course mth 271 „·F¢œDÜ轚íÑ´úæ¤c¿b“wRëï£Ì|±`¡¡assignment #011
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21:25:03 `qNote that there are 9 questions in this assignment. `q001. The most basic functions you studied precalculus were: the power functions y = x^n for various values of n, the exponential function y = e^x, the natural logarithm function y = ln(x), and the sine and cosine functions y = sin(x) and y = cos(x). We have fairly simple rules for finding the derivative functions y ' corresponding to each of these functions. Those rules are as follows: If y = x^n for any n except 0, then y ' = n x^(n-1). If y = e^x then y ' = e^x (that's right, the rate of change of this basic exponential function is identical to the value of the function). If y = ln(x) then y ' = 1/x. If y = sin(x) then y ' = cos(x). If y = cos(x) then y ' = - sin(x). There are also some rules for calculating the derivatives of combined functions like the product function x^5 * sin(x), the quotient function e^x / cos(x), or the composite function sin ( x^5). We will see these rules later, but for the present we will mention one easy rule, that if we multiply one of these functions by some constant number the derivative function will be the derivative of that function multiply by the same constant number. Thus for example, since the derivative of sin(x) is cos(x), the derivative of 5 sin(x) is 5 cos(x); or since the derivative of ln(x) is 1 / x, the derivative of -4 ln(x) is -4 (1/x) = -4 / x. Using these rules, find the derivatives of the functions y = -3 e^x, y = .02 ln(x), y = 7 x^3, y = sin(x) / 5.
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RESPONSE --> y'=-3e^x y'=.02/x y'=21x^2 y'=cos(x)/5 confidence assessment: 3
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21:25:09 The derivative of y = -3 e^x is -3 times the derivative of y = e^x. Since by the given rules the derivative of e^x is e^x, the derivative of y = - 3 e^x is y ' = - 3 e^x. The derivative of y = .02 ln(x) is .02 times the derivative of y = ln(x). Since the derivative of ln(x) is 1 / x, the derivative of y = .02 ln(x) is y ' = .02 * 1 / x = .02 / x. The derivative of y = 7 x^3 is 7 times the derivative of x^3. Since the derivative of x^n is n x^(n-1), the derivative of x^3 is 3 x^(3-1), or 3 x^3. The derivative of y = 7 x^3 is therefore y ' = 7 ( 3 x^2) = 21 x^2. The derivative of y = sin(x) / 5 is 1/5 the derivative of sin(x). The derivative of sin(x), according to the rules given above, is cos(x). Thus the derivative of y = sin(x) / 5 is y ' = 1/5 cos(x), or y ' = cos(x) / 5.
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RESPONSE --> yup self critique assessment: 3
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21:26:11 `q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10?
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RESPONSE --> y'=5/t y'=.5cm/s confidence assessment: 3
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21:27:04 The time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of ln(t) is 1 / t, we have rate = y ' = 5 * 1 / t = 5 / t. Since the rate is y ' = 5 / t, when t = 10 the water is rising at rate y ' = 5 / 10 = .5. If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .5 cm/sec.
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RESPONSE --> y'=e^2/10 y'=.7389 cm/s self critique assessment: 3
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21:28:10 `q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2?
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RESPONSE --> y'=e^2/10 y'=.7389 cm/s confidence assessment: 3
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21:28:18 The time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of e^t is e^t, we have rate = y ' = e^t / 10. Since the rate is y ' = e^t / 10, when t = 2 the water is rising at rate y ' = e^2 / 10 = .73, approx. If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .73 cm/sec.
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RESPONSE --> yup self critique assessment: 3
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21:28:50 `q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15?
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RESPONSE --> y'=36t^2 y'=8100 ft/s confidence assessment: 3
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21:28:59 The time rate at which altitude is changing is the derivative y ' = dy / dt of the depth function y. Since the derivative of t^3 is 3 t^2, we have rate = y ' = 12 * (3 t^2) = 36 t^2. Since the rate is y ' = 36 t^2, when t = 15 the altitude is changing at rate y ' = 36 * 15^2 = 8100, approx. If y is altitude in feet and t is clock time in seconds, then the rate is y ' = dy / dt = 8100 ft/sec.
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RESPONSE --> yup self critique assessment: 3
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21:30:55 `q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and what rate is position changing when t = 0, when t = `pi/2, and when t = 4?
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RESPONSE --> y'=.35cos(0) y'=.35 y'=.35cos(pi/2) y'=0 y'=.35cos(4) y'= -.2288 confidence assessment: 3
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21:31:16 The time rate at which position is changing is the derivative y ' = dy / dt of the position function y. Since the derivative of sin(t) is cos(t), we have rate = y ' = .35 cos(t). Since the rate is y ' = .35 cos(t), When t = 0 the position is changing at rate y ' = .35 cos(0) = .35. When t = `pi/2 the position is changing at rate y ' = .35 cos(`pi/2) = 0. When t = 4 the position is changing at rate y ' = .35 cos(4) = -.23. If y is position in cm and t is clock time in seconds, then the rates are .35 cm/s (motion in the positive direction), -.35 cm/s (motion in the negative direction), and -.23 cm/s (motion in the negative direction but not quite as fast).
