Saturday

April 19, 2014

April 19, 2014

Number of results: 2,608

**calculus help thanks!**

PV = nRT V dP/dt + P dV/dt = nR dT/dt Now just plug in the data: (14)(0.13) + (8.0)(-0.17) = (10)(0.0821) dT/dt dT/dt = 0.56 °K/s
*Wednesday, April 2, 2014 at 4:35pm by Steve*

**physics(URGENT!!!!!)**

dx/dt = .075 cos T (dT/dt) is: [ .075 cos T ] {dT/dt) BOTH of those are changing with time form is f(t) = a(t) b(t) so f'(t) = a db/dt + b da/dt so d/dt(dx/dt) = [ .075 cos T ]d^2T/dt^2 +{dT/dt) [- .075 sin T ] (dT/dt) or ax = what is said
*Monday, December 9, 2013 at 3:08am by Damon*

**calculus**

Well, we know x=3 and y=8 and dx/dt=4. We need to find z (the distance from the particle to the origin) dy/dt, and finally our answer dz/dt. First solve for dy/dt y=2(3x+7)^(1/2) dy/dt=[2(1/2)(3x+7)^(-1/2)](3dx/dt) dy/dt=[(3x+7)^(-1/2)]3dx/dt dy/dt=[(3(3)+7)^(-1/2)](3(4)) dy/...
*Tuesday, October 26, 2010 at 6:52am by Stephanie*

**calculus**

xdy/dt + ydx/dt - 5dx/dt + 6x^2dx/dt = 0 2dy/dt - 2(-5) - 5(-5) + 6(25)(-5) = 0 solve for dy/dt
*Sunday, February 24, 2008 at 10:18pm by Reiny*

**calculus**

s^2 = 3^2 + 4^2 - 2(3)(4)cos T 2s ds/dt = 24sin T dT/dt s ds/dt = -12sin T dT/dt when T = 90°, s = 5, sinT = 1 , dT/dt = 2 rad/sec 5 ds/dt = -12(1)(2) ds/dt = -24/5 = -4.8 inches/sec at that precise instant, the third side is decreasing at 4.8 inches/sec
*Saturday, March 5, 2011 at 4:35pm by Reiny*

**differentiate**

A = 1/2 ab Sinè How is dA/dt related to da/dt, db/dt and dè/dt dA/dt= 1/2 b SinTheta da/dt + 1/2 a SinTheta db/dt + 1/2 ab cosTheta dTheta/dt This is a U = xyz form dU/dt= xy dz/dt + ...
*Saturday, December 9, 2006 at 1:14pm by Jen*

**chemistry**

well yea, but the only responses listed are as follows: -d(Cl2)/dt = 2[-d(F2)/dt] -d(Cl2)/dt = 3[-d(F2)/dt] 2[-d(Cl2)/dt] = -d(F2)/dt 3[-d(Cl2)/dt] = -d(F2)/dt -d(Cl2)/dt = -d(F2)/dt and none of those make sense.
*Friday, May 29, 2009 at 7:33am by uci student*

**chemistry(check my answer)**

Which equation describes the relationship between the rates at which Cl2 and F2 are consumed in the following reaction? Cl2(g) + 3F2(g) yield 2ClF3(g) a)-d(Cl2)/dt = -d(F2)/dt b)-d(Cl2)/dt = 2[-d(F2)/dt] c)3[-d(Cl2)/dt] = -d(F2)/dt d)2[-d(Cl2)/dt] = -d(F2)/dt e)-d(Cl2)/dt = 3...
*Friday, May 14, 2010 at 2:27am by eng*

**chemistry**

Which equation describes the relationship between the rates at which Cl2 and F2 are consumed in the following reaction? Cl2(g) + 3F2(g) yield 2ClF3(g) a)-d(Cl2)/dt = -d(F2)/dt b)-d(Cl2)/dt = 2[-d(F2)/dt] c)3[-d(Cl2)/dt] = -d(F2)/dt d)2[-d(Cl2)/dt] = -d(F2)/dt e)-d(Cl2)/dt = 3...
*Friday, May 14, 2010 at 11:06pm by eng*

**chemistry(check my answer)**

Which equation describes the relationship between the rates at which Cl2 and F2 are consumed in the following reaction? a)-d(Cl2)/dt = -d(F2)/dt b)-d(Cl2)/dt = 2[-d(F2)/dt] c)3[-d(Cl2)/dt] = -d(F2)/dt d)2[-d(Cl2)/dt] = -d(F2)/dt e)-d(Cl2)/dt = 3[-d(F2)/dt] is it e
*Thursday, May 13, 2010 at 1:50am by eng*

**calculus**

A= 1/2 bh dA/dt= 1/2 hdh/dt + 1/2 bdb/dt compute, you have h, dh/dt, b, db/dt
*Thursday, October 14, 2010 at 7:40pm by bobpursley*

**calculus**

let the width be x let the length be y let the height be h V= xyh dV/dt = xy dh/dt + xh dy/dt + yh dx/dt given x=20, y=12, h = 8 dx/dt = 5 dy/dt = -4 dh/dt = 0 Plug in those values into dV/dt
*Wednesday, January 5, 2011 at 9:05pm by Reiny*

