Derivative+Applications

=__**Derivative Applications**__= = =
 * A derivative has many uses that can be examined.**

 **So it's velocity can be modeled by the derivative, f'(x)=12x^(2) + 6x + 4**  **And if you take yet another derivative, the acceleration is modeled by f"(x)=24x + 6** **Now let's examine the relationships between these functions.**  **When f(x) is increasing, f'(x) is positive. So if position is increasing, then the object will have a positive velocity.**  **And when f(x) is decreasing, f'(x) is negative. So if the position is decreasing, then the object will have a negative velocity.**  **When f"(x) is positive, the original f(x) is concave up.**  **When f"(x) is negative, the original f(x) is concave down.**
 * **If a function, f(x), models the position of an object at a given time, we can use this to find it's velocity and acceleration. Remember that if we take a derivative of the function f(x), then f '(x) is the object's velocity. If we go even further to take the derivative of the velocity, then we have the acceleration, denoted f "(x).**
 * **Example: A car's position is modeled by the equation f(x)=4x^(3) + 3x^(2) + 4x **


 * If [[image:http://upload.wikimedia.org/wikipedia/en/math/f/9/e/f9e4836997912186459bbcbc6e27b746.png caption=" f^{primeprime}(x) < 0"]] then [[image:http://upload.wikimedia.org/wikipedia/en/math/6/1/6/616fb717ed0ab1dbf5ded834a72ae83e.png caption=" f"]] has a local maximum at [[image:http://upload.wikimedia.org/wikipedia/en/math/b/c/5/bc518bd42e0cfa424357a26712828988.png caption=" x"]].
 * If [[image:http://upload.wikimedia.org/wikipedia/en/math/e/0/0/e000178f7f6b9c3307e0f9e4e9e077e9.png caption=" f^{primeprime}(x) > 0"]] then [[image:http://upload.wikimedia.org/wikipedia/en/math/6/1/6/616fb717ed0ab1dbf5ded834a72ae83e.png caption=" f"]] has a local minimum at [[image:http://upload.wikimedia.org/wikipedia/en/math/b/c/5/bc518bd42e0cfa424357a26712828988.png caption=" x"]].
 * If [[image:http://upload.wikimedia.org/wikipedia/en/math/9/1/0/91069bea30cf990f5a96b148822cf83b.png caption=" f^{primeprime}(x) = 0"]], the second derivative test says nothing about the point [[image:http://upload.wikimedia.org/wikipedia/en/math/b/c/5/bc518bd42e0cfa424357a26712828988.png caption=" x"]], a possible [|inflection point].

**Graph I, in blue is the original function. It has a local maximum at x=0 because its derivative, graph III goes from positive to negative. Now notice graph II, the second derivative. Where it first crosses the x axis is where the original function, graph I, changes concavity from concave up to concave down. The next time graph II crosses the x axis, going from negative to positive, graph I goes from concave down, back to concave up. These two points are called Points of Inflection (Where the second derivative changes sign).**



Now here is a series of questions to help practice this concept. The graph of the first derivative f ' of function f is shown below



__Solutions to Questions a-g __

 * ====a) f is increasing for values of x for which f ' is positive. The set of these x values is given by the intervals (-infinity, -6.6) , (0 , 3.6) ====
 * ====b) f is decreasing for values of x for which f ' is negative. The set of these x values is given by the intervals (-6.6, 0) , (3.6 , +infinity) ====
 * ====c) f has a local maximum at values of x for which f '(x) = 0 and f ' changes from positive to negative. Hence f has a maximum at x = -6.6 and x = 3.6 ====
 * ====f has a local minimum at values of x for which f '(x) = 0 and f ' changes from negative to positive. Hence f has a maximum at x = 0 ====
 * ====d) The concavity is determined by the sign of f". Using the graph of f ' shown above, we can deduce the following: 1 - for the values of x in the interval (-4, 2) f ' is increasing and therefore f" is positive. Hence f is concave up for values of on (-4 , 2) 2 - for the values of x in the intervals (-infinity , -4) and (2 , +infinity) f ' is decreasing and therefore f" is negative. Hence f is concave down for values of on (-infinity , -4) U (2 , +infinity). ====
 * ====e) Again using the graph of f ', f"(x) = 0 for x = -4 and x = 2 since f ' has extrema at these two values of x. These are also the values for which f" changes and are therefore points of inflection. ====

<span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">**The graph below is a graph of f(x)on the interval of 0 to 3. Directly below it is a graph of f'(x) on the same interval. Notice that f increases until x=0.5, where it reaches a local maximum. At this point, the derivative of f is equal to 0, and is changing from positive to negative. The graph of f then begins to decrease until x=2. Notice that at this point, f' is equal to 0 again, but this time it is changing from negative to positive. This is called a local minimum.**





