Saturday
March 8, 2014

# Search: dt

Number of results: 2,557

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

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

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

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

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

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