course mth172 see ya tommorrow. i'm kinda really tired.
......!!!!!!!!...................................
20:12:25 Note 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.
......!!!!!!!!...................................
RESPONSE --> y = -3 e^x y' is the same as the function. y = .02 ln(x) y' = .02(1/x) y = 7 x^3 y' = 21 x^2 y = sin(x) / 5 y' = cos(x) / 5
.................................................
......!!!!!!!!...................................
20:14:23 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.
......!!!!!!!!...................................
RESPONSE --> ok.
.................................................
......!!!!!!!!...................................
20:28:50 `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?
......!!!!!!!!...................................
RESPONSE --> y' = 5/(t) y' =2 units per time
.................................................
......!!!!!!!!...................................
20:29:48 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.
......!!!!!!!!...................................
RESPONSE --> oohhh crap. i did the math backards'.
.................................................
......!!!!!!!!...................................
20:35:05 `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?
......!!!!!!!!...................................
RESPONSE --> y = e^t / 10 y' = e^t / 10 y' = .693 / 10 y' =.0693
.................................................
......!!!!!!!!...................................
20:36:26 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.
......!!!!!!!!...................................
RESPONSE --> oohh crap. i did nat. log not e. ok. i got it.
.................................................
......!!!!!!!!...................................
20:43:10 `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?
......!!!!!!!!...................................
RESPONSE --> y = 12 * t^3 y' = (12) 3* t^(3-1) y' = 36* t^2 y' = 36* (15)^2 y' = 540^2 y' = 291600
.................................................
......!!!!!!!!...................................
20:45:03 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.
......!!!!!!!!...................................
RESPONSE --> dammit. order of operations. i'm tired. i had to work on my day off. i'll finish it tomorrow.
.................................................