January Regionalcalculus Team Solutions

January RegionalCalculus Team Solutions

1. 46

First find y-coordinate. . Taking the derivative of at will give us the slope of the line tangent to at . . So, the slope of the normal is . We can use these values to give us our equation of . .

2.

The first two derivatives are and . Looking at , we can see that it will always be increasing. So our answer merely depends on the downward concavity of . Therefore, by looking at the second derivative, negative values (and consequently downward concavity) will occur at .

3.

All four parts use the same equation to solve the problem, only with different numbers. , where . Using this equation four times, we get , , , and . Therefore, .

4.

The Left- and Right-hand Riemann sums differ only in that while the Right-hand sum ignores the leftmost data, the Left-hand sum ignores the rightmost data. So the positive difference between the two sums is equivalent to the positive difference between the areas of the rectangles at either end of the data. So, . The trapezoidal approximation is = . For the acceleration, a simple slope-approximation from to will suffice. . Finally, .

5.

For A, , , . For B, , . For C, , , . For D, , . Therefore, , and so our answer is . (There is a rule that does state that , which shows why A and B are equivalent).

6.

The ladder forms a right triangle with the wall and the ground, so we can use the Pythagorean Theorem as our primary equation. The ladder is ft, and the base’s length is increasing (the ladder is being pulled away) at feet per second. Since we want to solve for (with three different values each time), (the ladder c is constant, however, so when we take the derivative of this, we get . Plugging values into this equation for the three sets of values, we get , , and . Finally,

7. 443

A and B are both definitions of the derivative of a given function at . For A, , , and . For B, , , and . Part C is the negative of a definition of the derivative at . So, , , and . Finally, . (L’Hopital’s Rule may be used to evaluate these limits as well, if it is known to the student).

8. 2

By drawing the picture, we see that and . The area of a given rectangle is . To maximize the area we take the derivative and set it equal to zero. In doing this we get , so . We can plug in the value of x into the equation of the semicircle to get y; . Finally, .

9. AD

For all positive values of x, the first equation and all of its derivatives will be positive. So, A is true. B is false because the piecewise function at can only increase if (the problem states for all values of m). C is false because, though the top function has a zero at , it is not a part of its domain; the lower function can only have a zero at if . D is true; set the two functions equal to each other at their domain division (), and we find that . E is false, however, because in order for a value of m to exist, we must be able to attain a value of when we set the derivatives equal to each other (since to be differentiable a function must be both differentiable and continuous). Since m will equal 6 and not 1, then no value of m exists. Finally, F is false; the one sided limit will be 4, but m is equal to 1.

10.

A = .

B = . For C, the maximum displacement is equivalent to the amplitude of y, which is the square-root of the sum of the squares of the individual trigonometric functions. (this can also be solved by taking the derivative, finding the x-coordinate of the maximum, and plugging this value into the original equation). For D, the period of the overall function is the least common multiple of the individual periods. Here, since both have a period of , then the overall function has a period of . Therefore, .

11.

, , and . However, note that since the first few terms of compose of the linear function, and . So our answer only depends on E, since the other four letters will cancel each other out. , and so .

12. 8

For A, since the two functions in question are inverses, the expression can be written as , which is 1. For B, even though neither nor are defined, the derivative of any constant is zero. For C, the determinant of the matrix is equal to 7, and the derivative of is 7. Therefore, .

13.

MVT states that, if a function is continuous on , then there is a point within that interval where . So, . Rolle’s Theorem is a special condition of the MVT, where and consequently . Since , then . The Extreme Value Theorem states that c is a critical point of if or undefined., so , and . Finally, .

14. 9

; B is simply the horizontal asymptote, or . For C, the variable we want to assess the limit of is not the same as the variable that appears in the given function. So we can treat the entire function as a constant; . For D, since is a continuous function where it exists, . Finally, .

15. 36

First we need points of intersection. . So and . Now we evaluate the integral of the difference between the two functions from . ,

which is our final answer.