I'm having trouble with these Calculus problems...?
a. You will need the (u/v) rule for derivatives. Usually with these you might first try just converting everything to sines and cosines but this is not always necessary since formulas exist for csc, etc. but if you don't remember these trying sines and cosines might be the best bet. cot x/ (1 + csc) cot(x) = cos(x)/sin(x) csc(x) = 1/sin(x) cot x/ (1 + csc) = [cos(x)/sin(x)] / [1 + 1/sin(x)] cot x/ (1 + csc) = cos(x)/(sin(x) + 1) (d/dx)(u/v) = (vu' - uv')/v^2 u = cos(x) and u' = -sin(x) v = sin(x) + 1 and v' = cos(x) And so on, just some trig. b. this is really just u^2 + v^2 where u = sin(x) and v=cos(x) df/dx = 2u (du/dx) + 2v(dv/dx) du/dx = cos(x) and dv/dx = -sin(x) OK? c. f(x)= (sinx secx)/ (1 + x tan x) that xtan(x) stuff first note that sec(x) = 1/cos(x) so the numerator is sin(x)/cos(x) = tan(x) again this is a (u/v) with. (d/dx)(u/v) = (vu' - uv')/v^2 u = tan(x) and u'= sec^2(x) v = 1 + xtan(x) and v' = tan(x) + xsec^2(x) (vu' - uv')/v^2 = [(1 + xtan(x))(sec^2(x)) - (tan(x))(tan(x) + xsec^2(x))] / [1 + xtan(x)]^2 Looks a little messy so I will try to work it out. Look at the numerator: Note tan^2(x) = sin^2(x)/cos^2(x) = sin^2(x)sec^2(x) [(1 + xtan(x))(sec^2(x)) - (tan(x))(tan(x) + xsec^2(x))] sec^2(x)[(1 + xtan(x)) - (sin^2(x) + xtan(x))] sec^2(x)[1 - sin^2(x)] = sec^2(x)cos^2(x) = 1 So the numerator is just 1 and finally: f'(x) = 1/[1 + xtan(x)]^2 d. as with any problem involving physical units you must keep track of just what you are calculating. In this case you calculated out to feet/radian and not feet/degree. If you convert then you will have the right answer. To do that you must multiply by pi and divide by 180. Units work like this: 1/degrees = 1/(radians*(degrees/radian)) degrees/radian = 180/pi 1/degrees = (pi/180) (1/radians)
Easy Calculus Problems- Help?!?
I'm having some trouble with solving (or even setting up) with 3 calculus problems, so any help is appreciated! Thanks! 1)If h(t) represents the height of an object above ground level at time t and h(t) is given by: h(t)= -16t^2+11t+1 find the height of the object at the time when the speed is zero. 2) Let f(x)=x^2+ x+ 13. What is the value of x for which the tangent line to the graph of y=f(x) is parallel to the x -axis 3)Let f(t) =8t^2+4t+1. Find the value t for the tangent line to the graph of f(t) has the slope 1.
I'm having trouble with this calculus problem?
dy/dx = 2y/x^2 --> dy/y = 2dx/x^2 = 2x^-2 dx Integrate both sides to get ln y = 2[-x^-1] + c Exponentiate both sides to get y = k * e^[-2/x] for some constant k. We have the point (1,4) so 4 = k * e^[-2/1] = k * e^-2 = k * 0.135335283 --> k = 29.5562244 y = 29.5562244 * e^[-2/x]
Calculus problem?
Assume that the best path is from oil well to a point on coast x km down toward the refinery. Distance in ocean = sqrt(4 + x^2) km by pythagoras Distance on land = 4 - x km Cost = 2*sqrt(4 + x^2) + (4 - x) Differentiate this and equate to zero in the usual way. Use chain rule on the sqrt part written as (4 + x^2)^(1/2). Can you take it from there? x is between 1 and 1.5 You can check that making x both slightly larger and slightly smaller than the value you find will increase the cost. Edit. Derivative is 2x/sqrt(4 + x^2) - 1 leading to x = 2/sqrt3.
