Saturday

April 19, 2014

April 19, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Pre-Calculus**

I know that sin (π/6) or sin 30° = .5 so (π/4)(x-6) = π/6 or (π/4)(x-6) = π - π/6 = 5π/6 (in II) case 1 times 12 and divide by π 3(x-6) = 2 3x - 18 = 2 3x=20 x = 20/3 case 2 times 12 , and divide by π 3(x-6) = 10 3x - 18=10 3x...
*Saturday, January 11, 2014 at 9:02pm*

**Pre-Calculus**

Here is correct question: Solve sin (pi/4*(x-6))=0.5, 0<x<360
*Saturday, January 11, 2014 at 8:10pm*

**Pre-Calculus**

Find an equation of the line that satisfies the given conditions Through (-2,4); slope -1 How would you solve that by using the point slope form and writing it in the general equation of a line? ax +by + c =0
*Saturday, January 11, 2014 at 7:30pm*

**Pre-Calculus**

y = -1 x + b 4 = -1(-2) + b 2 = b so y = -x + 2
*Saturday, January 11, 2014 at 7:21pm*

**Pre-Calculus**

Find an equation of the line that satisfies the given conditions Through (-2,4); slope -1
*Saturday, January 11, 2014 at 7:10pm*

**Pre-Calculus**

maybe you mean sin [pi(x-6)/4] sin (pi/6) = .5 (x-6)/4 = 1/6 6 x - 36 = 4 x = 40/6 = 20/3
*Saturday, January 11, 2014 at 5:43pm*

**Pre-Calculus**

Solve sin (pi/4(x-6))=0.5, 0<x<360
*Saturday, January 11, 2014 at 4:18pm*

**Calculus**

The following table displays the number of HIV diagnoses per year in a particular country. Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 Diagnoses 2512 2343 2230 2113 2178 2495 2496 2538 2518 a) Using Curve Expert or another curve modelling program, determine an equation ...
*Saturday, January 11, 2014 at 12:17am*

**calculus**

y= (x) / ((x^2)-1)^(-1/2) do you have a typo? That is the same as: y = x (x^2-1)^(+1/2) dy/dx = x(1/2)(x^2-1)^-.5(2x) +(x^2-1)^.5 = x^2(x^2-1)^-.5 + (x^2-1)^.5
*Friday, January 10, 2014 at 5:58pm*

**calculus**

there are tutoring chatrooms, but not here. To play around with derivatives and much more, visit wolframalpha.com. For this problem, enter derivative (x) / ((x^2)-1)^(-1/2) and it will show you (2x^2-1)/√(x^2-1) and you can click on the "step-by-step solution" ...
*Friday, January 10, 2014 at 5:49pm*

**calculus**

i need to find the derivative using chain rule and show steps: y= (x) / ((x^2)-1)^(-1/2) and is it possible to have free tutoring online like through a chatroom? i am in need of help
*Friday, January 10, 2014 at 5:44pm*

**Pre calculus**

since distance = speed * time, 50x + 40(1-x) = 48
*Thursday, January 9, 2014 at 6:16pm*

**4 CALCULUS study problem**

Never mind! I solved them all and got them right. :)
*Thursday, January 9, 2014 at 2:14pm*

**4 CALCULUS study problem**

1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/...
*Thursday, January 9, 2014 at 1:23pm*

**Pre calculus**

A woman went for a 48km drive in her car, partly in town in the country. She drove at an average of 40km/hr in town and an average speed of 50km/hr in the country. If the drive took her one hour altogether then how long did she drive in the country?
*Thursday, January 9, 2014 at 12:53pm*

**calculus**

f'(x)=[-5* 1/2]*[1/sqrt(x^2-4x+1)] * (2x-4)
*Wednesday, January 8, 2014 at 10:22pm*

**algebra**

Since you have no calculus (the easy way), lets do it this way. Obviously, it is a parabola going up, then down. So look for the half way point between both x intercepts. So find the x intercepts. 0=-480(F-3.25)^2 + 5270 F^2+6.5F+3.25^2-5070/480=0 check that, then put it in ...
*Wednesday, January 8, 2014 at 10:21pm*

**calculus**

updating previous question so easier to understand im using chain rule to answer this question: f(x) = -5 sqrt(x^2 - 4x + 1) what is f'(x)? tried to use calculator to get the answer to base my answer off of but this is a nonreal number. does this mean i will have to use ...
*Wednesday, January 8, 2014 at 9:57pm*