**math**

Well, since the ship is equidistant from A and B, it lies on the perpendicular bisector of AB, so if A is at (0,0) and B is at (6,0), the ship is at (3,4). If the ship is at (x,y), the distances u from A and v from B are u^2 = x^2 + y^2 v^2 = (x-6)^2 + y^2 2u du/dt = 2x dx/dt...
*Monday, July 15, 2013 at 9:04am by Steve*

**CALCULUS**

Pretty similar to the truck and the police car: Area = A = pi r^2 dA/dt = 2 pi r dr/dt (which by the way is the circumference times the outward speed logically enough.) part 2 is rate of volume change Vol = V = pi r^2 h where h is thickness dV/dt = pi r^2 (dh/dt) + h (2 pi r ...
*Thursday, April 8, 2010 at 2:20am by Damon*

**Calculus**

If x = t^2 + 1 and y = t^3, then d^2y/dx^2 = I know I can solve for t in terms of x and substitute that into y = t^3 and find the double derivative. I also know that I can take the derivative of x and y then divide dy/dt by dx/dt. Then take the derivative to get the answer. ...
*Thursday, May 15, 2008 at 5:02pm by Anonymous*

**calc**

Draw your little triangle. cos a = x/s where s is the string length (the hypotenuse) and x is the horizontal distance from the flyer. x = s*cos a dx/dt = ds/dt cos a - s*sin a da/dt At our moment, s=100, so 8 = ds/dt cos a - 100 * sin a da/dt Still not home free. What are ds/...
*Friday, December 16, 2011 at 4:41pm by Steve*

**Calculus**

let the base be x let the height be y then A = xy dA/dt = x dy/dt + y dx/dt but dA/dt = 0 x dy/dt = -y dx/dt given: dx/dt = 3, find : dy/dt when x = y x dy/xt = -x(3) dy/dt = -3 So at the moment when it turns into a square, the height is decreasing at 3 inches/second
*Thursday, March 21, 2013 at 10:12pm by Reiny*

**Calculus**

1)Suppose Q = x^3 y^2 . If [dQ/dt] = −1 and [dx/dt] = 2, find [dy/dt] when x = 4 and y = 5 . [dy/dt] =______ 2)Let A be the area of a circle with radius r. If [ dr / dt ] = 2, find [ dA / dt ] when r = 3
*Tuesday, November 3, 2009 at 9:43pm by Mulan*

**Calc**

distanceapart^2=distaneAfromcollision^2+distanceBfromcollision^2 Take the derivative.. s ds/dt=2xdx/dt + 2y dy/dt where x, y are the distances from the intersection point, and s is sqrt(x^2+y^2) you are given dx/dt, dy/dt, solve for ds/dt
*Tuesday, October 1, 2013 at 11:27am by bobpursley*

**Calculus**

write the functions. revenue= number books sold * price book dRev/dt=n dp/dt + p dn/dt you are given n, p and dp/dt, and dn/dt calculate. For the second part, set dRev/dt to 5000 find dn/dt given n, p, dp/dt I will be happy to critique your work.
*Sunday, November 25, 2007 at 5:26pm by bobpursley*

**math**

given: dx/dt = 5 , dy/dt = -6 when x=3 and y = 4 , hypotenuse = 5 , (the infamous 3-4-5 right-angled triangle) H^2 = x^2 + y^2 2H dH/dt = 2x dx/dt + 2y dy/dt dH/dt = (x dx/dt + y dy/dt)/H = (3(5) + 4(-6))/5 = -9/5 at that moment of time, the hypotenuse is decreasing at a rate ...
*Monday, November 26, 2012 at 4:46pm by Reiny*

**chemistry**

Which of the following equations correctly describes the relationship between the rate at which NO2 and Cl2 are consumed in the following reaction? 2 NO2(g) + Cl2(g) → 2 NO2Cl(g) A. -d(NO2)/dt = 1/2 [d(Cl2)/dt] B. -d(NO2)/dt = 2 [d(Cl2)/dt] C. -d(NO2)/dt = 2 [-d(Cl2)/dt...
*Monday, May 16, 2011 at 8:19pm by dude*

**science**

Which of the following equations correctly describes the relationship between the rate at which NO2 and Cl2 are consumed in the following reaction? 2 NO2(g) + Cl2(g) → 2 NO2Cl(g) A. -d(NO2)/dt = 1/2 [d(Cl2)/dt] B. -d(NO2)/dt = 2 [d(Cl2)/dt] C. -d(NO2)/dt = 2 [-d(Cl2)/dt...
*Monday, May 16, 2011 at 8:19pm by dude*

**calculus**

p^2 + 2 q^2 = 1100 2 p dp/dt + 4 q dq/dt = 0 dp/dt = -(q/p) dq/dt R = p q price times quantity dR/dt = p dq/dt + q dp/dt = A dp/dt so A is just q when q = 20 p^2 + 2(400) = 1100 p^2 = 300 p = 10 sqrt 3 dR/dt = q dp/dt = 20*4 = 80
*Saturday, October 27, 2012 at 11:31pm by Damon*

