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About Josh
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When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.

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Experience:
I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.

Supplementary Website:
See a selection of past questions in my maths repository under "Question Archive"

Education Credentials:
Bachelor degree in Engineering Science.

"Everyone struggles with something."

You are here: Experts > Science > Math for Kids > Basic Math > urgent math

Basic Math - urgent math

Expert: Josh - 11/3/2009

Question
Hey there I have a test tomorrow nigh for Calc and I just can't seem to get the answers for these questions, i've been trying hard, but always come out wrong. Please reply by tomorrow any time if you can it will be GREATLY appreciated. PLEASE IF YOU CAN"T GET ALL SHOW ME HOW TO DO ANY THAT YOU CAN DO. IF YOU THINK ITS TOO MUCH PLEASE DO WHAT YOU CAN.

1.A baseball diamond is a square with side 90 ft. A batte1r hits the ball and runs toward first base with a speed of 22 ft/s.

a)At what rate is his distance from second base decreasing when he is halfway to first base?

b)At what rate is his distance from third base increasing at the same moment?

2.Whats the 30th derivative of cos(2x)?

3. An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle x with the plane, then the magnitude of the force is given by the following equation, where μ is a constant called the coefficient of friction.

μW / μsin(x)+cos(x)

a.Find the rate of change of F with respect to x.

b.When is this rate of change equal to 0?

c.If W = 60 lbs and μ = 0.8, draw the graph of F as a function of x and use it to locate the value of x for which dF / dx = 0. (Round the answer to two decimal places.)

4.Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the square is 49 cm^2? (in cm^2/s)

5.Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment in years.

6.If h(x) is given below, where f(3) = 7 and f '(3) = 5, find h'(3)

h(x)=sqrt(7+6f(x))

7.Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 300 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant? cm^3/min

THANKS SO MUCH! George

Answer
Hi George,

It is always stressful having so many unanswered questions the day before an exam. Let's see if we can handle these one by one. With your exam in mind, I hope to convey a sense of what general strategy can be used rather than worry too much about the numerical details.

1. The baseball pitch is drawn as a square instead of a diamond. In the diagram, the vertical axis is represented by OF. As the batter runs to first base (F), the distance |OP|=y from the origin becomes greater. In fact, y(t)=v*t, where velocity v=22. At any point in time, the distance between the person and the second base is given by the Pythagoras distance: D=|SP|=sqrt(|SF|^2+(|FO|-|OP|)^2). Since |SF|=|FO|=90, D(t)=sqrt(2*(90)^2-180*y+y^2). Expressing this as a function of time (using y(t)=v*t), D(t)=[2*(90)^2-180*v*t+v^2*t^2]^(1/2).

S-----------F
|
|
|
|-----------P
|
T-----------O

The rate at which the distance d diminishes is given by the first derivative D'(t)=(1/2)*[2*(90)^2-180*v*t+v^2*t^2]^(-1/2) *(-180*v+2*v^2*t).

| Note: Here, we have simply used the chain rule,
| where dD(f(t))/dt = (dD/df)(df/dt).
| Given D=[f(t)]^(1/2), dD/dt = (1/2)[f(t)]^(-1/2) * f'(t).

The time taken to reach the half way point between O and F is given by T=0.5*|OF|/v, where |OF|=90, v=22. Substituting t=T into D'(t) gives us the answer.

b) Use the same procedure as in part (a), but |PT| has a different expression which you can work out from the geometry.

2) The key to this is recursion. Observe the following relations. For any integer k (i.e., k can be 1, 2, 3 and so forth), d[cos(kx)]/dx = -k*sin(kx). Similarly, d[sin(kx)]=k*cos(kx). Thus, for an even number of differentiations, these two operations will simply be repeated in cascade. Let n=1, say, then 2n times differentiation of cos(kx) yields the nested calculation d{d[cos(kx)]/dx}/dx = d{-ksin(kx)}/dx = -k d[sin(kx)]/dx = -(k^2)*cos(kx). This may look complicated at first sight, but we can understand this in two parts. The exponent for the constant factor k is just 2n. Also, each cycle of even differentiation introduces a minus sign (factor of -1). Thus, for 30=2n differentiations, we will get d^n[cos(kx)]/dx^n = (-1)^n*(k)^(2n)*cos(kx). Here, k=2; n=15 implies an odd number of double differentiation.

3) I'm not sure about the expression because missing parenthesis makes this a little ambiguous. But working from first principle, the static friction coefficient (μ) between two solid surfaces is defined as the ratio of the tangential force (T) required to produce sliding divided by the normal force between the surfaces (N): i.e., μ = T/N. For an object lieing on a horizontal surface, the weight (W) of the object (viz., W=N=mg, where m=mass, g=gravitational acceleration) accounts for the normal force. Using a vector addition diagram, the horizontal force component T=F*cos(x). Thus, we have μ = F*cos(x)/(mg), F=μW/cos(x).

a) We can use, for instance, the quotient rule of differentiation d[f(x)/g(x)]/dx = [g(x)f'(x)-f(x)g'(x)]/[g(x)]^2 to find F'(x). Treating f(x)=μW as a constant (since it is independent of x), F'(x)=[cos(x)*0-μW*-sin(x)]/[cos(x)]^2 = μW*sin(x)/[1-[sin(x)]^2].

b) To solve F'(x)=0, find the value of x (between 0 and pi/2) which sets the numerator to zero.

c) This is a graphing exercise. F(x) is minimized when dF/dx=0. The force required to pull the object (overcome static friction) is smallest when it is parallel to the surface. I think the expression you were given is different to the one we are using here. There are other aspects being modeled by your equation which makes the solution x non-zero. The way to solving this problem is still the same.

4. The initial dimensions are unknown. So, suppose the initial length is x. The initial area is then x^2. If length grows at a rate of v=8 cm/s, then the area at time t is A(t) = (x+v*t)^2 = x^2+2*v*x*t+v^2*t^2. dA/dt = 2[v*x+v^2*t]. Note: v is given, x is either given or assumed to be zero; t is the only unknown quantity in A(t) and A'(t). As in Q1, solve for t when area A=49, then substitute this value of t into A'(t).

5. Set up an exponential decay equation of the form: y(t)=Ao*exp(-kt), where Ao is the full amount (we can let this be 1 for simplicity). The half time is T=5730. So, by definition, y(T)=Ao/2. We use this to work out the constant k. Using the fact that -log(0.5)=log(2), k = ln(2)/T (...a few steps omitted). Finally, set y(t)=0.75*Ao and solve for t.

6. Recall the chain rule mentioned in Q1. dh(f(t))/dt = (dh/df)(df/dt). Given h(x)=sqrt(7+6f(x))=[7+6*f(x)]^(1/2), h'(x)=(1/2)[7+6*f(x)]^(-1/2) * 6*f'(x). Using the data given, knowing f(3) and f'(3) you can find h'(3).

7. One way to proceed is to make V the subject of the equation, treating P as a variable, C as a constant. With V=C/P, knowing that P(0)=150, V(0)=300, you can work out the value of C. To find dV/dt, again use the chain rule. dV(P,t)/dt = (dV/dP)(dP/dt). dP/dt is known (40 kPa/min). dV/dP=-C/(P^2). I'll let you finish the rest. Remember, the value of P at t=0 is given.

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