**calculus**

For the life of me, I dont see a question. Am I missing something? I assume the power of the quadratic is 1/2, it is hard to tell from your grouping symbols.
*Wednesday, January 8, 2014 at 9:30pm*

**calculus**

im using chain rule to answer this question: -5 ((x^2) - 4x + 1)^(1)/(2) tried to use calc to get answer to base off of but this is a nonreal number. does this mean i will have to use imaginary numbers? please help
*Wednesday, January 8, 2014 at 9:07pm*

**Pre Calculus**

Been there, done that a few minutes ago http://www.jiskha.com/display.cgi?id=1389226155
*Wednesday, January 8, 2014 at 8:47pm*

**Pre Calculus**

Find the exact value of sin(1/2arcsin(-7/25))
*Wednesday, January 8, 2014 at 8:23pm*

**Pre-Calculus**

sin A = x/sqrt(x^2+4) draw that right triangle side adjacent to A is sqrt (x^2+4-x^2) = sqrt 4 = 2 so cos A = 2/sqrt(x^2+4) sec A = 1/cos A = sqrt(x^2+4)/2
*Wednesday, January 8, 2014 at 7:47pm*

**Calculus**

I bet you mean y = x/(x+3) The denominator is 0 when x = -3 so x = -3 is not in our domain. Otherwise the domain is all real numbers. Inverse x = y/(y+3) x y + 3 x = y y(x-1) = -3 x y = -3x/(x-1) domain is all real numbers except x = 1 f[f^(-1)] = f[-3x/(x-1) ] = [-3x/(x-1...
*Wednesday, January 8, 2014 at 7:41pm*

**Calculus**

Let f(x)=(x/x+3). Find Domain Find f^(-1) and its domain Verify f*f^(-1)=f^(-1)*f=x Find (f(x+h)-f(x))/h.
*Wednesday, January 8, 2014 at 7:15pm*

**Pre-Calculus**

Write an algebraic expression in x for sec(arcsin(x/sqrt(x^(2) +4)))
*Wednesday, January 8, 2014 at 7:10pm*

**Pre-Calculus**

y = a x^2 + b x + c if x = 1 3 = a + b + c if x = 2 5/2 = 4 a + 2 b + c if x = -1 1 = a - b + c we have 3 linear equations with 3 unknowns. There are many ways to solve but lets try to eliminate c first 1 a +b + c = 3 4 a+2b + c = 5/2 ------------------ subtract - 3 a - 1 b = ...
*Wednesday, January 8, 2014 at 1:55pm*

**Pre-Calculus**

Find a quadratic function f such that f(1)=3, f(2)=5/2, f(-1)=1 I came out with 3 a+b+c=3, 4a+2b+c=5/2, a-b+c=1. And I know that I should do synthetic division but I am stuck.
*Wednesday, January 8, 2014 at 1:21pm*

**Calculus**

You have to specify your domain. For log(x-1) you need x>1 For log(1-x) you need x<1 May times you will find it written that ∫ dx/x = log |x| + C just for this reason.
*Wednesday, January 8, 2014 at 5:51am*

**Calculus**

Integrate x dx/(1-x). I have proceeded thus- Int xdx/(1-x)=int -(x-1+1)/(x-1) =-Int[1+ 1/(x-1)]dx =-Int dx-Int dx/(x-1) =-x-log(x-1). On differentiating, we get original expression- d/dx[-x-log(x-1)]=-1-1/(x-1)=-x/(x-1)=x/(1-x). However, the answer in the book is -x-log...
*Wednesday, January 8, 2014 at 4:04am*

**AP Calculus AB**

y´=3(cosx)²(-sinx) -3(sinx)²cosx= = - 3 sinx cosx(cosx+sinx)
*Friday, January 3, 2014 at 3:13pm*

**AP Calculus AB**

Find the derivative of: y=(cosx)^3-(sinx)^3
*Friday, January 3, 2014 at 12:23pm*

**AP Calculus AB**

1. assign x and y values to the length and width of each box within the pens, and use the formula 600=9x+8y to isolate the y value (i got the 8 and 9 by counting the number of length and width values existed in the diagram) you should get y=(600-9x)/8 2. then, since they asked...
*Thursday, January 2, 2014 at 5:09pm*

