| This page is for problems from Pre-cal and Cal |
sifmaAug 2, 2008 2:21 PM
| from limits and derivatives.... to vector integrals and Jacobians. All is welcome here.... |
bla bla bla The endless debate between fans and haters. At one point, after spending a lot of time on MAL, you just realize it's totally pointless.Niko-kun said: On MAL, everyone who has used the lame rating system becomes a critic and an intellectual by default, haven't you heard? |
| How do you solve this? "If sin (Theta)= 3x/2 and 0< (Theta)< (pie)/2, express (theta)/4 - sine 2(theta) as a function of x" Answer: So far I do not understand how sin2(arcsin (3x/2)) turns into (3x((4-9(x^2))^(1/2))/2 |
Cortez321 said: How do you solve this? "If sin (Theta)= 3x/2 and 0< (Theta)< (pie)/2, express (theta)/4 - sine 2(theta) as a function of x" Answer: So far I do not understand how sin2(arcsin (3x/2)) turns into (3x((4-9(x^2))^(1/2))/2 trigonometric identity: sin(2theta)= 2*sin(theta)*cos(theta) then, cos(theta)= sqrt(1-(sin(theta))^2) this is not calculus by the way.... |
bla bla bla The endless debate between fans and haters. At one point, after spending a lot of time on MAL, you just realize it's totally pointless.Niko-kun said: On MAL, everyone who has used the lame rating system becomes a critic and an intellectual by default, haven't you heard? |
wakka9ca said: Cortez321 said: How do you solve this? "If sin (Theta)= 3x/2 and 0< (Theta)< (pie)/2, express (theta)/4 - sine 2(theta) as a function of x" Answer: So far I do not understand how sin2(arcsin (3x/2)) turns into (3x((4-9(x^2))^(1/2))/2 trigonometric identity: sin(2theta)= 2*sin(theta)*cos(theta) then, cos(theta)= sqrt(1-(sin(theta))^2) this is not calculus by the way.... It is pre-cal |
| What do I do when something like sin (arctan x) or cos (arcsin X) appear? |
Cortez321 said: What do I do when something like sin (arctan x) or cos (arcsin X) appear? For arctan x, since arctan x is an angle, x must be the trig. proportions between the sides. And tan is opp/adj, so opp=x, adj=1. Then, the hyp must sqrt(x^2+1) sin is opp/hyp, so sin(arctan x)= x/sqrt(x^2+1) Similarily, arcsin is the inverse function of sin. sin is opp/hyp, so opp=x, hyp=1 then, adj must be sqrt(1-x^2) since cos is adj/hyp, cos(arcsin x) = sqrt(1-x^2)/1 The important thing to notice is that the domain of sin, cos, tan, sec, csc, cot functions are ANGLES, and their range are TRIGONOMETRIC RELATIONSHIPS/PROPORTIONS/RATIOS. Picture a function like a transforming machine. The domain is what enters and the range is what comes out. For arctan, arcsin, the domain are TRIG. PROPORTIONS and the range are angles. So if you do a sin after doing a arccos, you will get back a trig. proportion ratio..... |
bla bla bla The endless debate between fans and haters. At one point, after spending a lot of time on MAL, you just realize it's totally pointless.Niko-kun said: On MAL, everyone who has used the lame rating system becomes a critic and an intellectual by default, haven't you heard? |
| How about in the case of a tansq(arctan x)? (in other words: tan(arctanx)^2) (The outside tan is squared (or at the power of two just in case))) |
| I have a question. But it's not an equation... This has to do with the Mean Value Theorem and concepts related to it. Two runners start at time 0. At some time t=a, one runner has pulled ahead, but the other runner has taken the lead by time t=b. Prove for some time t=c>0, the runners were going exactly the same speed. I assume that c happens at some time between a and b. I intuitively know that this is correct, but I don't know how to prove it... |
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aznHoopsfan said: I have a question. But it's not an equation... This has to do with the Mean Value Theorem and concepts related to it. Two runners start at time 0. At some time t=a, one runner has pulled ahead, but the other runner has taken the lead by time t=b. Prove for some time t=c>0, the runners were going exactly the same speed. I assume that c happens at some time between a and b. I intuitively know that this is correct, but I don't know how to prove it... First, you need the following conditions to apply the theorem: -The displacment function x(t) of the two runners are continuous function where the derivative exists and are continuous (speed function v(t)) then, for some reason (maybe long time since I did calculus proofs), it gets a bit complicated for me because I did not end up directly applying the theorem <_< (I feel I went around too much and the real solution is much simpler): -If runner 1 took the lead but runner 2 overrun him, then it must mean at a certain time t = d >0, x1(d) = x2(d) (they were at the same location at one time) This is true only if the x(t) functions are continuous. -create a function g(t) that is the RELATIVE distance between the two runner, where g(t) = x1(t) - x2(t). -Since x1(0) = x2(0) (they started at one place) and x1(d) = x2(d) for d>0 and the functions are both smooth and continuous, we know that g(0) = x1(0) - x2(0) = 0 (obviously since the distance between the two is zero at start) and g(d) = x1(d) - x2(d) = 0 (they end up meeting somewhere along the race as one overrun the other) -g(t) is smooth so apply Rolle version of MVT and we know that g'(c) = 0 for some t = c in interval [0,d]. Since g(t) = x1(t) - x2(t), then g'(t) = x1'(t) - x2'(t), which means: g'(c) = 0 g'(c) = x1'(c) - x2'(c) = 0 x1'(c) - x2'(c) = 0 x1'(c) = x2'(c) x1' and x2' are just the speed function of the two runner Q.E.D Hope it helps.... |
bla bla bla The endless debate between fans and haters. At one point, after spending a lot of time on MAL, you just realize it's totally pointless.Niko-kun said: On MAL, everyone who has used the lame rating system becomes a critic and an intellectual by default, haven't you heard? |
| @wakka9ca: Thanks for the reply! I was kind of stuck on it because I didn't know how to apply the Mean Value Theorem...but the way you explained it makes so much sense! |
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| How do I solve these problem? Suppose the position equation for a moving object is given by (3t^2)- 2t +5 where s is measured in meters and t is measured in seconds. Find the velocity of the object when t=2. and A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x= -3/2 ? |
| for the first problem, you're given s(t) = (3t^2)- 2t +5 is the position function. in order to find the velocity function, you must find the derivative of the given function, since s'(t) = ds/dt (change in position over change in time = velocity). once you find s'(t), plug in t = 2 to find the velocity at the specified time. the second problem is a bit trickier, since you're trying to find the change of x, or dx, with a given change in y, or dy, and position. you're given the position function again, and you need the velocity function. take the derivative of both sides with respect to x and you should get dy/dx = ______. solve for dx and substitute the given values you know. to find its value. |
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