What is the actual definition for how to find a derivative? A derivative is a rate of change. So to find a rate of change, you must have some sort of average value. The velocity is also known as the instantaneous rate of change. To find a derivative, you would use the general formula:

This means we are taking a value of a function, f(x), at a given x, and subtracting that value for f(x) from the value of the function if we were to add "h" to it. , h represents a really small value that is getting closer and closer to zero, as noted by the limit. The difference between these two is then divided by the value of h to obtain a velocity. So we can use this on any function to get an instantaneous rate of change at a given point. This is giving us the slope of the line tangent to the graph at that given point.
 * <span style="background-color: #ffffff; background-color: #ffffff; color: #454545; color: #454545; font-family: verdana,arial,sans; font-family: verdana,arial,sans; text-align: -webkit-left; text-align: -webkit-left;">For example, to calculate the average [|rate] of change between the points:

(0, -2) = (0, f(0)) and (3, 28) = (3, //f//(3)) >> <span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">where //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">f //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">( //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">x //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">) = 3 //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">x //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">2 + //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;">x //<span style="background-color: #ffffff; font-family: verdana,arial,sans; text-align: -webkit-left;"> – 2 we would: >> This is the average rate of change of the function between x=0 and x=3. >> The derivative could also be taken to solve a related rates problem. For example, if a container is being filled with water, and we know the shape of the container along with it's dimensions, we can find out how fast the water is increasing in volume, radius, or height. Say we have a spherical container. If water is pouring into the sphere and we know its dimensions, we can use the formula for volume of a sphere to get a rate at which a certain dimension is changing. If the volume is increasing at a rate of 4 cubic centimeters per second, how fast is the radius changing when r= 3 centimeters? Lets derive this to get the rate of change of the volume, dV/dt. >> If dV/dt=4, and r=3, then we can plug these in to find dr/dt, the rate at which the radius changes. >> 4 = 4π (3^2) (dr/dt) → 4 = 36π (dr/dt) → (4/36π) = dr/dt OR 1/9π centimeters per second >> Here are a couple of free response questions that make derivative applications using rates of change: >> >> 1. An 8 foot long ladder is leaning against a wall. The top of the ladder is sliding down the wall at the rate of 2 feet per second. How fast is the bottom of the ladder moving along the ground at the point in time when the bottom of the ladder is 4 feet from the wall. >> Solution: y = distance from the top of the ladder to the ground >> x = distance from the bottom of the ladder to the wall >> dy/dt = -2 Find dx/dt when x = 4 (and y = 4(31/2) by the Pythagorean Theorem) >> SOLUTION: x2 + y2 = 64 >> 2x dx/dt + 2y dy/dt = 0 >> 2x dx/dt - 4y = 0 >> dx/dt = 4y/(2x) = 2(31/2) ft/sec when x = 4 ft >> >> >>> 2. Water is being poured into a conical reservoir at the rate of pi cubic feet per second. The reservoir has a radius of 6 feet across the top and a height of 12 feet. At what rate is the depth of the water increasing when the depth is 6 feet? >>> Solution: h = depth of the water in the reservoir >>> r = radius of the water in the reservoir >>> The volume of the water in the reservoir (V) is given by >>> >>> V = (1/3)pi(r2h) and r = (1/2)h using similar triangles so >>> V = (1/12)pi(h3) >>> pi = dV/dt = (1/4)pi(h2)dh/dt >>> dh/dt = 4/h2 >>> dh/dt when h = 6 is 1/9 ft/sec >>> >>> ** Use of implicit differentiation to find the derivative of an inverse function. **
 * X || Y ||
 * 0 || -2 ||
 * 1 || 2 ||
 * 2 || 12 ||
 * 3 || 28 ||
 * 4 || 50 ||

Slope Fields If you are given a differential equation, say one what is already in terms of dy/dx, a slope field it an x,y plane with a line at each point. The angle of the line depends on what dy/dx is equal to when you plug in both the x and y values. If the differential equation yields an undefined fraction, no line is drawn. For example, if dy/dx = y/x At all points on the plane where x=0, you would not draw anything. At all points where y=0, you would have a flat line because the slope is zero.At (-2,1), dy/dx= -0.5 At (-1,1), dy/dx= -1At (1,1), dy/dx = 1At (1,2), dy/dx = 2 ..... and so on. You would then plot these mini tangents on a slope field at the appropriate point. Each value of dy/dx represents a slope of the tangent line of the original function at that corresponding point. For dy/dx= y/x it would look like: Now how do we find a solution to a differential equation? ***Once you have solved for Y, your new equation is the original function for which the tangent line at that specific point is the differential equation you started with!** __Works Cited**__ [] [] [] []
 * Separate the variables to get all x terms on one side and all y terms on the other.
 * Integrate both sides. Don't forget a "+ C" on one side
 * Plug in the points of the initial condition to solve for C
 * Incorporate your C value into your equation and solve for Y