**Math(Please help. Thank you)**

Is number 3 considered a related rates problem? I suppose so. I should have been more careful about chain rule to show you what I did. 6p + x + xp =94. 6 dp/dx + dx/dx + x dp/dx + p dx/dx = 0 times dx/dt 6 dp/dx*dx/dt + dx/dx*dx/dt + x dp/dx*dx/dt + p dx/dx*dx/dt = 0 6 dp/dt +...
*Sunday, May 1, 2011 at 9:41am by Damon*

**calculus**

first find dr/dt. The diameter changes twice as fast as the radius. v = (4/3) pi r^3 dv/dt = 4 pi r^2 dr/dt so 1 = 4 pi (r^2) dr/dt but r^3 = (3/4)(36)/pi so r = 2.05 ft so 1 = 4 pi (4.2) dr/dt so dr/dt = .019 ft/min D = 2 r dD/dt = 2 dr/dt = .038 ft/min another way rate of ...
*Wednesday, October 6, 2010 at 5:09pm by Damon*

**calculus**

the rate of ordinate change is dy/dt. x^2 + y^2 = 100 2x dx/dt + 2y dy/dt = 0 when x=8, y=6, so 2(8)(72) + 2(6) dy/dt = 0 12 dy/dt = -1152 dy/dt = -96 in/s
*Wednesday, November 7, 2012 at 4:16pm by Steve*

**diffirential calculus**

did you make a diagram? let her distance from the lightpole be x ft let the length of her shadow be y ft by similar triangles ... 6/y = 10/(x+y) 6x + 6y = 10y 6x = 4y 6 dx/dt = 4 dy/dt we are given: dx/dt + dy/dt = 2 ft/sec dx/dt = 2 - dy/dt so 6(2 - dy/dt) = 4 dy/dt 12 - 6dy/...
*Sunday, September 25, 2011 at 12:13am by Reiny*

**calculus (related rates)**

distance=sqrt(y^2 + x^2) ddistance/dt= 1/2distance *( 2ydy/dt+2xdx/dt) now examine y dy/dt= x ln(x) (ln(x)+1) dx/dt ddistance/dt=rateyou want=1/2distance*(2xlnx+1)17cm/sec + x dx/dt) check that, I typed it in a hurry.
*Tuesday, October 27, 2009 at 9:05pm by bobpursley*

**Math Calculus**

Reini - How did you ended up getting x dx/dt + y dy/dt=0? my professor said something with a^2 + b^2 = c^2 then differentiate it so 2a(da/dt) + 2b(db/dt) = 2c(dc/dt) so a^2+b^2=c^2 a^2 + 6^2 = 30^2 so a=sqrt(864) then I differentiate let da/dt = 1 b = 6 c=30 2a(da/dt)+2b(db/dt...
*Friday, July 8, 2011 at 10:06am by Meg*

**Math Calculus**

Reini - How did you ended up getting x dx/dt + y dy/dt=0? my professor said something with a^2 + b^2 = c^2 then differentiate it so 2a(da/dt) + 2b(db/dt) = 2c(dc/dt) so a^2+b^2=c^2 a^2 + 6^2 = 30^2 so a=sqrt(864) then I differentiate let da/dt = 1 b = 6 c=30 2a(da/dt)+2b(db/dt...
*Friday, July 8, 2011 at 10:06am by Meg*

**Classical Mechanics - Need Help Please**

F = m dV/dt m dV = F dt so change in momentum = integral of F dt F = 1.2 t^2 integral F dt = 1.2 t^3/3 = 50 at t = 5 yes for part b F = 1.2 t^2 - 10(3)(.2) = 1.2t^2-6 integral F dt = 1.2 t^2/3 - 6 t for part 3 P = F dx/dt F = 1.2 t^2 - 6 at 4 sec F = 13.2 Newtons to get dx/dt ...
*Sunday, December 8, 2013 at 3:38am by Damon*

**calculus**

No. You are given db/dt Using the pyth theorm c^2= b^2 + h^2 dc/dt=0= 2b db/dt + 2h dh/dt solve for dh/dt in terms of b,h, db/dt this is what related rates means.
*Monday, November 5, 2007 at 8:58pm by bobpursley*

**calculus**

multipy both sides by (x-y) x+y=x^3 -yx^2-xy^2-y^3 check that. dx/dt+dy/dt= 3x^2 dx/dt-2yx dx/dt-x^2dy/dt-y^2dx/dt-3y^2 dy/dt gather terms.... dx/dt(1-3x^2+2yx+3y^2)=dy/dt (-1-x^2-3y^2) check that carefully, it is easy to make an error in sign. solve for dy/dt
*Sunday, July 4, 2010 at 6:14pm by bobpursley*

**CALCULUS HELP!!**

a^2+b^2=c^2 ada/dt + bdb/dt = cdc/dt (6)da/dt+12(6) =13(0) da/dt =−12ft/sec
*Sunday, December 1, 2013 at 3:51pm by Kuai*

**calculus**

a particle moves along the curve xy=10. if x=2 and dy/dt=3, what is the value of dx/dt. i'm guessing this is something to do with parametric equations. if x is 2, then y=5. but how do i get dx/dt. you have to differentiate the equation with respect to t so xy=10 x(dy/dt) + y(...
*Saturday, April 21, 2007 at 3:11pm by Anonymous*

