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 E2P 3
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April 2, 2013 at 3:49 am #13256
Ezra HalleckParticipant3.6: Find coordinates of all points on , , where tangent line is horizontal.
April 4, 2013 at 1:12 am #16769
AbdoulayeParticipantTo answer this question, one must recognize that when the derivative to a function is zero at a certain point, then the tangent line to the graph of said function is going to be horizontal. Once, this is recognized, it becomes evident that the student has to find the derivative of the given function — y=sinx*cosx — and set the resulting function equal to zero.
Step 1: The Product rule must be used to find the derivative: y’=uv’+vu’ where u=sinx and v=cosx
Step 2: The derivative trignometric theorems must be applied: namely, the derivative of sinx= cosx and the derivative of cosx=sinx
Step 3: All the information for the product rule must be gather together: u=sinx, u’=cosx, v=cosx, v’=sinx
Step 4: These values are then substituted into the product rule equation: y’=uv’+vu’. Doing so yield: y’= sinx*sinx+cosx*cosx.
Step 5: The student must recognize, from basic algebra, that the previous result could be simplified to y’=sin^2x+cos^2x. Rearraning the terms will further clarify the equation: y’=cos^2xsin^2x.
Step 6: The derivative is then set to zero. Mind you, that the reason for doing this is because when the tangent line is horizontal, the derivative, i.e., the slope of the graph at a specific point, is equal to zero (This, of course, makes perfect sense, since the slope of a horizontal line is zero — the derivative is the slope, therefore it is zero). Doing so yields: cos^2xsin^2x=0.
Step 7: The next steps involve algebraic manipulation. The sin^2x should be transposed to both sides yielding cos^2x=sin^2x. Then, the sqaure root of both sides are taken in order to eliminate the squares, which yields the following: cosx=sinx.
Step 8: From here, my method of solving the problem may be unorthodox and may differ from the way others may solve it. It requires the use graphical trignometry and one of the special triangles, namely the one with angles of 45, 45, and 90 degrees and sides of 1,1, and sqrt(2), the latter being the hypothenuse.
Step 9: In the special triangle described one notices that the cosine of 45 degrees and the sine of 45 degrees are equal to one another. Indeed, this bit of information is very useful to remember (the memerization of the special triangles are also extremely useful). That the sine and cosine of 45 are equal could also be derived by graphing the functions of sine and cosine. Whichever way one wants to derive it, the result is the same: the sine and cosine of 45 degrees are equal. It is also important to recognize that 45 degrees expressed in radians is (pi/4), since radians is the standard unit for angles in calculus.
Step 9: After that recognition, one takes into account that the special triangle used was only in the range of (0,(pi/2) — the first quandrant of the Cartesian coordinate system. However, the question asked for values in the range of (0,2*pi). Thus, one has to the values of x that would satisfy the equation cosx=sinx for the three quadrants. Remember that the value for the first quadrant was already found using the special triangle stated i.e, 45 degrees of (pi/4). However, it is quite easy to find the values for the other three quadrants. One need just take pi/4 as a reference angle for each quadrant. Then the actual angle is found with respective to the positive xaxis. For example: To find the value satisfying the equation in the second quadrant one takes pi/4 as a reference angle in that quadrant and follows the procedure for finding the actual angle: pi(pi/4) or 18045. Doing so yields the angle (3*pi/4) or 135 degrees.
Step 10: The answers for all the quadrants are as follows: (pi/4), (3*pi/4), (5*pi/4), and (7*pi/4) or 45, 135, 225, 315 degrees.I hope this help.
Feel free to ask any questions about my method and process.
Abdoulaye
April 15, 2013 at 12:06 am #16843
Ezra HalleckParticipantYou eventually did find all the answers. However, when you took the square root of cos^2x=sin^2x, you should have introduced a plus or minus, e.g. x^2=9 has solution x=+ or – 3.
I suggest another path. divide both sides by cos^2x to get 1=tan^2x. Take square root of both sides to get tanx=+ or – 1. Basically, we are looking for the angles where the slope (tangent) is 1 (45 deg) or 1 (45 deg). The answer is what you found:
pi/4), (3*pi/4), (5*pi/4), and (7*pi/4).April 15, 2013 at 12:21 am #16845
H PardonMemberokay i’m good until we take the square root of both sides.
Do I plug in my value for X=0 into the equation?April 15, 2013 at 1:40 am #16854
DevindraParticipantOkay I’m getting it but kind of confused still
April 15, 2013 at 2:32 am #16859
IronManMemberI agree with Professor. While Abdoulaye’s method is totally good, dividing both sides by cos²x will give you a shorter path to the answer. From there, tan²x = 1, you take square root of both sides and end up with tan(x) = ±1. At this point you can just graph tan(x) and all the angles (0 to 2p) where tan(angle)=±1 will be the answers for the problem.
May 21, 2013 at 3:19 am #17056
B AmarsinghParticipantThat method works much better ^

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