© 2002 by Karl Hahn
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If you have followed along with all the sections prior to this one, you have completed your introduction to limits, continuity, and derivatives. Those topics form the part of calculus known as differential calculus. Here are presented for you a series of problems that are similar to what you might find on an exam. Each problem is accompanied by a "battle plan," which is an outline of what you should do to find the solution. Each also has listed with it a "par" time. In actual exam situations, of course, there is a time limit. The par times are meant to give you an idea of the rate at which you will have to work to complete a real exam in the time allowed, and the degree to which you need to study and practice more. You should not, however, sacrifice the quality of your work for speed. Try to do your best and most accurate work on the problems, and if you find you are over-par, it just means you need to practice a little more.
Problem 1) [par: 9 minutes] Find the equations of all the lines passing through the point, (1,4), that are tangent to the curve,
y(x) = x3 - 10x2 + 6x - 2as well as each point of tangency.
You do problems like this by solving for the x values of the points of tangency. At the point of tangency, the slope of the line, m, must be equal to y'(x) and the line itself must be equal to the function, y(x).
Your battle plan:- Find the derivative of y(x)
- Use the equation of the line in the form of y - 4 = m(x - 1), where you substitute the cubic given for y and its derivative for m (note that the y - 4 and the x - 1 come from the fact that the line has to pass through the point (1,4)).
- At this point you'll have an equation to solve for x. Find all the solutions for x.
- From each solution for x, you can find an m by plugging the x into the function you got for y'(x).
- Use each m you got to find a corresponding b to form the equation of the line with slope of m that passes through the point, (1,4). That is, solve the y intercept for each slope that makes the line pass through the desired point.
- Take each of your solutions for x and plug them back into the original cubic, y(x). Those (x,y) combinations are the points of tangency.
Problem 2) [par: 9 minutes] Use implicit differentiation to solve for y' in terms of x and y if
i) y3 - 3xy + 2x2 = 3
x
ii) |
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iii) xy2 - 3x2y + 4 = 0Your battle plan: In each example, use the rules of differentiation to take the derivative of both sides of the equation. Then use algebra to get the y' stand by itself on one side of the equation.
Problem 3) [par: 12 minutes] Johanas Kepler was the first to solve the riddle of how the planets orbit the sun -- that is that they follow elliptical paths, and that a planet sweeps equal area in equal time. But to find out where a planet actually is at a particular time, the equal area in equal time about an ellipse requires you to solve a nasty equation:
x = y - E sin(y) Kepler's equationwhere E is the eccentricity of the ellipse. Here x is known, but y is not. For any given value of x (which is proportional to the area the planet has swept), you need to be able to determine the value of y that solves Kepler's equation. The y turns out to be the angle that the planet has traversed in the time since its nearest approach to the sun. You can try all day to solve this thing using algebra and trig and you won't get anywhere. For that reason, we say that this equation is transcendental. But you can use Newton-Raphson iteration to rapidly approximate solutions to this equation. So for this problem, set up the iteration
Your battle plan: Recall that Newton-Raphson finds the place where a function is zero. Here we are looking for where the function, y - E sin(y) is equal to some given value for x. So munge it into something that is in the form of f(y) = 0. Then find the derivative of f with respect to y, because you'll need that to apply the Newton-Raphson formula. Finally, plug your f(y) and f'(y) into the formula, put in the numbers, and iterate it with your calculator. And don't be confused because I threw you a curve by reversing the x and y symbology. Just make the translation as you go.
Problem 4) [par: 12 minutes] Find the dimensions of the isosceles triangle whose area is maximum, with the constraint that its perimeter is equal to the constant, P.
Your battle plan: Note that the constraint gives you the perimeter, P, and nothing else. It is reasonable to assume the dimensions of the solution triangle will end up being in terms of P. Indeed, if you double the perimeter, it is reasonable to assume that the base and height of the triangle that maximizes area would also double. Keep that in mind as your solution emerges. It will guide you as to whether your answer is right.
This is partly a geometry problem, so recall what you learned in geometry about isosceles triangles. In particular, if you bisect one of the angles of an isosceles triangle, you cut it into two congruent right triangles. That bisector will also bisect the base. The length of the bisector gives you the height of the isosceles triangle. Draw the whole thing so that you can visualize it. Label each half of the base as having length, x. Label the other two sides as having length, s. Let the height be h. Write an equation that relates s and x to the perimeter, P. Write another equation that relates the height, h, to s and x. Use that to write the equation for the area of the triangle in terms of s and x. Now use the perimeter equation to substitute s with an expression of x and P. So now you will have an area equation in terms of only x and the constant, P. Find the derivative, dA/dx, of the area equation, and set it to zero. Now solve for x. Finally, use your solution for x to express the base and height of the triangle.