**math**

I have a hunch you actually mean: g(x)=x/ (x-7) because that is much more exciting than what you typed. Now clearly if x = 7 we have a serious problem with this function. It has a zero for its denominator and is therefore undefined at x = 7. Now what if x = 8 ? g(8) = 8/1 = 8 ...
*Friday, December 27, 2013 at 2:15pm*

**Pre calculus**

First step, add/subtract multiples of 180° until you have an angle between -180° and 180° If θ is in QI, you're done. If θ is in QIII, reflect through the origin. That is, add 180°. Done. If θ is in QIV, y<0, so reflect through the x-axis...
*Thursday, December 26, 2013 at 12:12am*

**Pre calculus**

1. Reference angle for -280 degrees? I do not understand why you add 360???? 2. Theta=? If reference angle =60 and theta terminates in quadrant 3. What does terminate mean? Please show work and explain why. My priority is to understand the concept ... 3. Exact value for sec ...
*Wednesday, December 25, 2013 at 10:30pm*

**calculus-changing voltage**

V = IR V is increasing at the rate of 1 volt/sec while I is decreasing at the rate of 1/3 amp/sec. Let t = time in sec What is dV/dt? What is dI/dt? Find the rate at which R is changing when V = 12 volts and I = 2 amp
*Monday, December 23, 2013 at 12:45pm*

**calculus**

if these are logs base ten 10^log x = x = 10^5.3027 = 10^5 * 10^.3027 = 10^5 * 2.007705 or 2.008 * 10^5
*Saturday, December 21, 2013 at 5:25pm*

**calculus**

Evaluate to four significant digits log x = 5.3027
*Saturday, December 21, 2013 at 3:48pm*

**CALCULUS - Check my answers :)**

#5 At a time of t seconds after the boy passed under the balloon, distance covered by boy = 15t height of balloon = 5 + 5t let the distance between them be d ft d^2 = (5t+5)^2 + (15t)^2 2d dd/dt = 2(5t+5)(5) + 2(15t)(15) dd/dt =( 5(5t+5) + 15(15t) )/d when t = 3, d^2 = 20^2 + ...
*Thursday, December 19, 2013 at 10:30pm*

**CALCULUS - Check my answers :)**

#4, the way you typed it, ... y' = 2x/3 when x = 8, y' = 16/3, which is not found in any of your choices as the slope if you meant , f(x) = x^(2/3) then y' = (2/3)x^(-1/3) = (2/3)(1/x^(1/3)) when x = 8 y' = (2/3)(1/2) = 1/3 the only one that has a slope of 1/2 ...
*Thursday, December 19, 2013 at 10:18pm*

**CALCULUS - Check my answers :)**

And 5 would be 25?
*Thursday, December 19, 2013 at 9:28pm*

**CALCULUS - Check my answers :)**

Is number 4 the last one?
*Thursday, December 19, 2013 at 9:24pm*

**CALCULUS - Check my answers :)**

Well Damon, calculus isnt supposed to be.. interesting. Haha
*Thursday, December 19, 2013 at 9:02pm*

**CALCULUS - Check my answers :)**

3. If 12 ft^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box. 2 ft^3 4 ft^3 5 ft^3 *8.5 ft^3 9 ft^3 ================================= side length s and height h A = 4 s h + s^2 = 12 4 s h = 12 - s^2 h = (3...
*Thursday, December 19, 2013 at 7:49pm*

**CALCULUS - Check my answers :)**

2. Determine the equation of the normal line to y = x^2 + 5 at the point (2, 9) y=-1/4x + 17/4 *y=1/2x + 8 y=4x+1 y=4x-34 y=-1/4x + 19/2 ================================== dy/dx = 2x at x = 2, slope = 2*2 =4 so m of normal= -1/slope = -1/4 y = -1/4 x + b 9 = -1/4 *2 + b 9 = -1...
*Thursday, December 19, 2013 at 7:36pm*

**CALCULUS - Check my answers :)**

1. Determine which of the following is true for the function f(x) = x^3 +5x^2 - 8x +3. I. f(x) has a relative minimum at x = 2/3. II. f(x) has a relative maximum at x = 2/3. III. f(x) has a zero at x = 2/3 I only *II only III only I and III only II and III only...
*Thursday, December 19, 2013 at 7:31pm*

**CALCULUS - Check my answers :)**

I'd appreciate corrections for the ones I get wrong. :) Thanks! 1. Determine which of the following is true for the function f(x) = x^3 +5x^2 - 8x +3. I. f(x) has a relative minimum at x = 2/3. II. f(x) has a relative maximum at x = 2/3. III. f(x) has a zero at x = 2/3 I ...
*Thursday, December 19, 2013 at 7:18pm*

