m just going to trip up simplifying something.
Anyway, I’m quite frustrated right now. I’ve tried a different question, similar to this one, and I’m stuck in the same place, one big ass equation with one variable.
Please, someone, show me how to solve:
0 = (SD - 6)[(49 + SD^2)^0.5] + SD[(SD^2 - 12SD + 61)^0.5]
Thanks.
Studi
Nevermind, I got it.
Thanks to all.
Studi