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April 20, 2014

Homework Help: Math: Calculus

Recent Homework Questions About Calculus

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Calculus
An observer stands at a point P, one unit away from a track. Two runners start at the point S in the figure and run along the track. One runner runs 2 times as fast as the other. Find the maximum value of the observer's angle of sight è between the runners. Thank ...
Saturday, November 23, 2013 at 10:31pm

Calculus
just as a check, the distance from point P(h,k) to the line ax+by+c=0 is |ah+bk+c|/√(a^2+b^2) In this case, that is |24+5+1|/√(36+25) = 30/√61 Hmmm. What went wrong above? Oops. Forgot to multiply -6 by 6
Saturday, November 23, 2013 at 7:17pm

Calculus
Let the point be P(x,y) , point (-4,1) be A Since the shortest distance must be where AP is perpendicular to the given line. slope of given line = 6/5 so slope of AP is - 5/6 equation of AP: y-1 = (-5/6)(x+4) times 6 6y - 6 = -5x- 20 5x + 6y = -14 , #2 6x - 5y = -1 , #1, the ...
Saturday, November 23, 2013 at 3:53pm

Calculus
Find the point on the line -6 x + 5 y - 1 =0 which is closest to the point ( -4, 1 )
Saturday, November 23, 2013 at 3:22pm

Calculus Area between curves
Actually your answer is correct if it's indefinite integral. :) Now that you got the answer, we can now evaluate it at the given bounds: -(1/3)(8-2x)^(3/2) , at x = -7 to 0 = -(1/3)(8-2(0))^(3/2) - [-(1/3)(8-2(-7))^(3/2)] = -(1/3)(8)^(3/2) - [-(1/3)(22)^(3/2)] = (22/3)*(22...
Friday, November 22, 2013 at 8:23pm

Calculus Area between curves
Evaluate the definite integral: sqrt(8-2x) lower limit=-7 upper limit=0 I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!
Friday, November 22, 2013 at 8:05pm

Calculus
Yes I figured it out thanks
Thursday, November 21, 2013 at 11:16pm

Calculus Help!!!
notice that one parabola opens up, the other down. their intersection ..... x^2 - c^2 = c^2 - x^2 2x^2 = 2c^2 x^2 = c^2 x = ± c ---- y = 0 in both they intersect at (-c,0) and (c,0) using symmetry area = 2∫(2c^2 - 2x^2) dx from 0 to c = 2(2c^2 x - (2/3)x^3) | from...
Thursday, November 21, 2013 at 11:16pm

Calculus-Area between curves
Thanks I figured it out my answer was 9, and it was correct
Thursday, November 21, 2013 at 11:15pm

Kelly - Amy - Chrissy Calculus
Why are you switching names ???? Follow the method I showed you in the other two questions of this type. Did you look at the solutions? Let me know how far you get.
Thursday, November 21, 2013 at 11:08pm

Chris - Amy -Calculus Area between curves
Why are you switching names ?? find the intersection; from the first: x = 3 - 3y into the 2nd: y^2 - (3-3y) = 1 y^2 + 3y - 4 = 0 (y+4)(y-1) = 0 y = -4 or y = 1 in x = 3-3y .... if y = 1, x = 0 ---->(0,1) if y = -4 , x = 15 , ---> (15, -4) Your sketch should look like ...
Thursday, November 21, 2013 at 10:56pm

Calculus-Area between curves
first of all reduce the equation to simplest form: y = 2√x y = 5 y + 2x = 4 Your sketch should look like a "triangle" with one side slightly curved see: http://www.wolframalpha.com/input/?i=plo​t+y%3D2*sqrt%28x%29+%2C+y%3D5+%2C+y%2B2x​%3D4 we need ...
Thursday, November 21, 2013 at 10:41pm

Calculus Help!!!
Find c > 0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 13.
Thursday, November 21, 2013 at 10:32pm

Calculus
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. x+y^2=42, x+y=0
Thursday, November 21, 2013 at 10:29pm

Calculus Area between curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2-x=1
Thursday, November 21, 2013 at 10:22pm

Calculus-Area between curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't ...
Thursday, November 21, 2013 at 10:16pm