**Calculus**

Since sinc(1) = cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n) the limit here is cos(1)sinc(1) = cos(1)sin(1) = 0.4546... You can read about sinc(x) in various places. This function doesn't usually pop up in introductory calculus courses.
*Thursday, December 19, 2013 at 3:52pm*

**Calculus**

what is the limit n to infinite of cos1*cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)
*Thursday, December 19, 2013 at 1:03pm*

**calculus**

pick any sides x and y. You will see that the largest area for a given perimeter is always a square. In this case, y = 250-x a = xy = x(250-x) = 250x - x^2 da/dx = 250-2x da/dx=0 when x=125 So, the rectangle is 125x125, a square A good check on all these fencing problems, even...
*Thursday, December 19, 2013 at 12:44am*

**calculus**

differentiate, not derive. f = ∫[√x,2] ln t dt = t(ln t - 1) [√x,2] = (2(ln2-1)-(√x)(1/2 lnx - 1) f' = logx / 4√x you can do it more directly; see wikipedia's article on differentiation under the integral sign.
*Thursday, December 19, 2013 at 12:41am*

**calculus**

domain of f is x >= 2 Note that the range of f inverse is also f >= 2, since x^2 >= 0 for all x. Since f(x) = √(x-2), f >= 0 because √4 = 2, not ±2! So, since the range of f is f >= 0, the domain of f inverse is also x >= 0. f inverse is ...
*Thursday, December 19, 2013 at 12:32am*

**calculus**

f(x)=sqrt(x-2) what is the inverse of f? I got inverse f(x)= x^2+2 however, you need to specify the domain of x. the answer key says when x=>0 can you please explain why? I thought it would be when x=>2 ?
*Wednesday, December 18, 2013 at 7:16pm*

**calculus**

if f(x)= the integral of lntdt from sqrt(x) to 2, what is f'(x)? Do you solve the integral first then derive it? And if so, wouldn't that just give you the function? Please help
*Wednesday, December 18, 2013 at 6:35pm*

**calculus**

I got half of this problem wrong and I DO NOT know where and how to fix. I cannot use my calculator and have to show my work. Question: You have a 500 metre roll of fencing and a large field. You want to construct a rectangular playground area. a.) using optimization ...
*Wednesday, December 18, 2013 at 6:27pm*

**calculus**

In the process of electrolysis, electrical power is used to separate water into oxygen and hydrogen molecules via the reaction: H2O --> H2 + ½O2 This is very much like running a hydrogen fuel cell in reverse. We assume that only the activation potential for the ...
*Wednesday, December 18, 2013 at 5:34pm*

**calculus**

Help on last question!
*Wednesday, December 18, 2013 at 5:26pm*

**calculus**

Your 1st two answers are correct, but I'm not up on specific charge stuff.
*Wednesday, December 18, 2013 at 4:55pm*

**calculus**

A competing technology uses a photovoltaic cell to charge a battery which can then be used to power a LED light. To compare these technologies, we make some reasonable assumptions. We assume the solar cell is 10 cm on a side, and that it has an efficiency of 15%. We assume the...
*Wednesday, December 18, 2013 at 3:57pm*

**Pre-Calculus**

It is going to be a -1/5
*Wednesday, December 18, 2013 at 1:09pm*

**Pre-Calculus**

y= ((2-3x^(1/x))/(2+3x^(1/x)) Is the horizontal asymptote y= 1 or y= -1?
*Wednesday, December 18, 2013 at 10:45am*

**Calculus (check my answer)**

Oh, sorry !
*Tuesday, December 17, 2013 at 8:15pm*

**Calculus (check my work)**

C = 2 pi r = 64 {circumference is 2 pi r} so r = 64/(2pi) = 10.19 dC/dr = 2 pi { just taking the derivative of 2 pi r} so dC = 2 pi dr {multiplied both sides by r} .9 = 2 pi dr (used .64 by mistake) {the .9 cm was given in the problem statement for dC} so dr = .1433 {divided ...
*Tuesday, December 17, 2013 at 8:13pm*

**Calculus**

dT/dx=2.5+x dT=2.5dx+xdx x=25, dx=1 solve for dT
*Tuesday, December 17, 2013 at 7:51pm*

