Carlson Jessica Sarah Raubinger Christie Donahue
Group Homework 10/19/06
56. Find the lines through the origin that are tangent to the parabola. y=x2-2x+4
X -3 -2 -1 0 1 2 3
Y 19 12 7 4 3 4 7
The derivative of the parabola is the slope of the tangent line at a given point. Using the Power Rule (d/dx(xn)=nxn-1), we can find the equation of y' by bringing down the power of 2 down in front of the x in the first term of the equation making the term 2x. The second term will become a 2 since the power on the x is a 1. The last term will drop out since there is the "power on the x" is a 0. y'=2x-2 As you can see in the graph above, two lines (one on each side of the y-axis) exist that are line tangent to the graph and that pass through the point (0,0). Since we know the slope of the line is equal to the derivative of the original equation at a given point, we can use the Slope Formula (y2-y1)/(x2-x1) to find the slope of a line that is tangent to the graph of the original function AND passes through the point (0,0). Since we don't know the exact point on the graph that the tangent will pass through, we can use the an arbitrary values of x and y for the graph of the function (x, x2-2x+4). Thus, our two points for the slope formula are (x, x2-2x+4) and (0,0). We used these points as (x1,y1) and (x2,y2) respectively (although they could be switched to obtain the same answer). These two points in the slope formula are equal to the derivative of the equation of the graph, which we found above. 0-(x2-2x+4) = 2x-2 0-x -x2+2x-4 = 2x-2 -x -x2 +2x-4 = -2x^2+2x x2 = 4 x= -2,2
Comment [CCD1]: Should we explain which values are y1 and y2 and x1 and x2 so we know which order to put them into the equation?
Jessica Carlson Sarah Raubinger Christie Donahue
Group Homework 10/19/06
This answer means that wheren x equals -2 and when x equals 2, there is a line that is tangent to the graph of the function that also passes through (0,0). We can plug these xcoordinates into the equation of the derivative of the function, y'=2x-2, which is the slope of each tangent line. f'(-2)=2(-2)-2 =-6 f'(2) = 2(2)-2 =2 Now knowing the value of the derivative at these points, we can use these slope values and the point (0,0) to find the equations of the tangent lines by using the Point-slope Formula (y-y1)=m(x-x1) We will call the tangent line that passes through (0,0) and x=-2 on the graph of the function h(x). h(x)-0=-6(x-0) h(x)=-6x We will call the tangent line that passes through (0,0) and x=2 on the graph of the function g(x). g(x)-0=2(x-0) g(x)=2x 3.2 38. f(t)=5.3(1.018)t t= number of years since 1990 f(0)=5.3(1.018)0 =5.3 The population is 5.3 billion people in the year 1990 (t=0). In order to find an equation for f'(t), we can use the formula for the derivative of ax, where a is a constant: d/dx (ax)=(ln a)(ax) f'(0)=5.3(ln(1.018)(1.018)0) =.09455 In 1990, if the population continued to change by the same amount, the population would increase by approximately .09455 billion, or 94.55 million people by 1991. f(30)=5.3(1.018)30 =9.0513 In 2020 (when t=30), the population will be 9.05 billion people. f'(30)=5.3ln(1.018)(1.018)30 =.16147
Jessica Carlson Group Homework Sarah Raubinger 10/19/06 Christie Donahue 2020, In the population will be increasing by .16147 billion, or 161.47 million people each year. In 2021, there will be approximately .16147 billion more people than in the year 2020, thus we can approximate that the world population at f(31) (in 2021) will be f(30) (population at 2020) plus 0.16147 billion, to total 9.2128 billion people. 56. f(v)=gas consumption in L/km v=km/hr f(80)=.05 f'(80)=.0005 Interpretation: At 80 km/hr the gas is being used at .05 L/km. At 80 km/hr the gas is being used at a rate of .0005 (L/km)/(km/hr)=.0005 (L*hr)/km2 a) Since g(v) is in units of (km/L), g(v) is the reciprocal of f(v). In other words, 1/f(v)=g(v). Now we have to find g(80). Since f(80) =0.5, this means that the car uses .05 L of gas per kilometer traveled. 1/f(80)=g(80)= 1/0.5= 20 km/L. Thus, the car can travel 20 km on one liter of gas. We know that g(v)=1/f(v), thus g'(v)= d/dv (1/f(v)). Using the quotient rule to solve: g'(v)= f(v)*0 1/f'(v) (f(v))2 = -f'(v) (f(v)) 2 To check units:
L * hr L * hr km 2 hr km 2 f' (v) 2 L km 2 L2 L km Graphing the functions we can prove the units.
Jessica Carlson Sarah Raubinger Christie Donahue Now we can go ahead and solve for g'(80). g'(80) = -f'(80) (f(80)) 2 = -.0005 .052 = -0.2 The units here are hours per liter. b. h(v) is gas consumption in L/hr. h'(v) = (L/hr)/(km/hr) = L/km. Thus, h'(v) = f(v).
Group Homework 10/19/06
h'(80) = f(80) = .05 L/km. h(v) = .05v Since we are given that h(v) is in L/hr, and we know that the units of the slope of the graph are in L/km, we can conclude that the graph is linear because we know that L L km . In other words, multiplying the slope velocity is measured in km/hr, and hr km hr by velocity will give you gas consumption in L/hr. Using graphs of the units we can further prove the units of the derivative of h(v).
We know that the graph is linear, but we cannot identify what the y-intercept is without knowing what gas consumption is when velocity equals zero. Assuming gas consumption is 0 L/hr when v=0: h(80)= .05 L/km *(80 km/hr) = 4 L/hr c. Gas consumption can be measured one of two ways. At a constant speed, we can measure how much gas is used for every kilometer driven or also how many kilometers a car will drive on a given amount of gas. Gas consumption isn't always the same, however. For instance, if you're going at a faster speed, such as on the highway, you can sometimes use less gas for every kilometer you drive, which is better for your wallet. On the other hand, city driving (such as stop and go traffic in rush hour) can use more gas for every kilometer driven. To conclude, gas consumption can vary with changes in velocity.
Jessica Carlson Sarah Raubinger Christie Donahue
Group Homework 10/19/06
66. a) The derivative of the function is equal to the function multiplied by a constant k. Thus, our formula is: P'(t)=kP(t) b) P'(t)=k(Aekt) Since in part a, we have that P'(t)=kP(t), we can substitute the given expression of P(t)= Aekt and show that it satisfies the original equation for any constant A. We can prove this by taking the derivative of P(t)=Aekt. We can do this by using the chain rule to find that P'(t)=A(k)(ekt), or P'(t)=kAekt. Thus, the function in part b satisfies the equation in part a.