Let us start by making a general equation for the situation. The parabola can only have an equation of the form y = kx²+4. The first-quadrant x-intercept would then be the square root of -4/k. Now we set up the equations. The integral of kx²+4 from x = 0 to x = sqrt(-4/k) would be equal to the integral of pi*x² from y = 0 to y = 4. In this case, x is solved in terms of y and then squared. If we go ahead and solve the equations, it would yield to k = -9pi²/4, which is approximately -22.21 and the volume - or area - would be 32/9pi or about 1.13. Wooh, a lot of work, wasn't it?

07/01/02) In a party, you can choose between two glasses, in which you can drink. One has a parabolic shape, the other a linear. Yet, both have the same height of 15 cm. If the parabolic glass is 8 cm in diameter on the top, 6 cm in diameter at half of its height and 3 cm in diameter on the bottom, whereas the linear one is 9 cm in diameter on top and 4 cm on the bottom. Comparing the volumes of each, which one should fill up completely and drink in order not to exceed a limit of 40 mL?

Solution:

Woo, a lot of information. Let us first investigate for a way to set up coordinates and make up equations for our glasses' shape. Doing this, we need to use points (2,0) and (9/2,15) for the linear glass and points (3/2,0), (3,15/2) and (4,15) for the parabolic one. These are obtained using the radii and heights that are given in the problem. Now, let us find equations. The general formulas for a line and a parabola eventually lead us to the linear equation of y = 6x-12 and the quadratic equation of y = x²+x/2-3. Not done yet! We have to find the volumes. One way is to use disks of radius x. This would get us a volume of about 36.69 mL for the parabolic glass and 50.85 mL for the linear one. Therefore, after all these calculations, we come up with the fact that the parabolic should be use. Don't drink and drive, no matter which one is chosen!!

Topic:

Applications of Integrals

______________________________________

06/24/02) Part I: The sum of coordinates of a function vary directly with a power n of its y-value at any point, n being a positive integer. Where does this function have extrema? Why?
Part II: The sum of coordinates of a function vary directly with a power n of its x-value at any point, n being a positive integer. For what values of n does this function have extrema?

Solution:

Part I: Mathematical Language requested! The question leads to the result that x + y = kyⁿ, by the defintion of direct relationships. Implicit differentiation would get us this:
dy/dx = 1/(nky^n-1-1), which is NEVER zero. Therefore, these functions do not have an extrema whatsoever.
Part II: Doing the same process would get us the following:
dy/dx = nkx^n-1-1. This can be zero. In fact, it is zero where x equals the (n-1)^th root of 1/(nk). This means that n has to be odd for the function to have extrema. Beautifully solved, right?

Topic:

Derivatives

______________________________________

06/17/02) It has been found that the acceleration and velocity of an object are related in the following equation: a = v²-t²-2bt where a is acceleration, v is velocity, t it time and b is a positive constant. Further information reveals that the object's velocity varies linearly where v = t+b. Calculate the displacement of the object from t = 0 to t = 20 and describe the motion.

Solution:

First let's find b. Knowing that velocity is t+b, we can plug that back into our a equation. So, a = (t+b)²-t²-2bt. Yet we know that the slope of v = t+b is 1 and so is a. Hence, 1 = (t+b)²-t²-2bt = b². Knowing that b is positive, we would get b = 1. Now, we know that v = t+1. The displacement would be the integral of this from 0 to 20, which is 220. The object is moving with a constant acceleration and increasing speed.

Topic:

Applied Calculus

______________________________________

06/10/02) A line is divided into equal parts, each being a diameter of a semicircle. If the process of dividing continues forever, to what value, if any, does the sum of all circumferences converge? What about the sum of areas? Going further, imagine we start adding the areas of the semicircles accumulatingly starting with 2 divisions and going on and on. Would the sum of all infinite areas converge to a value? Explain.

Solution:

We begin by the circumference. Each semicircle has a diamter of L/n where L is the line's length and n is the number of divisions. The circumferences of the semicircles would be (pi(L/n)*n)/2, since there are n semicircles. As you can see, this is always equal to pi(L/2). The area in terms of L and n would be (pi*L²)/8n. The limit of this as n goes to infinity is zero.
The second part of this problem asks for the accumulative area, which can be noted as sum of (pi*L²)/8n from n=1 to infinity. In the study of series, this is just a multiple of harmonic series, which would always diverge.