Calculus - Integrals
dE/dt = 6,000(t+1)^-3/2 E(t) = -12000(t+1)^-1/2 + c 3000 = -12000+c c = 15000 E(t) = 15000 - 12000/√(t+1) E(15) = 15000 - 12000/4 = 12000
Thursday, November 21, 2013 at 7:54pm

Calculus - Integrals
The projected rate of increase in enrollment at a new college is estimated by dE/dt = 6,000(t+1)^-3/2 where E(t) is the projected enrollment in t years. If the enrollment is 3,000 now (t=0), find the projected enrollment 15 years from now.
Thursday, November 21, 2013 at 5:51pm

Fundamental Theorm of Calculus
That would be ∫[3,6] 3x^2 - 2x dx --------------------------- 6-3
Wednesday, November 20, 2013 at 8:02pm

Fundamental Theorm of Calculus
that would be ∫[1,5] 7/x^2 dx ------------------------ 5-1
Wednesday, November 20, 2013 at 8:00pm

Fundamental Theorm of Calculus
Find the average value of f(x)=7x^{-2} on the interval [1,5]
Wednesday, November 20, 2013 at 7:59pm

Fundamental Theorm of Calculus
Find the average value of f(x)=3x^2-2 x on the interval [3,6]
Wednesday, November 20, 2013 at 7:36pm

Fundamental Theorm of Calculus
since the curve has x-intercepts at ±√7/2, we have ∫[-√7/2,√7/2] 7-4x^2 dx = 7x - 4/3 x^3 [-√7/2,√7/2] = 14√7 / 3
Wednesday, November 20, 2013 at 5:27pm

Fundamental Theorm of Calculus
Use a definite integral to find area of the region under the curve y=7-4x^2 and above the x-axis. Thanks in advance!
Wednesday, November 20, 2013 at 5:13pm

Calculus Fundamental Theorem
Hello,? By calculus time you should be well aware that |x-13| ≠ x+13 |x-13| makes a big difference. For x<13, |x-13| = -(x-13) For x>=13, |x-13| = x-13 So, that means that we have to break the interval into two parts: ∫[0,22] x-13 dx = ∫[0,13] -(x-13) ...
Wednesday, November 20, 2013 at 4:59pm

Calculus Fundamental Theorem
Thank you!
Wednesday, November 20, 2013 at 4:54pm

Calculus Fundamental Theorem
∫[-√2,√2] (t+8)(t^2+3) dt = ∫[-√2,√2] t^3 + 8t^2 + 3t + 24 dt = 1/4 t^4 + 8/3 t^3 + 3/2 t^2 + 24t [-√2,√2] = 176/3 √2
Wednesday, November 20, 2013 at 4:53pm

Calculus Fundamental Theorem
Original problem was: absolute value x-13, because of absolute value I made it x+13
Wednesday, November 20, 2013 at 4:52pm

Calculus Fundamental Theorem
I tried that way, my answer was wrong
Wednesday, November 20, 2013 at 4:51pm

Calculus Fundamental Theorem
∫[0,22] x+13 dx = 1/2 x^2 + 13x [0,22] = (1/2 * 22^2 + 13*22)-(1/2 * 0^2 + 13*0) = 528
Wednesday, November 20, 2013 at 4:50pm

Calculus Fundamental Theorem
Evaluate the definite integral. function: (t+8)(t^2+3) with respect to variable t lower limit: -sqrt(2) upper limit: sqrt(2)
Wednesday, November 20, 2013 at 4:49pm

Calculus Fundamental Theorem
Evaluate the definite integral. function: x+13 with respect to variable x lower limit:0 upper limit:22
Wednesday, November 20, 2013 at 4:47pm

Pre-Calculus
wow that was a silly mistake I didn't see the x next to 500 and thought I could just subtract it from 1000 thank you!
Wednesday, November 20, 2013 at 1:03pm

Pre-Calculus
well, you should see right off that since 51x^2 is always positive, you can never have 51x^2 = -500. So, let's see what happened. 50x^2 + 1000 = 500x - x^2 51x^2 - 500x + 1000 = 0 Now just use the quadratic formula to find the roots.
Wednesday, November 20, 2013 at 6:35am