**Calculus**

look at the triangle. c^2=b^2+h^2 b= base, h=height take derivitave d c^2/dt=0=2bdb/dt + 2h dh/dt b=4, db/dt=3ft/sec h=sqrt(18^2-4^2) find dh/dt
*Sunday, October 30, 2011 at 7:29pm by bobpursley*

**calculus**

v = u^5 v = 7sqrt(w) - wz dv/dt = 5u^4 du/dt = 7/(2sqrt(w)) dw/dt - (w dz/dt + z dw/dt)
*Tuesday, October 11, 2011 at 12:31pm by Steve*

**Calculus**

Let the distance of the bottom of the ladder from the wall be x, and let the distance of the top of the ladder from the floor be y. x^2 + y^2 = 256 You want to know dy/dt when dx/dt = 3 and x = 2. At that time, y^2 = 252 and y = 15.87 2x *dx/dt + 2y*dy/dt = 0 dy/dt = -(x/y)*dx...
*Saturday, March 12, 2011 at 12:20am by drwls*

**related rates problem-calculus**

The balloon is spherical, so V = (4/3)(pi)r^3 SA = 4(pi)r^2 Differentiate both with respect to time. dV/dt = 4(pi)r^2(dr/dt) dSA/dt = 8(pi)r(dr/dt) We're given the rate of change in SA and need to find the rate of change in volume. Let's write an equation for it. (dV/dt)/r = 4...
*Thursday, December 30, 2010 at 3:25pm by Marth*

**chemistry**

If aA + bB ==> cC + dD, rate of rxn = -(1/a)(dA/dt) = -1/b(dB/dt) etc. So wouldn't rate of reaction equal -(1)(dCl2/dt) = -(1/3)(dF2/dt) = (1/2)(dClF3/dt) Check me out.
*Friday, May 29, 2009 at 7:33am by DrBob222*

**Math**

xy = 5 , use the product rule and differentiate with respect to t x(dy/dt) + y(dx/dt) = 0 when x= 1 ---> (1)y = 5, so y=5 now sub in all that stuff in x(dy/dt) + y(dx/dt) = 0 (1)(dy/dt) + 5(3) = 0 dy/dt = -15
*Sunday, March 15, 2009 at 10:00pm by Reiny*

**Maths**

y= 4sqrt(4x+1) dy/dt= 4*1/2*1/sqrt(4x+1)*4dx/dt you are given dx/dt, solve for dy/dt now, since r=sqrt(x^2+y^2) dr/dt= 1/2 *1/sqrt(x^2+y^2)* (2xdx/dt+2ydy/dt) solve for dr/dt
*Tuesday, November 29, 2011 at 7:07pm by bobpursley*

**Calc 3**

Try a circle with parameter t=θ x=cos(t) y=sin(t) dx/dt=-sin(t) dy/dt=cos(t) at t=0, (dx/dt,dy/dt)=(0,1) at t=π/4, (dx/dt,dy/dt)=(-√2/2,√2/2) ...etc Here: x= e^t y=te^t z=te^(t^2) (1,0,0) Solve for t at the point (1,0,0) gives t=0 (the only way to get y=z=...
*Wednesday, November 10, 2010 at 11:24pm by MathMate*

**calculus**

Differentiate both sides with respect to t: dy/dt = d(sqrt(1+x^3))/dx * dx/dt =(3x²/(2sqrt(1+x³)))*dx/dt So given x=2,y=3 and dy/dt=3 cm/s substitute in formula above to solve for dx/dt. I get dx/dt=2.
*Monday, February 21, 2011 at 12:58pm by MathMate*

**MATH**

So we have 1/x + 1/y = 1/5 differentiate with respect to t (-1/x^2)dx/dt + (-1/y^2)dy/dt = 0 (-1/x2)dx/dt = (1/y^2)dy/dt dx/dt = (-x^2/y^2)(dy/dt) given: when y = 9 1/x + 1/9 = 1/5 1/x = 4/45 x = 45/4 and dy/dt = ???? , not clear what you mean by 16 CM^(-1) what ever it is, ...
*Sunday, May 5, 2013 at 8:42am by Reiny*

**physics(URGENT!!!!!)**

Oh, I see... So, you must derive function "sin/cos" and theta, as both change with time. Then (without l/2, for simplicity) x = sin(theta); dx/dt = cos(theta)*d(theta)/dt; d^2x/dt^2 = a*db/dt + b*da/dt; a*db/dt = cos(theta)*d^2(theta)/dt b*da/dt = (d(theta)/dt)^2*-sin(theta) ...
*Monday, December 9, 2013 at 3:08am by Anoninho*

**calculus 1**

if the angle is θ, and the string length is s, and the horizontal distance the kite has flown is x, sinθ = 100/s cosθ dθ/dt = -100/s^2 ds/dt x^2 + 100^2 = s^2 2x dx/dt = 2s ds/dt, so ds/dt = x/s dx/dt = cosθ dx/dt so, dθ/dt = -100/s^2 dx/dt so, ...
*Tuesday, February 26, 2013 at 6:18pm by Steve*