Topic:

Optimization

______________________________________

06/03/02) A rectangle is to be made in such a way that its base lays on the x-axis and its height is bound by a triangle of vertices (0,0), (6,0) and (4,2). Find the maximum area this rectangle could attain.

Solution:

Let's start by finding the equations of the lines making the triangle. One is y = x/2 and the other is y = -x+6. The base of the rectangle has a length of x₁-x₂ where x₁ is x-value of y = -x+6 and x₂ is the x-value of the other line. These x-values are 6-3y and 2y respectively. So the base of the rectangle is 6-y-(2y) or 6-3y. The height is simply y. The area, hence, is y(6-3y). Optimization tells us that the derivative of the area function has to be zero. Doing that we get y = 1. So the height should be 1 and base has to be 3, using 6-3y = 1. Now we are left with the maximum area of 3(1) or 3. Kinda neat, wasn't it?

Topic:

Optimization

______________________________________

05/27/02) A container is formed by rotating the first quadrant region of y = -x³+4x²+5x about the line x = -1. Find the volume of the container in liters, if the Cartesian coordinate unit is in centimeters.

Solution:

Let's start by the graph. The function is y = -x(x-5)(x+1) which intersects the x-axis in the first quadrant at x=5. Therefore the volume can be thought of as shells, with radius 1+x and height of y from x=0 to x=5. Hence, the integrand would be -2pi(1+x)x³+4x²+5x, from x=0 to x=5. Evaluating this would yield to 1125pi/2 or about 1167 cm³ or 1.167 liters. Not a lot of volume, is it?

Topic:

Volumes and Integrals

______________________________________

05/20/02) Find the area of the yellow region in the following picture. Four semicircles are inscribed in a square of side 10.

Solution:

I am still in wonder how to do this problem without calculus. I would be interested to know how to do this with geometry, but for the purpose of this site, we will do this with calculus. If we set up a coordinate system, a circle of radius 10 and center (-5,-5) intersects both the x- and y-axis and the area of the curve in the first quadrant would be a quarter of the wanted area. Taking the integral of y = sqrt(100-(x+5)²)-5 from x=0 to x=5(sqrt(3-1)). This would lead to an area of 25(pi/3+1-sqrt(3)) and the area asked in the problem would be 100(pi/3+1-sqrt(3)). This is close to 31.5, interesting right?

Topic:

Areas and Integrals

______________________________________

05/13/02) Two tangents are drawn at the function f(x)=(x-2)(x+3)(x+5) in points x=-4 and x=-1. The area enclosed by the two lines and the graph is divided into two parts by the function itself. What is the ratio of the larger portion to the smaller one?

Solution:

A lot of calculations in this, right? A combination of ideas. First we find the equations of the lines. One is y = -x+2 at x =-4; the other is y = -10x-34 at x = -1. The intersection of the lines is fortunately at x = -4 itself. Now the integral part. The integral of f(x)-(-10x-34)dx from -4 to -1, which turns out to be 27/4. The larger portion is the integral of -x+2-f(x)dx from -4 to 2, which yields to 108. The ratio is therefore found to be 16. Oh, this problem is not hard, but painstaking, huh?

Topic:

Applications of Derivatives

______________________________________

05/06/02) The largest cylinder that can be inscribed in a sphere is taken out of the sphere. What fraction of the original volume remains?

Solution:

This problem seems hard, but a drawing would help. The radius of the sphere, R, is constant. In terms of R and h, the height of cylinder, the radius of the cylinder is the square root of R²-h². The volume of the cylinder would then be pi*h*root(R²-h²). Applying derivatives, the maximum volume would occur when h equals 2R/root3. Now, we compare the volumes together. The volume of the sphere would be (4/3)pi*R³. The volume of the cylinder can be found in terms of R now that we know what h is in terms of R. As it turns out the ratio would be root3 to 1. Now this 1 is taken out of the sphere. So the fraction remaining is (root3-1)/root3. Neat, isn't it?

Topic:

Applications of Derivatives

______________________________________

04/29/02) Write a generalized condition or equation for a cubic function whose inflection point is also its extrema. Use your own notations.