Math (Calculus)
without a diagram, I assume point A is an anchor in between the towers. If its distance from T1 is x, then the cable length is d = √(60^2+x^2) + √(20^2+(140-x)^2) to find the minimum, we want dd/dx = 0 That happens when x=105
Wednesday, November 20, 2013 at 6:30am

Math (Calculus)
Tower 1 is 60 ft high and tower 2 is 20 ft high. The towers are 140 ft apart. A guy wire is to run from point A to the top of each tower. [See a picture of this situation on page 274 of the textbook.] How many feet from tower 1 should point A be so that the total length of guy...
Wednesday, November 20, 2013 at 2:30am

Pre-Calculus
A piece of equipment has cost function C(x)=50x^2 + 1000 and its revenue function is R(x)= 500x - x^2, where x is in thousands of items. What is the least number of items that must be sold in order to break even? 50x^2 + 1000 = 500x - x^2 50x^2 + x^2 + 500= 0 51x^2 + 500 = 0 ...
Wednesday, November 20, 2013 at 1:38am

Pre-Calculus
Wow, I didn't realize that finding (x-2) was already finding the equation of the line. Thanks!
Tuesday, November 19, 2013 at 11:00pm

Pre-Calculus
did you notice that f(x) = (x-9)(x-2)/(x-9) ?? = x-2, x ≠ 9 So for all values of x except x=9 the given function reduces to f(x) = x-2 which is a straight line. when x = 9, the original gives you 0/0 which is indeterminate, while the reduced function gives you (9,7) So ...
Tuesday, November 19, 2013 at 10:34pm

Pre-Calculus
The graph of f(x)= (x^2 - 11x + 18)/(x-9) consists of a line and a hole. Find the equation of the line and the coordinates of the hole. I really have no idea how to do this. I tried factoring it and ended up with (x-2), but I don't know how that helps me. please help
Tuesday, November 19, 2013 at 10:05pm

Calculus
f" = 5x+2sinx f' = 5/2 x^2 - 2cosx + c f'(0)=3, so 0-2+c=3 c=5 so, f' = 5/2 x^2 - 2cosx + 5 f = 5/6 x^3 - 2sinx + 5x + c f(0)=3, so 0-0+0+c=3 c=3 so, f=5/6 x^3 - 2sinx + 5x + 3 I assume you can now find f(3)
Tuesday, November 19, 2013 at 9:29pm

Calculus
f" = -4sin2x f' = 2cos2x + c 2+c = 4 c=2 so, f' = 2cos2x + 2 f = sin2x + 2x + c 0+0+c = -6 c = -6 so, f = sin2x + 2x - 6
Tuesday, November 19, 2013 at 9:26pm

Calculus
Given f''(x)= -4sin(2x) and f'(0)=4 and f(0) =-6. Find f(pi/2)
Tuesday, November 19, 2013 at 9:17pm

Calculus
Consider the function f(x) whose second derivative is f''(x)=5x+2sin(x). If f(0)=3 and f'(0)=3, what is f(3)?
Tuesday, November 19, 2013 at 7:21pm

12th Calculus
help me!
Tuesday, November 19, 2013 at 2:15pm

Pre-Calculus
oh yeah thanks I forgot that the reciprocal had to be negative
Tuesday, November 19, 2013 at 1:33pm

Pre-Calculus
since the line is perpendicular, their slopes are negative reciprocals of each other, that means, they must be OPPOSITE in sign so first slope = -3/2 2nd slope = +2/3 y-2 = (2/3)(x-4) 3y-6 = 2x-8 3y = 2x - 2 y = (2/3)x - 2/3
Tuesday, November 19, 2013 at 10:43am

Pre-Calculus
What is the equation of the line passing through (4, 2) and perpendicular to the line passing through the points (9,7) and (11,4)? Here's what I did: (7-4)/(9-11) = -3/2 y-2= (-2/3)(x-4) y-2=(-2x/3) + (8/3) y = (-2x/3) + 14/3 but the answer key says that it's y=(2x/3...
Tuesday, November 19, 2013 at 8:22am

Calculus
If the printed area has width x and height y, xy = 386 The total area is thus a = (x+8)(y+4) = (x+8)(386/x + 4) = 4x + 418 + 3088/x max area is where da/dx=0 da/dx = 4 - 3088/x^2 so, da/dx = 0 when x = 2√193 Now just evaluate y.
Tuesday, November 19, 2013 at 5:25am