**Calc**

1. Determine an expression for dy/dt if x^2+2y^2=8 and dx/dt=3. 2. A point is moving along the right branch of a hyperbola defined by 4x^2-y^2=64. What is dy/dt when the point is at (5, -6) and dx/dt=3? I will start you on the first, using implicit differention. x^2+2y^2=8 2x ...
*Tuesday, May 8, 2007 at 9:39pm by Icy*

**Calculus**

Let the independent variable be s = length of one side of the cube. As you mentioned, V(s) = s³ S(s) = 6s² dV/dt = dV/ds*ds/dt = 3s² ds/dt = 24 ... (1) dS/ds = dS/ds*ds/dt = 12s ds/dt = 12 ...(2) Solve for s and ds/dt by substitution or dividing (1) by (2). I ...
*Tuesday, December 7, 2010 at 8:40pm by MathMate*

**Calc**

V = Pi(r^2)(h) dV/dt = pi(r^2)dh/dt + pi(h)(2r)dr/dt sub in the known values -3 = pi(300^2)(-.0003) + pi(.004)(600)dr/dt solve for dr/dt now in A = pi(r^2) dA/dt = 2pi(r)dr/dt = 2pi(300)(above value for dr/dt) = ....
*Monday, February 25, 2008 at 9:06pm by Reiny*

**Calculus I**

You mean on this circle find vertical component of velocity. 2 x dx/dx + 2 y dy/dx = 0 y dy/dx = -x y dy/dx * dx/dt = -x * dx/dt dy/dt = -(x/y) dx/dt here x = 3 then y = +4 or -4 if y is +, dy/dt is + for -dx/dt if y is -, dy/dt is - for -dx/dt so dy/dt = -(3/4)(-8) = 6 in ...
*Wednesday, February 12, 2014 at 9:57am by Damon*

**Physics**

F = m a = m w^2 r where w = 2 pi f let u = tangential velocity let v = radial velocity = w r dv/dt = w dr/dt at = du/dt = d/dt(wr) = w dr/dt ar = d^2r/dt^2 = dv/dt= (w^2)r so w dr/dt = w^2 r dr/dt = w r or dr/r = w dt ln r = w dt + c' r = e^(wt + e^c') = C e^wt that is the ...
*Thursday, February 10, 2011 at 4:48pm by Damon*

**Calculus Please help!**

so, did you draw a diagram as suggested? a = 1/2 bh so, b = 2a/h da/dt = 1/2 (db/dt * h + b * dh/dt) Now just plug in your values: 3500 = 1/2 (db/dt * 7000 + 2*87/7 * 2500) db/dt = -7.88
*Friday, February 28, 2014 at 2:51pm by Steve*

**Physics**

I assume that Theta is up and down from horizontal calling theta = A d sin A /dA = cos A d sin A/dt = d sin A/dA * dA/dt so d sin A /dt = cos A (dA/dt ) similarly d cos A/dt = -sin A dA/dt That settled, let's look at the problem Force = m * a a = F/7 da/dt = (1/7) dF/dt F in ...
*Tuesday, January 21, 2014 at 3:05pm by Damon*

**Calculus I**

dR/dt = .3 ohms/s I = 18V/R d/dt(I) = d/dt(18V/R) dI/dt(I) = dR/dt(18V/R) dI/dt = dR/dt*(-18/R**2) dI/dt = .3*(-18/4**2) dI/dt = -.3375amp/s Hope you understand my work! Cheerz
*Monday, March 1, 2010 at 9:48pm by Zen*

**Calculus-edited**

Let angle of elevation=θ height of plane = H horizontal distance from observer = x tan(θ)=x/H Use implicit differentiation d(tan(θ))/dt = d(x/H)/dt sec²(θ)dθ/dt = (dx/dt)/H dθ/dt=(1/(Hsec²(θ))(dx/dt) dθ/dt=(cos²(&...
*Sunday, November 1, 2009 at 5:20pm by MathMate*

**Math(Please help. Thank you)**

3) Suppose that the price p (in dollars) and the demand in x(thousands) of a commodity satisfy the demand equation 6p + x + xp =94. How fast is the demand changing at times when x= 4 and p=9 and the price is rising at the rate of $2 per week. Would I take the derivative and ...
*Sunday, May 1, 2011 at 9:41am by Damon*

**Math - Limits/Derivatives**

just plug and chug: dC/dt = 20 dx/dt = 20*600 dR/dt = 300 dx/dt - x/10 dx/dt = (300-500)(600) dP/dt = dR/dt - dC/dt
*Saturday, October 19, 2013 at 12:51pm by Steve*

**Calc**

dV/dt= 4/3 PI 3 r^2 dr/dt solve for dr/dt given dv/dt Then dS/dt=4PI 2r dr/dt, knowing dr/dt solve for dS/dt and it is done.
*Sunday, March 15, 2009 at 10:03pm by bobpursley*