Solution:

Let us assume the cubic is ax³+bx²+cx+d. The derivative would thus be 3ax²+2bx+c and the second derivative would be 6ax+2b. Then, we equate each to zero and solve for x by using quadratic formula for the first one and simple algebra for the second one. Now we know the x has to be the same for both; so beautifully we just need to make sure the term under the square root is zero in our quadratic formula. Therefore, 4b²-12ac = 0, or more simply b²-3ac is equal to zero. This is the condition we need to have in order for the cubic to be as questioned. Yet, it is interesting that the extrema would not be a relative maximum or minimum, but just an inflection point, in which the tangent line has a slope of zero!

Topic:

First and Second Derivative

______________________________________

04/22/02) Find the derivative of y = (x-1)²:
i) with respect to 2x.
ii) with respect to ln(x).
iii) with respect to 2x³.

Solution:

Seems hard, doesn't it? Now, let's learn a trick. We know that (dy/dx)/(dt/dx) is the same as dy/dt since dx would never equal zero. Therefore, for finding the derivative of any function with respect to any t, we just use this formula. Let us apply it. In part i, we have (d(x-1)²/dx)/d(2x)/dx, which would yield to (2x-2)/2 or just x-1. In part ii, it would be (d((x-1)²/dx)/(d(ln x)/dx) that is (2x-2)/(1/x) or 2(x²-1). Finally using the same process part iii would yield to (2x-2)/6x² or (1/3x)-(1/3x²)

Topic:

Derivatives

______________________________________

04/15/02) AP Calculus Sample Problem From point (3,5) two tangents are drawn to y = x². Find the slopes of these two lines.

Solution:

This is not a very hard problem. You have to carefully realize that each point at the parabola has coordinates (h,h²). Now using derivatives, you know that the slope of any tangent at y would be 2x or in this notation 2h. Now we imply the slope formula. (5-h²)/(3-h)=2h. Solving for h, we get two values of 1 and 5. The slopes are 2h, so they are 2 and 10 respectively.

Topic:

Derivatives and Slopes

______________________________________

04/08/02) Part I: Is it legal to say that the integral of (1/x)dx is ln(kx)+C where k and C are constants? Prove your point from both a derivative and integral point of view. Emphasize on the k part!
Part II: Prove that the derivative of ln(sin x+cos x) is (1-sin(2x))/cos(2x).

Solution:

Part I: Yes, indeed it is. From Mr. derivative's point of view, the derivative of ln(kx)+C is (1/kx)*k which is 1/x. From Mrs. Integral's point of view, ln(kx₁)-ln(kx₂) is just ln(x₁)-ln(x₂) according to the rules of logarithm. Therefore, the Fundamental Theorem of Calculus would prove this true.
Part II: Oh, a little trig, ready? The derivative would be 1/(sin x+ cos x) times cos x-sin x, because of chain rule. Now playing around with this we can multiply the numinator and denominator by sin x-cos x. That would yield to (1-2sin x*cos x)/sin²x - cos²x). Using double angle formulas, this would be (1-sin 2x)/(cos 2x).

Topic:

Derivatives and Integrals

______________________________________

04/01/02) Find the integral of csc⁶(x)-cot⁶(x)dx. Show all your work.

Solution:

Let's do some trigonometry, without which it is not possible to solve this problem. The integrand can be rewritten as
(csc²(x))³-(cot²(x))³. Now we use the algebraic expansion for difference of cubes. According to that, this can be rewritten as
(csc²(x)-cot²(x))(csc⁴(x)+csc²(x)*cot²(x)+cot⁴(x)). The first term is 1, according to pythagorean theorem. The second term can be simplified even more to
(csc²(x)-cot²(x))²+3csc²(x)*cot²(x). Hence, the whole expression is simplified to 1+3csc²(x)*cot²(x). It is way easier to take the integral of this rather than the first expression. By substituting u for cot(x) we obtain the solution to be x-cot³(x)+C. Wasn't that sweet?

Topic:

Algebra, Trig & Calculus

______________________________________

03/25/02) A bowl's shape looks like the graph of y=(x⁸-x⁶)^1/4 rotated about the x-axis from x=1 to x=4. What would the volume of this bowl be? Show how you solve it.

Solution:

Integral problem right? The volume would be the integral of pi times y(x)²dx from x=1 to x=4. Or simply, integral of the square root of x⁸-x⁶ from 1 to 4. Now, the question is how you take the integral of that? By taking out an x⁶ out of the square root, we obtain x³ times the square root of x²-1. x³ is the same thing as x times x². Therefore, if we substitute u for x²-1, we have the du in the integral, which is a factor of x. We also have x² which can be written in terms of u. Now the integral is simply (1/2)*(u+1)*u^1/2 from u=0 to u=15. Taking the integral would yield to a volume of 50*root(15) or about 608.37 units cubed. Another good way to solve the integral is trigonometric substitution. Isn't it a big bowl?!