Calculus
if the rectangular window has width 2x and height h, then the area of the whole shape is a = 2xh + pi/2 x^2 You also know that x+2h+pi*x = 25, so solve for h and plug into the formula for area. The set da/dx = 0.
Tuesday, November 19, 2013 at 12:33am

Calculus
since F(1) = 0, we have -7 + 6/5 + c = 0 which you can plug in to evaluate F(2)
Tuesday, November 19, 2013 at 12:29am

Calculus
We need the whole area to be twice the area under the line y=b. y(1/2) = 1 y(√b/2) = b So, integrating along x, we need ∫[0,1/2] 1-4x^2 dx = 2∫[0,√b/2] b-4x^2 dx Now just evaluate the integrals and solve the polynomial for b.
Tuesday, November 19, 2013 at 12:24am

brief calculus
You're right.
Monday, November 18, 2013 at 11:30pm

brief calculus
x/(x − 6)^2 I get ln(x-6)- 6/(x-6)+ C but it isn't correct, what should the answer be?
Monday, November 18, 2013 at 11:22pm

Calculus
Write it as f(x) = 7x^-2 - 6x^-6 F(x)= -7x^-1 + (6/5)x^-5 + c F(2) = -7/2 + (6/5)/2^5 + c = ......
Monday, November 18, 2013 at 10:19pm

Calculus
let the side of the square be x let the side of the triangle be y 4x + 3y = 40 x = (40-3y)/4 height of triangle from the 30-60-90 triangle ratio = √3/2 y area of triangle = (1/2)(y)(√3/2 y) = (√3/4)y^2 area of square = [(40-3y)/4]^2 = (1600 - 240y + 9y^2)/16...
Monday, November 18, 2013 at 9:20pm

Calculus
Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 1 into two regions with equal area. (Round your answer to two decimal places.)
Monday, November 18, 2013 at 8:45pm

Calculus
Consider the function f(x)=(7/x^2)-(6/x^6). Let F(x) be the antiderivative of f(x) with F(1)=0. Then F(2) equals _____.
Monday, November 18, 2013 at 8:16pm

Calculus
A piece of wire 40 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much of the wire should go to the square to minimize the total area enclosed by both figures?
Monday, November 18, 2013 at 8:08pm

Calculus
A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 25 ft, find the dimensions of the window so that the greatest possible amount of light is ...
Monday, November 18, 2013 at 8:07pm

Calculus
The top and bottom margins of a poster are 2 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 386 square centimeters, find the dimensions of the poster with the smallest area.
Monday, November 18, 2013 at 8:06pm

Calculus
10x−6y√(x^2+1)dy/dx=0 6y√(x^2+1) dy/dx = 10x 6y dy = 10x/√(x^2+1) dx 3y^2 = 10√(x^2+1) + c Since y(0) = 4, 3*16 = 10+c c = 38 now you have it.
Monday, November 18, 2013 at 2:04pm

Calculus
Solve the separable differential equation 10x−6ysqrt(x^2+1)dy/dx=0. with the initial condition y(0)=4.
Monday, November 18, 2013 at 1:55pm

Calculus
I did plot the curve on Wolfram and got full curve across 3pi. Curve length is also shown full at 3pi but period is shown as 6pi.I can't understand this.
Monday, November 18, 2013 at 6:47am

Calculus
since the curve is a parabola, there is no inflection point. Since #1 #2 do not provide any useful data, the other items are hard to figure.
Monday, November 18, 2013 at 6:07am

Calculus
No idea. Did you visit wolframalpha.com and look at the graph? If you enter arc length sin^3(x/3), x = 0 .. 3pi it will give you the arc length, but that's only for 1/2 period.
Monday, November 18, 2013 at 6:04am

Calculus
For calculating complete length of curve r=sin^3(theta/3) I got right answer on integrating the arclength integrand from 0 to 3pi. How could that be?
Monday, November 18, 2013 at 12:48am

Calculus
2a is correct the period of sin^3(x) is the same as for sin(x). SO, sin^3(x/3) has period 2pi/(1/3) = 6pi sure. sin(x) does just that. sin(pi)=0, but it takes 2pi to complete the period. in polar coordinates, sin(theta/2) goes through the origin twice in its period of 4pi. ...
Monday, November 18, 2013 at 12:24am