**calculus**

For question a, y = 3x^2, so dy/dt = 6x dx/dt dy/dt = 6(2) * 3 dy/dt = 36 For question b, as t approacing infinity, x is also approaching infinity (the object is moving along the track of the function, remember(, then dy/dt is also approaching infinity in vertical sense... For...
*Sunday, October 18, 2009 at 6:16pm by Andre*

**Thermodynamics**

a) Since it is a constant pressure heating, dQ = M Cp dT = [K*M*R/((K-1)]*dT Use that to determine the temperature changle dT, and add that to 80 F. You need to specify the units of R. They should be Btu/lbm*degF b) dU = M Cv dT = [M/(K-1)]dT c) dH = Cp*dT = dQ d) dS = ...
*Saturday, March 6, 2010 at 11:53am by drwls*

**calculus**

dV/dt = kA. V = (4/3) π r^3 and A = 4 π r^2 dV/dt = d/dt((4/3) π r^3) = (4/3) π 3 r^2 (dr/dt) Now let's plug that into the first equation: (4/3) π 3 r^2 (dr/dt) = k A = k(4 π r^2) = 4 π k r^2 So when we simplify by dividing left and right ...
*Tuesday, October 18, 2011 at 3:01pm by Steve*

**AP calculus AB**

dV/dt = 9 = surface area * dh/dt 9 = (pi r^2)(dh/dt) but h = 3 r so dh/dt = 3 dr/dt so 9 = (pi r^2)(2 dr/dt) but r = 3 so 9 = 9 pi (2 dr/dt dr/dt = 1/2pi
*Wednesday, December 29, 2010 at 4:47am by Damon*

**Calculus**

V=1/2 PI r^2 h dv/dt= 1/3 PI (r^2 dh/dt + 2hr dr/dt) now relate dr/dt to dh/dt (proportion) then solve for dv/dt given dh/dt
*Saturday, September 24, 2011 at 2:52pm by bobpursley*

**calculus**

bobpursely is right, the area of 18 in^2 has nothing to do with it. Here is how I did it D^2 = 2s^2 D = √2 s dD/dt = √2 ds/dt 3 = √2 ds/dt ------> ds/dt = 3/√2 P = 4s dP/dt = 4ds/dt=4(3/√2) = 12/√2 in/min which is the same as bobpursley's...
*Thursday, October 18, 2007 at 9:18pm by Reiny*

**calculus**

y = x^2 + 8 dy/dx = 2 x dy/dt = dy/dx * dx/dt dy/dt = 2x * 8 = 16 x s = distance from origin s^2 = x^2+y^2 2 s ds/dt = 2 x dx/dt +2 y dy/dt so s ds/dt = x dx/dt + y dy/dt so s ds/dt = [8 x + (x^2+8)(16 x)] ds/dt = [8 x + (x^2+8)(16 x)]/[x^2 +(x^2+8)^2]^.5 = [16 x^3 + 136 x...
*Wednesday, October 6, 2010 at 3:42pm by Damon*

**calulus**

distance to origin=sqrt(x+ x^2) d distance/dt= 1/(2sqrt(x+x^2)*(dx/dt+ 2xdx/dt) you have x, dx/dt solve for ddistance/dt
*Thursday, November 19, 2009 at 7:35pm by bobpursley*

**calc**

I will assume you meant to type y = 2x^2 + 1 dy/dt = 4x dx/dt given dy/dt = -2 when x = 3/2 -2 = 4(3/2) dx/dt -2 = 6 dx/dt dx/dt = -2/6 = -1/3
*Tuesday, December 27, 2011 at 10:38pm by Reiny*

**Calculus**

The vloume increase rate is dV/dt = 500 V = (4/3) pi R^3 dV/dt = (12/3)pi R^2 dR/dt = 4 pi R^2 dR/dt You can use that equation to compute the dR/dt expansion rate for any value of R. Note that that dV/dt equals the instantaneous surface area times dR/dt.
*Sunday, November 1, 2009 at 8:39am by drwls*

**calculus**

Well I assume that z is the diagonal. x^2+y^2 = z^2 2 x dx/dt + 2 y dy/dt =2 z dz/dt 4 dz/dt + 3 k dz/dt = dz/dt 4 + 3 k = 1 k = -1
*Wednesday, April 27, 2011 at 2:14am by Damon*

**Calculus**

V = s^3 dV/dt = 3v^2 ds/dt when s = 5, dV/dt = 2 2 = 3(25) ds/dt ds/dt = 2/75 A = 6s^2 dA/dt = 12s ds/st = 12(5) (2/75) = 8/5 inches^2 / min
*Wednesday, January 30, 2013 at 4:37pm by Reiny*

**Engineering Kinematic**

When t = 1, dx/dt = 5t = 5 m/s From v = dx/dt = 5t, and the starting conditions, you can conclude that x = (5/2)t^2. x = 5/2 = 2.5 when t = 1 y = 0.5*(2.5)^2 = 3.125 when t = 1 The distance from the origin in sqrt[x^2 + y^2]. At t=1, this is D = sqrt[2.5^2 + 3.125^2]= 4.002 ...
*Saturday, October 16, 2010 at 4:49am by drwls*

