course Mth271 Nc|assignment #019
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13:27:00 2.6.12 2d der of -4/(t+2)^2 What is the second derivative of your function and how did you get it?
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RESPONSE --> f(x) = -4 / (t = 2)^2 g(x) = t + 2 g'(x) = 1 f(x) = 4 / x^2 f'(x) = 4 / 2x f'(x) = 4 / 2(t + 2) f''(x) = 1/2 confidence assessment: 2
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13:28:16 You need to use the Constant Rule. [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] ' By the chain rule, with g(z) = z^-2 and h(t) = t + 1 this gives us -4 [ h '(t) * g ' ( h(t) ] = -4 [ (t+2) ' * -2(t+2)^-3 ] = -8 ( t+2)^-3. So g ' (t) = -8 ( t+2)^-3. Using the same procedure on g ' (t) we obtain g '' (t) = 24 ( t + 2)^-4. **
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RESPONSE --> I didn't begin with the Constant rule to factor out the 4. self critique assessment: 2
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13:29:38 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> I didn't have 30, here is what I had for 31 f''''(x) = ( x + 1)^2 = x^2 + 1 f'''''(x) = 2x f''''''(x) = 2 confidence assessment: 2
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13:29:54 The fourth derivative f '''' is equal to the derivative of the third derivative. So we have f '''' = (f ''') ' = [ 2 sqrt(x-1) ] ' = 2 [ sqrt(x-1) ] '. Using the Chain Rule (noting that our function is sqrt(z) with z = x - 1, and that sqrt(z) ' = 1 / (2 sqrt(z) ) we get 2 [ (x-1)' * 1/(2 sqrt(x-1) ) ] = 2 [ 1 * 1/(2 sqrt(x-1) ) ] = 1 / sqrt(x-1). **
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RESPONSE --> self critique assessment: 3
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13:33:37 2.6.42 brick from 1250 ft
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RESPONSE --> I had number 43 to do. ds/dt = 90t / (t + 10) a = [(t +10) (90)] * [(90t) (1)] = 100 / (t + 10)^2 t 10 20 30 40 50 60 v 45 60 67.5 72 75 77.14 ds/dt .25 .11 .06 .04 .03 .02 As velocity increases, accelaration decreases. confidence assessment: 2
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13:33:51 The detailed analysis is as follows: The equation comes from a = -32; since a = v' we have v' = -32 so that v = -32 t + c. Then s' = v so s' = -32 t + c and s = -16 t^2 + c t + k. c and k are constants. If s(t) = -16 t^2 + c t + k and at t = 0 we have s = 1250, then s(0) = -16 * 0^2 + c * 0 + k so k = 1250 and s(t) = -16 t^2 + c t + 1250. If the ball is dropped from rest then the initial velocity is v(0) = 0 so v(0) = -32 t + c = 0 and -32 * 0 + c = 0 so c = 0. So s(t) = -16 t^2 + c t + k becomes s(t) = -16 t^2 + 1250. To find how long it takes to hit the sidewalk: Position function, which gives altitude, is y = -16 t^2 + 1250. When the brick hits the sidewalk its altitude is zero. So -16 t^2 + 1250 = 0, and t = + - `sqrt(1250 / 16) = + - 8.8, approx. The negative value makes no sense, so t = 8.8 seconds. To find how fast the brick was moving when it hit the sidewalk: velocity = -32 t so when t = 8.8 we have velocity = -32 * 8.8 = -280 approx. That is, when t = 8.8 sec, v = -280 ft/sec. **
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RESPONSE --> self critique assessment: 3
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O{s{Ϝ assignment #019 019. `query 19 Applied Calculus I 05-06-2009
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17:47:50 2.6.12 2d der of -4/(t+2)^2 What is the second derivative of your function and how did you get it?