Calculus
1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after theta=0, the function ...
Monday, November 18, 2013 at 12:18am

Pre-Calculus
thanks...though I asked for (-3,-1).
Sunday, November 17, 2013 at 10:06pm

calculus
f is a function from R to R such that f(a) is rational when a is irrational and f(a) is irrational when a is rational . prove that f cannot be continuous
Sunday, November 17, 2013 at 7:00pm

calculus
Find a function f from R to R such that f is continuous at only one point?
Sunday, November 17, 2013 at 6:52pm

Calculus
Suppose that a population develops according to the logistic equation dP/dt = 0.06P−0.0001P^2 where t is measured in weeks. 1) The carrying capacity is . 2) The growth rate k is . Use your calculator to sketch a direction field for this equation. Sketch the solutions ...
Sunday, November 17, 2013 at 5:27pm

Calculus
ohaganbooks is offering a wide range of online books, including current best-sellers. a colleague has determined that the demand for the latest best selling book is given by q=(-p^2)+33p+9 (18<p<28) copies sold per week when the price is p dollars. can you help me ...
Sunday, November 17, 2013 at 4:59pm

calculus
time to go a distance d is d/v so the cost is (d/v)(kv^3) = dkv^2 Looks like k=0 makes cost=0. I suspect a typo.
Sunday, November 17, 2013 at 4:42pm

calculus
(a) time = distance/speed, so a trip of 1000km takes 1000/v hours. So, the cost is c(v) = (1000/v)(160 + 1/100v^3) (b) Now just find minimum cost where dc/dv = 0 dc/dv = 20(v^3-8000)/v^2 so minimum cost when v=20 (c) Since min occurs at v=20, if top speed is 16, then minimum ...
Sunday, November 17, 2013 at 4:39pm

Calculus - Optimization
if the expensive side is x and the other dimension is y, then the cost c is c = 4(x+2y) + 12x But, we know the area is xy=800, so y = 800/x and the cost is now c = 4(x+1600/x) + 12x minimum cost when dc/dx=0, so we need dc/dx = -16(400-x^2)/x^2 dc/dx=0 when x=20, so the fence ...
Sunday, November 17, 2013 at 4:24pm

Calculus - Optimization
A fence is to be built to enclose a rectangular area of 800 square feet. The fence along 3 sides is to be made of material $4 per foot. The material for the fourth side costs $12 per foot. Find the dimensions of the rectangle that will allow for the most economical fence to be...
Sunday, November 17, 2013 at 3:38pm

calculus
The cost of running a ship at a constant speed of v km/h is 160 + 1/100*v^3 dollars per hour. a)Find the cost of a journey of 1000km at a speed of v km/h. b)Find the most economical speed for the journey, and the minimum cost. c)If the ship were to have maximum speed of 16 km/...
Sunday, November 17, 2013 at 8:37am

calculus
A traveller employs a man to drive him from Sydney to Melbourne. Running costs of the car, which are also paid by the traveller, are k*v^3 dollars per hour, v is the speed and k is a constant. Find the uniform speed that will minimize the total cost of the journey.
Sunday, November 17, 2013 at 8:33am

Pre-Calculus
in algebra I you learned that (a+b)^2 ≠ a^2 + b^2 That's still true, even if you're taking pre-cal! :-) 2(x^2+y^2)^2=25(x^2−y^2) 2 * 2(x^2+y^2) (2x+2yy') = 25(2x-2yy') y' = -x(4x^2+4y^2-25) -------------------------- y(4x^2+4y^2+25) So, at (-3,1...
Sunday, November 17, 2013 at 7:40am

Pre-Calculus
Find the slope of the tangent line to the curve 2(x^2+y^2)^2=25(x^2−y^2) at the point (−3,−1)? Here's what I did: 2(x^4 + y^4) = 25(x^2-y^2) 2x^4 + 2y^4 = 25x^2 - 25y^2 8x^3 + 8y^3(dy/dx) = 50x - 50y(dy/dx) d/dx(8y^3 + 50y) = 50x - 8x^3 d/dx = (50x-8x^3...
Sunday, November 17, 2013 at 2:31am

