Could you help with this calculus question?
[math]\int_{0}^{x_C} g(x)-f(x) dx[/math] where [math]g(x)[/math] is the funny curve ABC’s equation, [math]f(x)[/math] is the line with slope [math]\sqrt{3}[/math] (OC) and [math]x_C[/math] is the x coordinate of point C.That’s the easiest integral to set up and evaluate, you can think of summing vertical strips with width dx.
Need help with some calculus homework on finding lenght of curve. can anyone help?
The parametric equations for the curve are: x(t) = t² y(t) = 2 t z(t) = ln t (1 < t < e) We get x'(t) = 2t; y'(t) = 2; z'(t) = 1/t Length of the arc: . . . . . . . .e . . . . . . . . . . . . . .e . . . . . . . . . . . . . . .e L = ∫ds = ∫√(x'² + y'² + z'²) dt = ∫√[4t² + 4 + (1/t²)]dt = ∫√[2t + (1/t)]² dt . . (c) . . .1 . . . . . . . . . . . . . .1 . . . . . . . . . . . . . . .1 . . . e L = ∫[2t + (1/t)]dt . . .1 L = (e² - 1²) + (ln e - ln 1) L = e²
Then find the area S of the region. Calculus Hw Question help?
When you integrate these trig functions, you need to substitute u = 5x, and that gives you du = 5 dx, so you need to divide your integral (and so your answer) by 5. Everything else looks fine.
CALCULUS 2 HW HELP!!!!!!!!! PLEASE HELP!!!!!!!!!! D:?
1.) Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. y = 4x2, y = 24x − 8x2 2.) In the example we modeled the world population in the second half of the 20th century by the equation P(t) = 2560e0.017185t. Use this equation to estimate the average world population during the time period of 1950 to 2000. (Round your answer to the nearest million.) 3.) Find the numbers b such that the average value of f(x) = 3 + 10x − 9x2 on the interval [0, b] is equal to 4. 4.) Find the average value have of the function h on the given interval. h(x) = 7 cos4 x sin x, [0, π]
AP Calculus Integration Homework Help?
a) differentiate both sides with respect to x using the product rule and implicit differentiation: x(2y)(dy/dx) + y^2 -x^3(1)(dy/dx) - 3x^2(y) = 0 Solve algebraically for (dy/dx) to finish part a) b) xy^2 - x^3y = 6 plug in x =1, solve for y: y^2 - y = 6 (y-3)(y+2) =0 y = -2, 3 equation for a tangent line: y - y0 = m(x-x0) m = dy/dx (from part a) for (1, -2) m = [3*(1^2)(-2) - (-2)^2]/[2(1)(-2) - 1^3] = 10/3 y +2 = (10/3)(x-1) for (1, 3) m = 0 y + 2 = 0(x-1) y = -2 c) tangent line is vertical when (dy/dx) is infinite...or the denominator of (dy/dx) = 0 2xy - x^3 = 0 But the point must also be on the curve itself so this equation must all be satisfied. xy^2 - x^3y = 6 Use first equation to find y = x^2/2 Subsitute y = x^2/2 into 2nd eqn. x^5/4 - x^5/2 = 6 x^5 - 2x^5 = 24 x^5 = -24 Use calculator if you want a better numerical value.
Calculus Homework Help Show Steps?
4x2 + 9y2 = 36 8x + 18y dy/dx = 0 4x + 9y dy/dx = 0 dy/dx = -4x/9y b) 4x2 + 9y2 = 36 9y² = 36 - 4x² y² = 4 - 4x²/9 y = √(4 - 4x²/9) y = (4 - 4x²/9)^(1/2) dy/dx = (1/2)(4 - 4x²/9)^(-1/2)*(-8x/9) dy/dx = (4 - 4x²/9)^(-1/2)*(-4x/9) dy/dx = [1/√(4 - 4x²/9) ] * (-4x/9) dy/dx = -4x/(9√(4 - 4x²/9) Since y = √(4 - 4x²/9) dy/dx is also = -4x/9y which is the same as (a)
HELP!! Calculus homework!?
A) y = cos(6x + 2) Let y = cos u where u = 6x + 2 dy/du = -sin u and du/dx = 6 dy/dx = dy/du * du/dx dy/dx = -sin (6x + 2) * 6 dy/dx = -6 sin (6x + 2) B) let y = cos (2u) where u = 3x + 1 dy/dx = -2 sin (2u) and du/dx = 3 dy/dx = dy/du * du/dx dy/dx = -2 sin [2(3x + 1)] * 3 dy/dx = -6 sin (6x + 2) ... same as (A)
Calculus homework need help please!!?
I am not sure how to do # 1 and 2. Sorry. As for #3 I got the answer: dx/dy= (-2x^4y-12xy^2)/(4x^3y^2-4x^3y+4y^3) I did the derivative by using the product rule in all three sections. Then bring anything w/o dx/dy the the opposite side of the equation. Do the GCF of dx/dy and the remaining numbers in parenthesis you divide on the other side. Hope this helps!
Need help with Calculus homework. f(x) = sq. root (4x + 1).?
f(x) = (4x + 1)^0.5 find derivative function (so you can find slope of tangent)... f ' (x) = 0.5 * (4x + 1)^(-0.5) * 4 (chain rule) f ' (x) = 1/2 * 1/√(4x+1) * 4 f ' (x) = 2/√(4x+1) now that you've simplified, plug in x = 5 f ' (5) = 2/√(20 + 1) f ' (5) = 2/√21 Not a very pretty slope, but it'll work. Now let's find the y point where the tangent intersects f(x) f(5) = √(4(5) + 1) y = √21 So a point on the tangent is (5, √21). Here's an equation to plug the stuff into: y1 -y2 = m(x1 - x2) (y1 and x1 are the y and x for the tangent, so leave them) y - √21 = (2/√21)(x - 5) y = (2/√21)x - 10/√21 + √21 y = (1/√21)(2x - 10 + 21) (factoring...) y = (1/√21)(2x + 11) !
Calculus homework?
1. Find the length of the given graph over the indicated interval. r= 5 ( 1 + cos (delta)). Interval: 0< delta < 2 pie. 2. Convert the given rectangular equation to polar form: 4 ((x)^2) - 5 ((y)^2) - 36 y - 36 = 0. 3. lim (-x ln x)=? *note as limit approaches 0 from the positive side. 4. the area bounded by r^2 =sin (delta). 5. The area bounded by the small loop of r = 2-4sin(delta) . 6. the integral arctan (2x) dx, between the interval (0, 1/2). 7. if f(x) = 2 ^(sin(^(2)) (x)), then f' (pie/4) equals ?