Problem 5) [par: 8 minutes] Farmer Gray owns productive cropland that abuts a highway. But the land that is 25 meters or more from the highway is yields twice as much per square meter as the land that is within 25 meters of the highway. Farmer Gray wants to fence-in some of the land to keep the farm animals out. Zoning regulations require that one section of any fence be right along the highway. Farmer Gray has only 600 meters of fencing material. If he fences a rectangular area, what are the dimensions of the rectangle that maximizes the total yield of the fenced-in land (assume that the highway is straight).
Your battle plan: Draw a diagram of the rectangle. Divide it into the two regions -- that is less than 25 meters from the highway and more than 25 meters from the highway. Label the width and length of the rectangle. If b is the number of bushels produced per square meter of land less than 25 meters from the highway, then 2b is the number of bushels produced per square meter more than 25 meters from the highway. You don't ever need to know what b is equal to in order to complete this problem -- only that b is a constant. Write the total yield, Y, as a function of b, length, and width. You know that the perimeter of the rectangle is 600 meters, which is also twice the length plus twice the width. Use that fact to write an expression for length in terms of width or vice versa. Substitute that into the equation for yield. Now yield will be in terms of the constant, b, and the remaining variable (length or width). Find the derivative of Y with respect to the remaining variable, and set that derivative to zero. Use the resulting equation to solve for that remaining variable. Plug that solution back into the perimeter equation to get the other of the two dimensions.
Problem 6) [par: 10 minutes] Use combinations of the rules for finding derivatives to find the derivatives of the following functions:
___________
i) f(x) = cos(x)√sin2(x) + 1
cos(x2)
ii) f(x) =
x2 - 2x + 14
On the next one, find the second derivative of the function given:
iii) f(x) = e-x sin(1/x)
Your battle plan: On i), it is a product of two functions. The first factor is a pretty simple function whose derivative you should know. But the second factor is a composite of a composite (can you see why?). This suggests that you will have to apply the product rule once and the chain rule twice.
On ii), this one is a quotient, but the numerator is a composite. This suggests that you will have to use the quotient rule once and the chain rule once.
Number iii) is the hardest because you have to take the derivative twice (because it asks for a second derivative), and the first derivative you get is more complex than the original function. This one is the product of a simple function and a composite. That suggests that to get the first derivative, you will apply the product rule once and the chain rule once. But you are not done yet -- you still have to find the second derivative, which is the derivative of the result you just got. It probably looks nasty to you because it contains a product of three factors. But don't worry. You can take the derivative of a product of three factors by applying the product rule twice to any such product. Do observe that each of the products that constitutes the first derivative contains, as one of its factors, a composite. So you will still have to make use of the chain rule several times.
If you are ambitious, you could skip taking the first derivative altogether and go directly for the second derivative directly from the original function by applying Leibniz' rule.
Problem 7) [par: 10 minutes if you use the linear equation solver; 17 minutes without] Find the fourth degree (quartic) polynomial that passes through the origin and to which the line
x + 2y = 14is tangent at both x = 4 and at x = -2.
Your battle plan: First, put the equation of the line into standard slope-intercept form. From that you can establish both the slope of the line as well as the y value of the line at both of the x values specified in the problem. The rest of the problem you must establish the five coefficients of the polynomial. But one of them is already given by virtue of the polynomial having to pass through the origin. What is that? That leaves only four coefficients for you to establish. Call them A, B, C, and D. Write the polynomial, P(x), in terms of those unknown coefficients. Also write the polynomial's derivative, P'(x), in terms of those unknown coefficients. From here you can establish your first equation by replacing x in P(x) with the first of the x values given in the problem. What must P(x) be equal to at that x in order to be tangent to the line at that point? Equate your substituted polynomial to that value. Establish the second equation the same way, but using the second x value given in the problem. Establish the third equation by replacing x in P'(x) with the first of the x values given in the problem. What must P'(x) be equal to at that x in order to be tangent to the line? Establish the fourth equation the same, but using the second x value given in the problem. You will now have four linear equations in the four unknowns, A, B, C, and D.
Problem 8) [par: 8 minutes] Find the point (or points) closest to the origin on the conic,
x2 + xy = Cwhere C can be any constant.
Your battle plan: First, DON'T go replacing C with some number of your choosing just because the problem says it can be "any constant." You want C to occur in your solution so that you can substitute "any constant" into the solution when you're done with this. So solve it with the C in place.
This problem is no different from most of the optimization problems you have already done (like the Farmer Gray problem above). You have a constraint and a parameter to be optimized. In this case the constraint is that the point is on the conic, and the parameter is distance from the origin. To establish distance from the origin, you use the Pythagorean formula. Let s be the distance to the origin. Use Pythagoras to write the relationship between s, x, and y. Now use the constraint equation to either write x as an expression in y or vice versa (whichever is easier). Once you have that, use that expression to replace either x or y (depending on which path you chose for the previous step) in the distance relationship. Now find the derivative of the distance, s, with respect to the remaining variable. Set it to zero, and solve for the remaining variable (which will be either x or y). Use back substitution to find the other variable. At that point you will have solved both x and y, which are the coordinates of the point the problem asks for.