Calculus
height: 240 max height at t=11 the curve is just a parabola. Don't forget your algebra I just because you're taking calculus! :-)
Saturday, November 16, 2013 at 10:46pm

Calculus
I throw a ball off the roof. It travels s = 240+22t-t^2. S is the balls distance after I release it. How tall is the building? How high above the ground did the ball get?
Saturday, November 16, 2013 at 10:20pm

Pre-Calculus
thanks
Saturday, November 16, 2013 at 9:47pm

Calculus
(4n-7)/(4n+9) = [(4n+9) -16]/(4n+9) = 1 - 16/(4n+9) The limit thus exists and is equal to 1. Clearly, for every epsilon > 0, there exists an N such that A_n will be within epsilon of the limiting value for all n > N, taking N = 4/epsilon will do.
Saturday, November 16, 2013 at 7:43pm

Calculus
|a_n| = |sin(2/n)| Since |sin(x)| <= |x|, we have: |a_n| <= 2/n The limit if |a_n| for n to infinity is thus 0, which then implies that the limit of a_n is zero. You can easily make this rigorous, for every epsilon > 0, you can using the above inequality find an N ...
Saturday, November 16, 2013 at 7:36pm

Calculus - Optimization
Good catch, Steve. Thank you.
Saturday, November 16, 2013 at 6:29pm

Calculus
Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. A (sub n)=((4n-7)/(4n+9))
Saturday, November 16, 2013 at 5:42pm

Calculus
Determine whether the sequence is divergent or convergent. a(sub n)=(−1)^n(sin(2/n))
Saturday, November 16, 2013 at 5:39pm

Calculus - Optimization
but an 8x8x8 box has length+girth = 8+32 = 40 inches, so it will not work. We need to optimize s^2(24-4s) since a square has 4 sides. v = 24s^2 - 4s^3 v' = 48s - 12s^2 v'=0 when s=4 So, a 4x4x8 box has max volume. Do (B) similarly
Saturday, November 16, 2013 at 5:25pm

Calculus - Optimization
Let the size of the square (cross-section) be s. Then we need to maximize V=s²(24-2s) with respect to s. First find the derivative and equate to zero: dV/ds = 48s-6s²=0 means s=0 or s=8 s=0 corresponds to a minimum volume and s=8 corresponds to a maximum volume. So ...
Saturday, November 16, 2013 at 5:18pm

Calculus - Optimization
A parcel delivery service a package only of the length plus girth (distance around) does not exceed 24 inches. A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has a maximum volume. What is the maximum ...
Saturday, November 16, 2013 at 4:24pm

Physics
So f = (9.8)(153) and d=.794? And then do I solve for x? How would I solve for x? ( never taken calculus)
Friday, November 15, 2013 at 8:07pm

Physics
So f = (9.8)(153) and d=.794? And then do I solve for x? How would I solve for x? ( never taken calculus)
Friday, November 15, 2013 at 7:58pm

Calculus
c(x) = (10 - x) + 1.4√(x^2 + 25) dc/dx = -1 + 1.4(2x)/2√(x^2+25) = 1.4x/√(x^2+25) - 1 because d/dx √u = 1/2√u du/dx
Friday, November 15, 2013 at 5:41pm

Calculus
What would the derivative of c(x) = (10 - x) + 1.4*sqrt((x^2) + 25)
Friday, November 15, 2013 at 5:29pm

Calculus
8yy' = x 8y dy = x dx 4y^2 = 1/2 x^2 + c 4(16) = 1/2 (64) + c c = 32 4y^2 = 1/2 x^2 + 32 8y^2 - x^2 = 64 y^2/8 - x^2/64 = 1
Friday, November 15, 2013 at 1:43pm

Calculus
y' = (x-7)e^(-2y) e^2y y' = x-7 e^2y dy = x-7 dx 1/2 e^2y = 1/2 x^2 - 7x + c e^2y = x^2 - 14x + c 2y = log(x^2-14x+c) y = 1/2 log(x^2-14x+c) Pick a c that fits a particular solution
Friday, November 15, 2013 at 1:40pm

Calculus
Solve the seperable differential equation 8yyŒ =x. Use the following initial condition: y(8)=4. Express x^2 in terms of y.
Friday, November 15, 2013 at 1:37pm

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