Math 114, Calculus I, Laboratory 7
1. Recall the function TA that transforms m, the mole fraction of element A, into the temperature at which a solution made up of m mole fractions of A will freeze. This function is described below.
4 3 2 5m 23.6m 146m 129m 17 0 m 0.76 TA(m) 1/ 3 45.056(m 0.76) 11.98132480 0.76 m 1
You determined in Lab 6 that if n represents the mole fraction of element B present in your solution, then m = f(n) = 1 – n. The function that transforms n into the temperature at which a solution made of n mole fractions of B freezes is described by TB(n) = TA(f(n)) = (TA@f)(n).
SUBMIT (a) a plot of TA and (TA@f) on the same coordinate axes (see graph on pre-lab 7) along with an explanation of why the horizontal axis has two labels: m and n. (b) the mixtures where the freezing point increases as m increases, (c) the mixtures where the freezing point decreases as n increases, (d) your physical interpretation (including units) for (TA)(m) and (TA@f)(n) and (e) how the two plots enable you to see that (TA@f)(n) = (TA)(f(n)) f (n).
2. Marc’s 3 year old son Brian brought a snowball into his house over last spring break. Suppose that a spherical snowball melts in such a way that the rate of decrease (measured in cubic inches per hour) is proportional to its surface area. At 10 a.m. its volume is 500in3 and at 11 a.m. its volume is 250in3.
SUBMIT when the snowball melts. (The answer is not noontime.)
3. Many curves that arise in mathematics and its applications are described by relations which do not give one variable in terms of another. An example of such a curve is known as the Devil’s Curve studied by Gabriel Cramer (born in Geneva in 1704), given by y 4 x 4 ay 2 bx 2 0. (A cool thing about these things is that different choices of a and b, called parameters, give you different looking curves. To explore a lot more about algebraic and geometric curves you can visit http://www-history.mcs.st-and.ac.uk/history/Curves/Curves.html). So how do we graph such curves? MapleV does it by selecting a rectangular region of the Cartesian plane, and parsing it into a grid. It then proceeds grid point by grid point to light up points that satisfy the given relation. You first must type with(plots): and then type implicitplot(y^4 - x^4 .6*y^2 + .75*x^2=0, x=-1.5..1.5,y=-3..3,grid=[100,50]);
SUBMIT (a) a graph of the Devil’s curve, (b) two ordered pairs within the viewing rectangle that are certainly on the curve and two that are certainly not on the curve. (c) How many tangent lines are there to this curve at at point with first coordinate x = 0.1? (d) Find a viewing rectangle including x = 0.1 where the graph of this curve has only one tangent line and (e) by localizing your graph, estimate the equation of your tangent line and (f) carefully sketch and label this tangent line on your graph from (a).
4. Continue with the curve from lab activity 3. Let’s focus our attention on the segment of the curve in the viewing rectangle .2 x .2 and .6 y .8 . In this window we see that y is defined as a function of x, since to each value of x there is only one point on the curve with that value of x as its first coordinate. We say that y has been implicitly defined as a function of x in this case. If we can write down an expression for y in terms of x, then we say that y is explicitly defined as a function of x. In this case this is not too difficult, since the relation describing the Devil’s Curve is quadratic in y 2 . So by a couple applications of the quadratic formula and paying careful attention to the correct signs, we would get the 1 explicit expression y 30 10 9 100x 4 75x 2 . (MapleV can also come in pretty 10 handy here, try ex := y^4 - x^4 + a*y^2 + b*x^2: and solve(ex=0,y);. Probably the first of the resulting expressions will be the expression above.)
SUBMIT a plot of the original relation for the Devil’s curve in the window .2 x .2 and 3 y 3, (b) a plot of the original relation for the Devil’s curve in the window .2 x .2 and .6 y .8 , and (c) a plot of the explicit expression for y given above, over .2 x .2 . (d) Also differentiate the explicit expression for y above with respect to x, and use the resulting expression to find the slope of the tangent line to the Devil’s curve at the point (0.1,0.7664217) (this point is not actually on the curve, but it is pretty close). (e) Show using the method of activity 3(e) above that your result in 4(d) makes sense.
5. To find the slope of the tangent line to the Devil’s curve at the point (0.1,0.7664217) by the method of implicit differentiation we should differentiate the relation y 4 x 4 ay 2 bx 2 0, with respect to x, assuming that y is a function of x. (After all it is, if we restrict our attention to an appropriate region of the plane.) Type the following in MapleV: assume(f,procedure); ex:=f(x)^4 - x^4 -.6*f(x)^2 + .75*x^2; diff(ex=0,x); solve( diff(ex=0,x), diff(f(x),x) );
SUBMIT (a) your hand implicit differentiation of the relation which gives the same result as the third MapleV command above (replace by d to get a more familiar notation for the derivative). (b) What do you think f(.1) should be and why? (c) Using this result and the result of the fourth MapleV command above, obtain the slope of the tangent line to the Devil’s Curve at the point (0.1,0.7664217). We note that on many occasions the approach to finding the slope of the tangent line to a curve must follow along the lines outlined in this activity. It is often impossible to write an explicit formula for the function that describes y as a function of x such as we did in activity 4. See the middle of page 161 for an example of such a relation.
6. Look at exercise #52 on page 168.
SUBMIT your answers to parts (a)-(c).