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RESPONSE --> oops i forgot the units self critique assessment: 3
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21:33:32 `q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5?
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RESPONSE --> y'=12x^2-14x+6 y'=4cos(x)+(8/x0 y'=5e^x +15x^-6 confidence assessment: 3
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21:33:41 Since y = 4 x^3 - 7 x^2 + 6 x is the sum of the functions 4 x^3, -7 x^2 and 6x, whose derivatives are 12 x^2, -14 x and 6, respectively, we see that y ' is the sum of these derivatives: y ' = 12 x^2 - 14 x + 6. Since y = 4 sin(x) + 8 ln(x) is the sum of the functions 4 sin(x) and 8 ln(x), whose derivatives are , respectively, 4 cos(x) and 8 / x, we see that y ' is the sum of these derivatives: y ' = 4 cos(x) + 8 / x Since y = 5 e^x - 3 x^-5 is the sum of the functions 5 e^x and 3 x^-5, whose derivatives are, respectively, 5 e^x and -15 x^-6, we see that y ' is the sum of these derivatives: y ' = 5 e^x + 15 x^-6. Note that the derivative of x^-4, where n = -4, is n x^(n-1) = -4 x^(-4-1) = -4 x^-5.
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RESPONSE --> yup self critique assessment: 3
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21:43:32 `q007. The rule for the product of two functions is a bit surprising: The derivative of the product f * g of two functions is f ' * g + g ' * f. What are the derivatives of the functions y = x^3 * sin(x), y = e^t cos(t), and y = ln(z) * z^-3?
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RESPONSE --> y'=3x^2sin(x)+x^3cos(x) y'=e^t(cos(t)-sin(t)) y'=z^-4(1-3ln(z)) confidence assessment: 3
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21:43:37 The derivative of y = x^3 * sin(x), which is of form f * g if f = x^3 and g = sin(x), is f ' g + g ' f = (x^3) ' sin(x) + x^3 (sin(x)) ' = 3x^2 sin(x) + x^3 cos(x). The derivative of y = e^t cos(t), which is of form f * g if f = e^t and g = cos(t), is f ' g + g ' f = (e^t) ' cos(t) + e^t (cos(t) ) ' = e^t cos(t) + e^t (-sin(t)) = e^t [ cos(t) - sin(t) ]. The derivative of y = ln(z) * z^-3, which is of form f * g if f = ln(z) and g = z^-3, is f ' g + g ' f = (ln(z)) ' z^-3 +ln(z) ( z^-3) ' = 1/z * z^-3 + ln(z) * (-3 z^-4) = z^-4 - 3 ln(z) * z^-4 = z^-4 (1 - 3 ln(z)).
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RESPONSE --> yup self critique assessment: 3
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21:55:25 `q008. The rule for the quotient of two functions is perhaps even more surprising: The derivative of the quotient f / g of two functions is [ f ' g - g ' f ] / g^2. What are the derivatives of the functions y = e^t / t^5, y = sin(x) / cos(x) and y = ln(x) / sin(x)?
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RESPONSE --> y'=(e^t(t-5))/t^6 y'=1/cos(x) y'=1/(xsin(x))-ln(x)cos(x))/(sin(x))^2) confidence assessment: 3
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21:55:31 The derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is (f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2 = (e^t * t^5 - e^t * 5 t^4) / (t^5)^2 = t^4 * e^t ( t - 5) / t^10 = e^t (t-5) / t^6.. The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is (f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 ' = (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 = ( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 = 1 / cos(x)^2. Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1.
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RESPONSE --> yup self critique assessment: 3
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21:55:49 The derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is (f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2 = (e^t * t^5 - e^t * 5 t^4) / (t^5)^2 = t^4 * e^t ( t - 5) / t^10 = e^t (t-5) / t^6.. The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is (f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 ' = (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 = ( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 = 1 / cos(x)^2. Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1. The derivative of y = ln(x) / sin(x), which is of form f / g if f = ln(x) and g = sin(x), is (f ' g + g ' f) / g^2 =( (ln(x)) ' sin(x) - ln(x) (sin(x)) ' ) / (sin(x))^2 = (sin(x) * 1/x - ln(x) * cos(x) ) / (sin(x))^2 = ( sin(x) / x - ln(x) cos(x) ) / (sin(x))^2 = 1 / ( x sin(x)) - ln(x) cos(x) / (sin(x))^2. Further simplification using the tangent function is possible, but the answer here will be left in terms of the sine and cosine functions.
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RESPONSE --> yup self critique assessment: 3
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21:59:16 `q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5.
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RESPONSE --> y'=4/(xsin(x))-4ln(x)cos(x)/(sin(x)^2-(cos(x))^2) y'=3e^t(t-1)/t^2+6/t y'= -25t^4ln(t)+5t^4/(ln(t))^2+cos(t)/5 confidence assessment: 3
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21:59:22 Since the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) = 4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2. Further rearrangement is possible but will not be done here. The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore y ' = 3 e^t ( t - 1) / t^2 + 6 / t. Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 = -25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5.
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RESPONSE --> yup self critique assessment: 3
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