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RESPONSE --> g(x) = t + 2 g'(x) = 1 f(x) = 4 / (t + 2) f'(x) 4 / (2x) g'(t) = 4 / [2 ( t+2) = 4 / (2t+4) g''(t) = 1/2 confidence assessment: 2
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17:48:47 You need to use the Constant Rule. [ -4(t+2)^-2 ] ' = -4 [ (t+2)^-2 ] ' By the chain rule, with g(z) = z^-2 and h(t) = t + 1 this gives us -4 [ h '(t) * g ' ( h(t) ] = -4 [ (t+2) ' * -2(t+2)^-3 ] = -8 ( t+2)^-3. So g ' (t) = -8 ( t+2)^-3. Using the same procedure on g ' (t) we obtain g '' (t) = 24 ( t + 2)^-4. **
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RESPONSE --> I have a problem with remembering to enter the Constant Rule to begin things with. I need to remember to do that. self critique assessment: 2
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17:49:55 2.6.30 f'''' if f'''=2`sqrt(x-1)
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RESPONSE --> I had number 31 to do. f^(4)x = (x+1)^2 = x^2 + 1 f^(5)x = 2x f^(6(x = 2 confidence assessment: 2
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17:49:57 The fourth derivative f '''' is equal to the derivative of the third derivative. So we have f '''' = (f ''') ' = [ 2 sqrt(x-1) ] ' = 2 [ sqrt(x-1) ] '. Using the Chain Rule (noting that our function is sqrt(z) with z = x - 1, and that sqrt(z) ' = 1 / (2 sqrt(z) ) we get 2 [ (x-1)' * 1/(2 sqrt(x-1) ) ] = 2 [ 1 * 1/(2 sqrt(x-1) ) ] = 1 / sqrt(x-1). **
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RESPONSE --> self critique assessment: 1
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17:52:58 2.6.42 brick from 1250 ft
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RESPONSE --> I had number 43 to do ds/dt = 90t/ (t +10) a = [(t +10) (90) - (90t) (1)] / (t +10)^2 t 10 20 30 40 50 60 v 45 60 67.5 72 75 77.14 ds/dt .25 .11 .06 .04 .03 .02 As velocity increases, accelleration decreases. confidence assessment: 2
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17:52:59 2.6.42 brick from 1250 ft
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RESPONSE --> confidence assessment: 1
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17:53:06 The detailed analysis is as follows: The equation comes from a = -32; since a = v' we have v' = -32 so that v = -32 t + c. Then s' = v so s' = -32 t + c and s = -16 t^2 + c t + k. c and k are constants. If s(t) = -16 t^2 + c t + k and at t = 0 we have s = 1250, then s(0) = -16 * 0^2 + c * 0 + k so k = 1250 and s(t) = -16 t^2 + c t + 1250. If the ball is dropped from rest then the initial velocity is v(0) = 0 so v(0) = -32 t + c = 0 and -32 * 0 + c = 0 so c = 0. So s(t) = -16 t^2 + c t + k becomes s(t) = -16 t^2 + 1250. To find how long it takes to hit the sidewalk: Position function, which gives altitude, is y = -16 t^2 + 1250. When the brick hits the sidewalk its altitude is zero. So -16 t^2 + 1250 = 0, and t = + - `sqrt(1250 / 16) = + - 8.8, approx. The negative value makes no sense, so t = 8.8 seconds. To find how fast the brick was moving when it hit the sidewalk: velocity = -32 t so when t = 8.8 we have velocity = -32 * 8.8 = -280 approx. That is, when t = 8.8 sec, v = -280 ft/sec. **
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RESPONSE --> self critique assessment: 1
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z܌Мɒevq assignment #020 020. `query 20 Applied Calculus I 05-06-2009
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17:56:59 2.7.16 dy/dx at (2,1) if x^2-y^3=3
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RESPONSE --> x^2 - y^3 = 3 d/dx[ x^2 - y^3] = d/dx(3)\ d/dx(2x) - (3y^2) = 0 d/dx = 3y^2) / 2x (2,1) d/dx = 3(1^2) / 2(2) d/dx = 3/4 confidence assessment: 1
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17:57:23 The derivative of x^2 with respect to x is 2 x. The derivative of y^3 with respect to x is 3 y^2 dy/dx. You can see this by realizing that since y is implicitly a function of x, y^3 is a composite function: inner function is y(x), outer function f(z) = z^3. So the derivative is y'(x) * 3 * f(y(x)) = dy/dx * 3 * y^3. So the derivative of the equation is 2 x - 3 y^3 dy/dx = 0, giving 3 y^2 dy/dx = 2 x so dy/dx = 2 x / ( 3 y^2). At (2,1), we have x = 2 and y = 1 so dy/dx = 2 * 2 / (3 * 1^2) = 4/3. **
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RESPONSE --> I mixed up the numerator and the denominator, I need to pay close attention to the order of things. self critique assessment: 2
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17:59:31 2.7.30 slope of x^2-y^3=0 at (1,1) What is the desired slope and how did you get it?