**Calculus**

The total stopping distance T of a vehicle is T=2.5x+0.5x^2 where T is in feet and x is the speed in miles per hour. Approximate the change and percent change in total stopping distance as sped changes from x=25 to x=26 miles per hour. I started taking the derivative, but that...
*Tuesday, December 17, 2013 at 7:47pm*

**Calculus (check my work)**

note that if I had done the second part first I would have known that dA/A = 2 (dC/C) dA/A = 2(.9/64) = .028 which is 2.8 percent which I got after much algebra
*Tuesday, December 17, 2013 at 7:41pm*

**Calculus (check my work)**

You realize I'm a terrible Calculus student? I don't understand the concepts that the other students can pick up on right away. I don't really get why you did any of what you did, and there aren't any examples like this one in our book for me to look back on.
*Tuesday, December 17, 2013 at 7:27pm*

**Calculus (check my work)**

well, let me do the second part slowly C = 2 pi r dC = 2 pi dr so dC/(2 pi r) = dr/r or dC/C = dr/r which any architect will tell you dA = C dr dA/A = C dr/A dA/A = C dr/(pir^2) dA/A = 2 pi r dr/(pi r^2) dA/A = (2)dr/r SO dr/r = (1/2) dA/A Important if dA/A = 3% then dr/r = ...
*Tuesday, December 17, 2013 at 7:26pm*

**Calculus (check my work)**

I made that a lot harder than it needed to be but follow it and you will be able to shorten it.
*Tuesday, December 17, 2013 at 7:18pm*

**Calculus (check my work)**

Okay, I really don't understand how you did any of that. What's the point of half of it? I tried to do some of what you did, but I got different numbers for my answers.
*Tuesday, December 17, 2013 at 7:14pm*

**Calculus (check my work)**

part 2 (dA/A)100 = 3 (2 pi r dr /pi r^2)100 = 3 or dr/r = .015 but dr = dC/(2pi) so dC/2 pi r = .015 or dC/C = .015 so 1.5 % error in C gives 3% error in A
*Tuesday, December 17, 2013 at 7:11pm*

**Calculus (check my work)**

C = 2 pi r = 64 so r = 64/(2pi) = 10.19 dC/dr = 2 pi so dC = 2 pi dr .64 = 2 pi dr so dr = .10186 A = pi r^2 = pi (10.19)^2 = 325.95 dA/dr = 2 pi r (of course:) dA = 2 pi r dr dA = 2 pi (10.19)(.10186) dA = 6.52 percent error = (6.52/325.95)100 = 2%
*Tuesday, December 17, 2013 at 7:02pm*

**Calculus (check my work)**

Also, it says to estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed 3%. How do I do this?
*Tuesday, December 17, 2013 at 6:55pm*

**Calculus (check my answer)**

No, this is a different part of that same problem. It asks for the surface area this time, not volume.
*Tuesday, December 17, 2013 at 6:53pm*

**Calculus (check my work)**

The measurement of the circumference of a circle is found to be 64 centimeters, with a possible error of 0.9 centimeter. Approximate the percent error in computing the area of the circle. MY ANSWER 8.83%
*Tuesday, December 17, 2013 at 6:52pm*

**Calculus (check my answer)**

No, it is 20.25 like you got in the first place.
*Tuesday, December 17, 2013 at 6:46pm*

**Calculus (check my answer)**

30) The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube. MY ANSWER: 20.25 ----------- v = x^3 dv = 3 x^2 dx dv = 3 (225)(....
*Tuesday, December 17, 2013 at 6:45pm*

**Calculus (check my answer)**

Hey, I did this
*Tuesday, December 17, 2013 at 6:44pm*

**Calculus (check my answer)**

The measurement of the edge of a cube is found to be 15 inches, with a possible error of 0.03 inch. Use differentials to approximate the maximum possible propagated error in computing the surface area of the cube. I got 5.4 inches squared. Is this right?
*Tuesday, December 17, 2013 at 6:39pm*

**Calculus (check my answers)**

Thank you soooo much!
*Tuesday, December 17, 2013 at 6:35pm*

**Calculus (check my answers)**

10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01 MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do...
*Tuesday, December 17, 2013 at 6:08pm*

**Calculus**

Hold on...I think I just managed to kind of figure it out!
*Tuesday, December 17, 2013 at 6:01pm*

**Calculus**

Wait, I'm not understanding how you get the tanx -x part...
*Tuesday, December 17, 2013 at 5:58pm*

