pleifer 0 #1 February 24, 2004 All right I have two problems that are worth a good chunk of my grade for a business calc class If there is anyone who feels like using there brain to double check me..... I all ready did them I just want to make sure they are right..... #1 Use implicit defferentiation to find the equation of the tangent line to the graph of the indicated eqation at the point with the givin value of x. xy-3x+2=0 where x= (-1) #2 A Point is moving on the graph of y^2- 4x^2=12, so that it's x coordinate is decreasing at 2 units per second when (x,y) = (1,4), find the rate of change of the Y coordinate..... Disclamer for the flamers- and for those that think this is just for the answer 90% of the grade is the work up to the answer.... see if my answer is wrong than the work leading up to it is wrong.... if it is right then I proly got the whole thing right....... _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
koz2000 1 #2 February 24, 2004 I'll leave this one to Kallend ______________________________________________ - Does this small canopy make my balls look big? - J. Hayes - Quote Share this post Link to post Share on other sites
CrazyThomas 0 #3 February 24, 2004 QuoteAll right I have two problems that are worth a good chunk of my grade for a business calc class If there is anyone who feels like using there brain to double check me..... I all ready did them I just want to make sure they are right..... #1 Use implicit defferentiation to find the equation of the tangent line to the graph of the indicated eqation at the point with the givin value of x. xy-3x+2=0 where x= (-1) I'll take a shot at this. It's been awhile since implicit differentiation, but it is something like treating y like a function of x or something. where d/dx xy would be found using the product rule, which is (1st * d/dx 2nd) + (2nd * d/dx 1st). Where the first function is x, hence making the derivative 1, and the second function is y, making the derivative dy/dx. Applying the product rule, we should get (x * dy/dx) + (1 * y) The second part, (-3x) is just -3 for the derivative, and the constant 2 has derivative of 0. so, supposedly (take with a grain of salt), the first equation ( xy-3x+2) should have d/dx of x*dy/dx + y - 3. And that is the slope of the line you are looking for. something about plugging in x = -1, and finding the rest of the function gets hazy. I hope that was along the lines of the way you were working out your implicit differentiation problem. Good luck, and maybe someone else can step in and clarify. Thomas Quote Share this post Link to post Share on other sites
pleifer 0 #4 February 24, 2004 all right I will post my work and see if anyone can spot mistakes.... #1 xy-3y+2=0 dy/dx(xy-3y+2=0) so we get: xy'+y-3=0 xy'=3-y y'=(3-y)/x so at y'l(-1,5) = 2 (get the 5 solving the original for y) so to get the tangent line in y=mx+b form (y-5)=2(x+1) so tangent line is y=2x+7 ??? any problems???/ _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
pleifer 0 #5 February 24, 2004 allright for the second one.... and to reply to myself step one take the derivative... leaving 2y-8x=0 so when dx/dt is (-2) ('cause its decreasing right?) dydt= -1 when x=1 and y=4 so the rate of change of the y coordinate is -1 meaning that the y coordinate is decreasing 1 unit at the point (1,4) any problems here??? _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
CrazyThomas 0 #6 February 24, 2004 looks pretty sweet to me. I am unsure of the tangent line part, but it looks like you did the differentiation part correctly. what about the second problem. that is basically the same thing right, finding the derivative, and then substituting it into y = mx + b or something. lemme look at it again (cheap post-whore attempt) Thomas Quote Share this post Link to post Share on other sites
CrazyThomas 0 #7 February 24, 2004 Quote #2 A Point is moving on the graph of y^2- 4x^2=12, so that it's x coordinate is decreasing at 2 units per second when (x,y) = (1,4), find the rate of change of the Y coordinate..... CAREFUL THERE. Implicit differentiation can catch you like that. Maybe it is not needed there, but I have a feeling it is. what about this: 2yy' -8x = 12 and then something about substitution of tangent line formulas Of course, taking the derivative of both sides of the equation will turn 12 into 0, cause the d/dx of a constant is 0. I think implicit might be used in that problem, but I am unsure. Maybe others can chime in. Thomas Quote Share this post Link to post Share on other sites
pleifer 0 #8 February 24, 2004 the first one we are finding the tangent line the second one we just need to know the rate of change... ie in phisics it would be the instantaious velocity _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
SarahC07 0 #9 February 24, 2004 Well, while we have a math thread going... I got a problem... It's also from a business calc class... it's off of the latest exam that we ALL failed, so we get to do test corrections....WeeeeeSimplify the following: 4^(2log(base4)9) I had an answer and apparently it was WRONG... So I created another way to solve the problem... but I still think I got the wrong answer.... Thanks ~Sarah Quote Share this post Link to post Share on other sites
pleifer 0 #10 February 24, 2004 I don't know what your instructor is looking for there if the ^4 was on the other side it would be just 9 squared but wrote as 4 to the power of... being there is no exponents... I would just solve it using the change of base formula and sticking it in the calc _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
SarahC07 0 #11 February 24, 2004 I used the change of base formula... and I got 80.995 BUT it was wrong... Quote Share this post Link to post Share on other sites
mcneill79 0 #12 February 24, 2004 Question: what are you taking the derivative with respect to? I don't see how you get 2y-8x = 0 I would envision doing something like this: d(y^2 -4x^2)/dt = 0 2y*dy/dt - 8x dx/dt =0 then substituting all the values in 2*4dy/dt - 8*1*-2 = 0 Therefore dy/dt = -2 Just my $.02 Disclaimer: I haven't taken calc in a many years, but I used to be good at it. Quote Share this post Link to post Share on other sites
pleifer 0 #13 February 24, 2004 thanks yep that is what i did, I saved time with the y prime,, also that is how the book shows... but yes it is dy/dt on the exam thanks for checking _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
mcneill79 0 #14 February 24, 2004 The only thing is I got a different answer than you. You say -1... I say -2 Quote Share this post Link to post Share on other sites
pleifer 0 #15 February 24, 2004 sweet and I just found a small error, from what you did thanks _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
swilson 0 #16 February 24, 2004 Using some math software I've got here at home I tried all the permutations of the change of base formula. I get 80.999...to more decimal places than is reasonable. If you change the 2 to a 2x, then take the limit as x approaches 1, I get 81 which is pretty damn close to what you got...I'd be requesting the prof to work it out for you if he/she still insists it's wrong. Please post when you find the answer. S Quote Share this post Link to post Share on other sites
SarahC07 0 #17 February 24, 2004 Yeah, I put 80.995 as my answer ... with work to back it up and i lost all 4 of the available 4 points on that problem.... I'm kinda ticked... I'm thinking if the answer were 81... then I would of received partial credit and not lost all of the possible points.... Who knows? The test corrections are due tomorrow... I won't see my prof before then... ~Sarah Quote Share this post Link to post Share on other sites
Viking 0 #18 February 24, 2004 *walks away rubbing temples* why did i click on this thread!I swear you must have footprints on the back of your helmet - chicagoskydiver My God has a bigger dick than your god -George Carlin Quote Share this post Link to post Share on other sites
pleifer 0 #19 February 24, 2004 Quote*walks away rubbing temples* why did i click on this thread! sorry to hurt your brain..... especialy on a Monday nun the less _________________________________________ The Angel of Duh has spoke Quote Share this post Link to post Share on other sites