**Calculus**

height y = 12 - 16t^2 shadow at x, similar triangles, so x/12 = (x-4)/y or, xy = 12x - 48 y dx/dt + x dy/dt = 12 dx/dt y=8 when t=1/2 x=6 when y=8 (12 - 16t^2) dx/dt + x (-32t) = 12 dx/dt at t=1/2, 8 dx/dt - 96 = 12 dx/dt dx/dt = 96/-4 = -24 ft/s
*Tuesday, April 24, 2012 at 12:05am by Steve*

**calculus**

V=PI*r^2 h dv/dt=0=PIr^2 dh/dt + 2PI r h dr/dt solve for dr/dt, you are given dh/dt.
*Saturday, March 16, 2013 at 10:57pm by bobpursley*

**Calculus**

You mean 18 / 6 = (x + y) / y we know dx/dt, we need dy/dt then the tip moves at dx/dt + dy/dt 18 y = 6x + 6 y 12 y = 6 x 12 dy/dt = 6 dx/dt dy/dt = .5 dx/dt so dy/dy = 3/2 = 1.5 and the sum 3+1.5 = 4.5 ft/sec
*Sunday, February 6, 2011 at 6:12pm by Damon*

**Math**

y = 3x^2 - 5x = 3(g(t)^2 - 5g(t) dy/dt = 6 g(t) d(g(t))/dt - 5 d(g(t))/dt = 6(g(14)) d(g(14))/dt - 5 d(g(14))/dt = 6(-4)(2) - 5(2) = -58
*Wednesday, March 23, 2011 at 10:19pm by Reiny*

**Pure Mathematics**

1) x = 1/t ... x^5 d^2z/dx^2 + (2x^4 - 5x^3) dz/dx +4xz = 6x +3 can be reduced to.... d^2z/dt^2 + 5 dz/dt + 4z = 3t+6 2) use the substitution y=x^2 to show that... x d^2x/dt^2 + (dx/dt)^2 + 5x(dx/dt) + 3x^2 = sin2t + 3cos2t can be converted to d^2y/dt^2 + 5dx/dt + 6y = 2sin2t...
*Wednesday, February 19, 2014 at 9:57pm by Kerry-Ann English*

**Calculus**

I will do one, and be happy to check your thinking. q=x^3 y^2 dq/dt= 2x^2 y dy/dt+ 3y^2x^2 dx/dt put in dq/dt, dx/dt, and find dy/dt ..
*Tuesday, November 3, 2009 at 9:43pm by bobpursley*

**Calculus**

There is 1 error in the above answer. 11dy/dt = 6 dx/dt is true dx/dt = 6 is also true but dy/dt is not 6/11 ft/sec the 6 must be multiplied by the 6 from dx/dt giving dy/dt = 36/11
*Thursday, September 24, 2009 at 3:40am by Jarred*

**math**

dV/dt = 3x^2 dx/dt 96 = 3(4^2)dx/dt dx/dt = 96/48 = 2 cm/second A = 6x^2 dA/dt = 12x dx/dt dA/dt = 12(4)(2) = 96 cm^2/sec
*Wednesday, August 17, 2011 at 10:28pm by Reiny*

**Calculus**

A line through the origin, rotates around the origin in such a way that the angle, ¦È, between the line and the positive x-axis changes at the rate of d¦È/dt for time t¡Ý0. Which expression gives the rate at which the slope of the line is changing? a) d¦È/dt b) cos¦È*d¦È/dt c...
*Wednesday, December 8, 2010 at 10:39pm by Anna*

**chemistry**

rate of change is dA/dT = (1/4)dB/dT = (1/2)dC/dT = 0.72 Solve for dA/dT, dB/dT, dC/dT
*Thursday, February 28, 2013 at 8:05pm by DrBob222*

**calculus**

V = pi r^2 L ln V = ln pi + 2 lnr + lnL d/dt (lnV) = 0 + (2/r) dr/dt + (1/L) dL/dt dL/dt = - (2L/r) dr/dt When r = 1.8, 128 pi = pi*(1.8)^2 L L = 39.51 inch You know dr/dt. Use the dL/dt formula
*Tuesday, December 27, 2011 at 10:05pm by drwls*

**physics**

wr = v a = r dw/dt 1.4 = r dw/dt dw/dt = angular acceleration = 1.4/r = 2.8/.57 assume constant angular deacceleration radial acceleration = w^2 r tangential acceleration = r dw/dt (we know dw/dt from above) w r = v = 3 m/s at t = 0 wo = 3/r w = wo - (dw/dt)t that gives you w^...
*Friday, November 26, 2010 at 5:02pm by Damon*

**Calculus**

dy/dt = 6x(dx/dt) so at (2,12) dx/dt = 3 then dy/dt = 6(2)(3) = 36 If dx/dt is always 3, then dy/dt = 18x so as x --> + infinity , 18x ---> + inf. and dy/dt --> + inf. for c) I think you meant to say, "then what happens to dx/dt as t --> +infinity?" if dy/dt = c ...
*Sunday, October 18, 2009 at 3:24am by Reiny*