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RESPONSE --> x^2 - y^2 = 0 d/dx(x^2 - y^2) = d/dx(0) d/dx (2x - 2y)dy/dx = 0 -2y (dy/dx) = -2x dy/dx = -2x/-2y = x/y (1,1) dy/dx = 1/1 = 1 confidence assessment: 2
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18:00:11 The derivative of the equation is 2 x - 3 y^2 dy/dx = 0. Solving for dy/dx we get dy/dx = 2x / (3 y^2). At (-1,1) we have x = 1 and y = 1 so at this point dy/dx = 2 * -1 / (3 * 1^2) = -2/3. **
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RESPONSE --> I know what I did, I thought the problem was y^2, instead of actual y^3. Sorry. self critique assessment: 2
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18:04:43 2.7.36 p=`sqrt( (500-x)/(2x))
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RESPONSE --> d/dx(x^2 +y^2) = d/dx(9) 2x + 2y (dy/dx) = 0 2y(dy/dx) = -2x dy/dx = -2x/2y Pt: (0,3) -2(0) /2(3) = 0 (y + 3) / (x-0) = 0 y + 3 = x y = x - 3 Pt: (2, (sqrt(5) -2(2) / 2*sqrt(5) = -4 / 2*sqrt(5) y + sqrt5 / x + 2 = -4/(2*sqrt5) y + sqrt5 = -4(x+2) / 2*sqrt5 = -4x - 8 / 2*sqrt 5 y = (-4x - 5 - sqrt5) / 2*sqrt5 confidence assessment: 2
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18:05:21 You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 - dx / dp = 0 (2 p^2 - 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 - 1) **
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RESPONSE --> I could have messed up on which problem was asked for. This isnt the problem I thought it was. Sorry. self critique assessment: 1
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ӇξjScΔ assignment #021 021. `query 21 Applied Calculus I 05-06-2009
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18:11:33 **** Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 **** What is dy/dt if x=3, y=4 and dx/dt = 8?
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RESPONSE --> a) y^2 = -x^2 + 25 d/dt(y^2) = dx/dt(x^2+25) 2y(dy/dt) = dx/dt(2x) dy/dt = 2x/2y = x/y(dx/dt) dt/dt =- 3/4 (8) dy/dt = 6 b) x^2 = y^2 + 25 dy/dt(x^2) = dx/dt(y^2 + 25) dy/dt (2x) = -2y(dx/dt) dy/dt = (-2y /2x)dx/dt dy/dt = -y/x (dx/dt) dy/dt = -4/3(-2) dy/dt = 7/3 confidence assessment: 1
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18:12:00 At (3,4) you are given dx/dt as x ' = 8. Since 2x dx/dt + 2y dy/dt = 0 we have 2(3) * 8 + 2 * 4 dy/dt = 0 so dy/dt = -48/8 = -6. At (4,3) you are given dy/dt as y' = -2. So you get 2 * 4 dx/dt + 2 * 3 * -2 = 0 so 8 dx/dt - 12 = 0 and therefore 8 dx/dt = 12. Solving for dx/dt we get dx/dt = 12/8 = 3/2. **
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RESPONSE --> I left out a negative sign on my 6. self critique assessment: 2
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18:14:38 **** Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?
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RESPONSE --> r = 6 v = 4/3 *pi*r^3 dV/dt = 2 d/dt(V) = d/dt (4/3 *pi*r^3) dV/dt = 4/3 *pi*(3r^2) (dr/dt) 1/4*pi*r^2 dVdt = dr/dt [1/4(pi)(6^2)] *(2) = dr/dt dr/dt = [1/144*pi] * 2 dr/dt = 2/288*pi = 1/144*pi confidence assessment: 1
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18:15:32 The shape is a sphere. The volume of a sphere, in terms of its radius, is V = 4/3 `pi r^3. Taking the derivative with respect to t, noting that r is the only variable, we obtain dV/dt = ( 4 `pi r^2) dr/dt You know that r increases at a rate of 2 in / min, which means that dr/dt = 2. Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx. Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **
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RESPONSE --> I think I made this problem harder than it actually was. I don't really know why I brought it below. But I can see now how easy this problem is. self critique assessment: 2
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