Calculus Principles and Applications Quiz

Answer: -,+,-

Find the derivative: f1(x)=3ax^2+2bx+c, so f1(0)=c, and no restrictions are given on c so f1(0) could be positive, negative, or 0. Next, find the second derivative: f2(x)=6ax+2b, and f2(0)=2b; since b has no restrictions, f2(0) could be positive, negative, or 0. Finally, find the third derivative: f3(x)=6a, but the restriction given in the problem was that a>0, so it follows that 6a>0 and f3(0)>0. Thus, the only choice that is not suitable is the one where f3(0) is not positive. Quick note: I apologize for the notation in this question, as it is obviously not standard calculus notation.

There was a glitch in the system that prevented me from using normal notation for this question for some reason.

Answer: cos(x)=sin(x)

Find the derivative using the quotient rule:
f'(x)=[e^(x)cos(x)-e^(x)sin(x)]/e^(2x)

Since there is a minimum at P, f'(x) at the x-coordinate of P is 0 or undefined. Here, it cannot be undefined, because e^(2x) (in the denominator) is never 0. So f'(x)=0 at P, and we have e^(x)(cos(x)-sin(x))=0, and since e^(x) is never 0, we have cos(x)-sin(x)=0, so cos(x)=sin(x).

Answer: 10

We need a formula for how much we travel after going n half-distances. We will call this Dn. Then,

Dn=5+5/2+5/4+5/8+...+5/2^(n-1)

If we factor out the 5, we get

Dn=5(1+1/2+1/4+1/8+...+1/2^(n-1))

=5[(1-1/(2^n))/(1-1/2)]

The limit of Dn as n goes to infinity is 5(1/(1/2))=10, so the answer is 10km.

In practical terms, this is a mathematical refutation of the paradox. If we travel "forever," we do in fact cover the distance in its entirety.

Answer: Sequence I converges, II sometimes converges, and III converges to 0.

Sequence I converges because the limit as n approaches infinity of cos(n*pi) is 0. Sequence II converges only if r is less than 1. If we take the limit of III as n approaches infinity, we find that the denominator approaches infinity, so the sequence itself converges to 0. Note that these are all sequences and not series.

Answer: 0.043

This is a standard related rates problem. From the given information, we can write C=17.4(B)^1/4, where B is the body mass and C is the circulation time. Then dC/dt=17.4((1/4)*B^(-3/4)*dB/dt). Substituting in B=55kg and dB/dt=0.2 kg/month gives the answer: dC/dt=0.043 seconds/month.

Answer: f has both a global maximum and a global minimum on [a,b].

This is a closed interval that the function is defined over, so it will have both a global maximum and a global minimum on that interval. These will occur either at the respective maximum/minimum or at the endpoints. Note that if we consider the function f(x)=x^2, it will not have a global maximum because its domain is the open interval of all real numbers. If we restrict its domain to some closed interval, however, it will have both a global maximum and a global minimum on that interval.

Answer: 0

The function is decreasing, so its derivative is always negative. The largest quantity is therefore the only nonnegative one, or 0.

Answer: (3^(5/3), 3^(1/3))

We start by differentiating implicitly:

3y^2(y')-(y+xy')=0

So y'=y/(3y^2-x), which is undefined when 3y^2-x=0 (remember that the derivative is the slope of the tangent line, and we want this slope to be undefined for the line to be vertical).

So we have x=3y^2.

Substituting into the original equation yields:

y^3-(3y^2)(y)=-6

-2y^3=-6

y^3=3

y=3^(1/3)

Since x=3y^2, we have x=3(3^(2/3))=3^(1+2/3)=3^(5/3)

So the point is (3^(5/3), 3^(1/3)).

Answer: -3.88

We are told that g is an antiderivative of f. So g'(x)=f(x).

This means that g(x) is the definite integral, from a to x, of (t^2+4t)^(1/3)dt.

This integral cannot be solved with any standard techniques (parts, partial fractions, etc.). We have to get creative.

Consider a function h(x) such that h(x)=g(x)-7. Then h(x) is also an antiderivative of f(x) and h(5)=0 (vertical shift 5 units down).

So, we can write that h(x) is the definite integral, from 5 to x, of (t^2+4t)^(1/3)dt. Notice that substituting 5 for x yields 0, as required. Plugging in x=1, we can evaluate that integral on a calculator with 1 and 5 as our limits, and the answer is -10.88222. Then, since h(x)=g(x)-7, g(x)=h(x)+7, and g(1)=h(1)+7=-10.88222+7=-3.88222. Rounding to 2 decimals gives g(1)=-3.88.

Answer: 0

Suppose that x is the part of the string that I will use for the triangle (that is, I make the cut x feet from the left side and use the part that is left of my cut). Each angle of an equilateral triangle is pi/3 radians, and each side is x/3. If the height is h, we have sin(pi/3)=h/(x/3), so h=x*sqrt(3)/6. The area of the triangle is thus At=(1/2)(x)(xsqrt(3)/6)=x^2(sqrt(3)/12).

Now, we need a formula for the area of the rectangle. Since x is how much we used for the triangle, 50-x is how much we use for the rectangle (it is the perimeter). If l is length and w is width, we have

2l+2w=50-x

Since we know that length is twice the width, l=2w

So 2(2w)+2w=6w=50-x

And w=(50-x)/6, so l=(50-x)/3 and the area is Ar=lw, so

Ar=(50-x)^2/18.

The total area is therefore A=At+Ar=x^2(sqrt(3)/12)+(50-x)^2/18

Now this becomes a standard optimization problem. Finding the derivative and setting it equal to zero yields x=100/(2+sqrt(3)). Notice, however, that the second derivative is always positive, so this x value can only be a local minimum. This, the maximum that we need--the global maximum--must happen at an endpoint, x=0 or x=50. At x=0, A=136.89, and at x=50, A=120.28. Thus, the string should not be cut. The maximum total area occurs when the area of the rectangle is 136.89 feet and the area of the triangle is 0 ft. The string should just be bent into a rectangle.