Calculus Help (Need help with 1 problem)?
difference quotient: f'(x) = lim h->0 (f(x+h)-f(x))/h with f(x) = 4x^2 - 5x f(x+h) = 4(x+h)^2 -5(x+h) so the quotient: = [4(x+h)^2 -5(x+h) - (4x^2 - 5x)] /h = [4(x^2+2hx + h^2) - 5x - 5h - 4x^2 + 5x]/h = [4x^2 + 8hx + 4h^2 - 5h - 4x^2]/h = 8x + 4h - 5 with lim h->0 f'(x) = 8x - 5 b) tangent line y=mx+b to f(x) at point (2,6) find the slope, m = slope f(x) = f'(x) at x=2 m = f'(2)= 16-5 = 11 y=11x +b point (2,6) x=2,y=6 is on y : 6 = 22 + b b = -16 so the line y: y= 11x - 16
Calculus 1/ Physics Problem?
The length contraction is a characteristic of the universe, as explained by the Special Theory of Relativity. It says that if any object is traveling through space, it appears shorter to an outside observer. At speeds that we are used to, the effect is extremely small, if it is measurable at all. However, at speeds that are a significant percentage of the speed of light, the observer sees the moving object as being shorter and shorter until, at light speed, the object appears to have no length at all! a) What is the observed length L of the ship if it is traveling at 50% of the speed of light? Below is the length contraction formula: L = L0√[1 - (v²/c²)] L0 = observed length of the spaceship when it is stationary L = observed length of the spaceship when it is moving c = speed of light = 300,000,000 m/s v = velocity of the spaceship (0.5c) = 150,000,000 m/s L = 0.866025404 L0 When the spaceship is moving at 50% of the speed of light, its length, as measured by the stationary observer is 0.866L0 or 87% of the object's stationary length. b) What is the observed length L of the ship if it is traveling at 75% of the speed of light? v = velocity of the spaceship (0.75c) = 225,000,000 m/s L = 0.661437828 L0 When the spaceship is moving at 75% of the speed of light, its length, as measured by the stationary observer is 0.661L0 or 66% of the object's stationary length. c) Comparing parts (a) and (b), what happened to the the observed length of the space ship? The observed length of the spaceship varies inversely with the spaceship's velocity as a fraction of the speed of light. In other words, the faster the spaceship goes, the shorter that it appears to the observer.
What is your favorite calculus 1 problem and why?
I have never taken calculus. The furthest I got was trigonometry and I failed out the First semester with a 68. I have no formal training in advanced mathematics. The problems in calculus that I do figure out, I work them out in my head, using the process of elimination when dealing with what numbers i plug into the equation.
Calculus optimization problem?
Good to hear that you could do optimization problems. Back to your problem. Let say m is the slope of the straight line that goes thru the point (6,10). The equation of the line would be y = m(x-6) + 10 or y = mx - 6m + 10. The straight line intersects the two coordinate axes at (0, -6m+10) and ((6m-10)/m, 0) consecutively. Therefore, the area of the triangle that the line cut off from the first quadrant is A(m) = (1/2)*( -6m+10)*(6m-10)/m = -(6m-10)^2/(2m) Then using the first derivative test (or a graphing calculator), you can determine that the area would be smallest if m = - 5/3, Therefore, the equation of that line must be y = -5x/3 +20
I'm panicking about this one calculus problem, please help me..?
I wouldn't say this is calculus. However here's how you find the answer. Think of it as a parabola. If it took 6 seconds to land, then each side of the parabola is 3 units long. If it took 3 seconds for the ball to make it from the maximum height to the catchers mit, and gravity makes objects fall at a rate of 9.8 Meters per second squared. all you do is multiply 3 by 9.8 m/s^2 So the maximum height was approximately 30 meters.