**calc 3**

there are diagonals along each face, but thety probably want the diagonal of the brick: d^2 = l^2 + w^2 + h^2 2d dd/dt = 2l dl/dt + 2w dw/dt + 2h hd/dt when (l,w,h) = (7,5,5) d = sqrt(99) = 3sqrt(11) 3sqrt(11) dd/dt = 2*7*6 + 2*5*6 + 2*5*(-4) 3sqrt(11) dd/dt = 104 dd/dt = 104...
*Friday, September 21, 2012 at 4:34pm by Steve*

**calculas**

A = Lw dA/dt = L dw/dt + w dL/dt but dL/dt = -2 and dw/dt = 2 dA/dt = 2L - 2w = 2(L-w) So it depends on the size of the rectangle, If the rectangle is a square (L=w), then area would remain constant, if L > w, the area is increasing, and if L < w ......
*Tuesday, December 7, 2010 at 9:39pm by Reiny*

**Math Calc**

x^2 + y^2 = 25 2x dx/dt + 2y dy/dt = 0 2(-4) dx/dt + 2(3)(9) = 0 -8 dx/dt + 54 = 0 dx/dt = 54/8 = 27/4 = 6.75 makes sense; the particle is moving up to the right.
*Sunday, November 25, 2012 at 10:07pm by Steve*

**math**

x^2 + y^2 = 100 , where x is the base, y the height 2x dx/dt + 2y dy/dt = 0 given: dx/dt = 2 , and x = 6 then 36+y^2 = 100 y^2 = 64 y = 8 in 2x dx/dt + 2y dy/dt = 0 2(6)(2) + 2(8)dy/dt = 0 dy/dt = -24/16 = -3/2 it is sliding down at 3/2 m/sec
*Friday, May 3, 2013 at 8:35am by Reiny*

**calculus**

make a sketch At a time of t hrs, let the distance the police car is from the intersection be x miles let the distance the speeding car is from the intersection be y miles let the distance between them be d miles d^2 = x^2 + y^2 2d dd/dt = 2x dx/dt + 2y dy/dt d dd/dt = x dx/dt...
*Tuesday, January 17, 2012 at 11:53am by Reiny*

**calculus**

take the derivative with respect to t 2x dx/dt+2ydy/dt=0 dx/dt= -y/x dy/dt so when y=8, find x, or x=+-sqrt(100-64) x=+-6 dx/dt=put the values in and compute. Yes, you have two solutions, the x^2+y^2 is not a function, it does not obey simple rules. Think about it. Your ...
*Thursday, November 1, 2012 at 8:17pm by bobpursley*

**calculus**

make a diagram showing a right angled triangle with hypotenuse of 8 let the base be x and the height be y given : dx/dt = 1 find: dy/dt when x=5 a) when x=5 , 2+ y^2 = 64 y = √39 x^2 + y^2 = 64 2x dx/dt + 2y dy/dt = 0 dy/dt = -x dx/dt/y = -5(1)/√39 = - .8 ft/sec (...
*Wednesday, October 6, 2010 at 9:49pm by Reiny*

**calculus**

from xy = x + 4 x dy/dt + y dx/dt = dx/dt + 0 when x = 4 , in original 4y = 4+4 y = 2 so when x=4, y=2 and dx/dt = -5 4dy/dt + 2(-5) = -5 + 0 4dy/dt = 5 dy/dt = 5/4 which is c)
*Thursday, September 22, 2011 at 10:56am by Reiny*

**Calculus**

draw the triangle (lets W, S) label the West leg W km, South leg Skm Distance between ship x. x= sqrt (W^2+S^2) dx/dt= 1/2 *1/sqrt( ) * (2w *dw/dt + 2S ds/dt) find dx/dt Caculate S, W from 1/2 hr at given speeds. you know dw/dt, ds/dt
*Sunday, November 6, 2011 at 5:58pm by bobpursley*

**Calculus**

If V = IR, then dV/dt = IdR/dt + RdI/dt plug in the given values, and solve for dV/dt (You will have to find R from the first equation)
*Monday, December 14, 2009 at 4:11am by Reiny*

**Chemistry**

I can't write delta on my keyboard; therefore, d = delta, t = time rate = -1/2(d[C4H10]/dt) = -1/13(d[O2]/dt) = 1/8(d[CO2]/dt) = 1/10(d[H2O]/dt)
*Sunday, November 27, 2011 at 11:04am by DrBob222*

**calculus**

PV^k=C take the derivative kPv^(k-1) dV/dt+V^k dP/dt=0 solve for dV/dt dV/dt= -V/Pk * dP/dt check that.
*Thursday, April 1, 2010 at 4:22am by bobpursley*

**chemistry**

a. I can't write a delta symbol. delta T will be dT. rate = d[N2]/2*dT = d[CO2]/dT b and c are done the same way except the disappearance is shown with a - sign. rate = d[CO2]/dT = -d[NO]4*dT
*Monday, October 15, 2012 at 6:56pm by DrBob222*

**calc**

Volume= 4/3 PI r^3 dV/dt=4PI r^2 dR/dt you are given dV/dt as 14ft^2/hr you are looking for dr/dt so what is the radius at t=3? 210-14*3=4/3 PI r^3, solve for r. put that r into the dV/dt equation, and solve for dR/dt
*Monday, February 14, 2011 at 9:07pm by bobpursley*

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