**Calculus (check my answers)**

I already did 20, scroll down
*Tuesday, December 17, 2013 at 5:55pm*

**Calculus (check my answers)**

Never mind number 20! Someone already helped me with it!
*Tuesday, December 17, 2013 at 5:54pm*

**Calculus (check my answers)**

10) Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x=dx=0.01 MY ANSWER: dy=-.32 and delta y=-.3224. Are they supposed to be negative, and how do I compare those in words? I don't even understand what I was supposed to do... 12) Find the ...
*Tuesday, December 17, 2013 at 5:53pm*

**Calculus**

[(x^2+1)d(sec^2x)-sec^2xd(x^2+1)]/(x^2+1)^2 [(x^2+1)2secxdsecx-sec^2x(2xdx)]/(x^2+1)^2 [2(x^2+1)sec^2xtanxdx-2xsec^2xdx]/(x+1)^2 2 sec^2 x dx [(x^2+1)tan x - x ]/ (x+1)^2
*Tuesday, December 17, 2013 at 5:47pm*

**Calculus**

Find the differential dy of the given function. y=sec^2x/(x^2+1) My answer was [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2 My friend (who's a lot smarter than me) got something wayyy different, so I'm not sure if I'm doing this wrong...?
*Tuesday, December 17, 2013 at 5:35pm*

**Calculus**

4 rectangles means 0.5 interval width. (Assuming they are all the same width) f(0.5) = -3.25 f(1.0) = -2.00 f(1.5) = -0.25 f(2.0) = +2.00 So, the area is 0.5(-3.25-2.00-0.25+2.00) = -1.75 The area is negative since it's mostly below the x-axis.
*Tuesday, December 17, 2013 at 12:41pm*

**Calculus**

Consider the function x^2+x-4 Estimate the area between the graph and the x-axis between x=2 and x=4 using four rectangles and right end points.
*Tuesday, December 17, 2013 at 12:27pm*

**Calculus**

oops. I read x^2-4, not 2-x^4. So, that would be -14, not 0. Sorry... Dang dyslexia (not really, just careless)
*Tuesday, December 17, 2013 at 12:35am*

**calculus**

If one train is x miles from the point of closest approach, and the other is y, then dx/dt = dy/dt = -40 the distance z between the trains is z^2 = (x+y)^2 + 2^2 So, when they are 8 miles apart, (x+y)^2 + 4 = 64 x+y = √60 z dz/dt = (x+y)(dx/dt+dy/dt) 8 dz/dt = √60...
*Tuesday, December 17, 2013 at 12:30am*

**calculus**

The parallel tracks are 2 miles apart
*Monday, December 16, 2013 at 10:46pm*

**Calculus**

Delta y is the Actual change of y when x changes from x to x+delta x. (I capitalized Actual so that you'd see the top part of A looks like delta - a way to remember what delta y is.) dy is the estimated change in y when x changes from x to x+delta x when you use the ...
*Monday, December 16, 2013 at 10:44pm*

**Calculus**

A=.5bh dA=.5b*dh +h*.5db Error = dA= .5(36)(.25)+.5(50)(.25)=10.75 Note: with given measurements, A = .5*36*50=900 If you are off by .25, A=.5*36.25*50.25=910.78125. Therefore you'd be off by 10.78125. Does this help to see how your estimate with differentials makes sense?
*Monday, December 16, 2013 at 10:22pm*

**calculus**

Hmmm. They are coming toward each other , distance decreasing, and you want to know when the are 8 miles apart when they started 2 miles apart? Makes no sense. relative velocity=80mph that is how fast the distance between them is decreasing at any time. This problem is not ...
*Monday, December 16, 2013 at 10:13pm*

**calculus**

two trains are on parallel tracks that are 2 miles apart. If both trains are traveling at 40 mph, how fast is the distance between the trains decreasing when they are 8 miles apart?
*Monday, December 16, 2013 at 10:09pm*

**Calculus**

What I can do is to lead you to some pages that explain the difference in dy and ∆y I thought this youtube did a good job http://www.youtube.com/watch?v=xoAnLgs6hiM There are also numerous other youtube links listed in the right column btw, I agree with your answers.
*Monday, December 16, 2013 at 7:57pm*

**Calculus**

The measurements of the base and altitude of a triangle are found to be 36 and 50 centimeters, respectively. The possible error in each measurement is 0.25 centimeter. Use differentials to approximate the possible propagated error in computing the area of the triangle.
*Monday, December 16, 2013 at 7:45pm*

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