Topic:

Applications of Integrals

______________________________________

03/18/02) Find the result of dividing the 99^th derivative of f(x)=x¹⁰⁰+1/x by 99!.

Solution:

Let us look at this problem piece by piece. The 99^th derivative of x¹⁰⁰ is simply 100!x, since each time the power is being multiplied by the function and decreased by 1. The other piece, 1/x, has also an interesting pattern. Its derivatives alternate between positive and negative and the power of x in the denominator is increased by 1. Therefore, its 99^th derivative would be -99!/x¹⁰⁰. Now when we divide our whole solution by 99!, we obtain 100x-1/x¹⁰⁰ which is indeed the answer to this problem. Take a look at the problem 12/31/01 in the solutions for more explanation.

Topic:

Derivatives

______________________________________

03/04/02) How does the graph of this equation look like: (y²+(x-3)²-4)² + (y²+x²-1)² = 0 ? At what value(s) for x does the derivative equal to zero? Does it have any inflection point? Explain.

Solution:

A very sweet idea underlines this problem. When the sum of the squares to two statements equals zero, they BOTH have to be zero, or the equation wouldn't work. Hence, y²+x²-4 and y²+x²-1 have to be zero. The two equations lead to two circles tangent at each other at x = 1. However, the equation does NOT work if the two quantities for x and y are not equal in the two graphs. Therefore, the only possible place for the function is at x = 1, or the point(1,0). This graph is lovely; IT IS JUST A POINT! There is no change in the y-value, so there is no derivative and no inflection point. Love that, right?

Topic:

Implicit Equations and Derivatives

______________________________________

02/25/02) The tangent line of y = x³-x²-4x+4 at x = 0 crosses the graph at point M (except for x = 0). Find the area between the graph and the tangent line between x = 0 and x = M.

Solution:

At x = 0, the slope of the tangent line is -4. Therefore, the equation of the line is y = -4x+4. By setting this equal to the function y, we get two solutions: x = 0 and x = 1. Hence M is 1. Now we have to take the integral of y minus the line's equation from 0 to 1. Whatever the absolute value of that value equals to is the area of the question. Mathematically, it is integral of
x³-x²-4x+4 - (-4+4) or x³-x² from 0 to 1. The answer is 1/12. It is a very tiny area, isn't it?

Topic:

Derivatives and Integrals

______________________________________

02/18/02) Find the limit of f(x) as x approaches infinity, if f(x) is the square root of x²+4x+10 minus the square root of x²-6x+17.

Solution:

As complicated as it sounds, a simple idea underlines this problem. As x approaches infinity the whole numbers do not count as much. Therefore, we can simply change 10 to 4 and change 17 to 9. This would make each of the terms under the square root a perfect square. Hence, the whole thing can be written as (x+2)-(x-3) which equals 5. Indeed, 5 is the limit of this function as x approaches infinity. Wasn't that sweet?

Topic:

Limits

______________________________________

02/11/02) Find the derivative of the following function with respect to x.

Solution:

Derivative and Integral cancel each other, so we plug each limit of integration into the function and by chain rule, we multiply each term by its derivative. Hence, we would obtain:
(1/sin(x))*cos(x)-(1/cos(x))*-sin(x) which simplifies to cot(x)+tan(x). This can be simplified even more to 1/(cos(x).sin(x)) by using common denominator. However, either way is correct!

Topic:

Derivative of Integrals

______________________________________

02/04/02) What is the slope of the line tangent to the point of inflection of f(x)=xⁿ + x ?

Solution:

First mission is to find the function's point of inflection, which is the place where its second derivative equals zero. The first derivative is n*x^n-1 +1 and the second derivative is (n-1)*x^n-2. The second derivative is zero only when x=0. Therefore, zero is its point of inflection. Now, since the slope of the tangent line or DERIVATIVE of that point is needed, we plug 0 into our first derivative function and we obtain 1. So the answer is 1, wasn't that easy?

Topic:

Derivatives

______________________________________

01/28/02) If f(x+1)=x²-5x, what is the local extrema of f(x)?

Solution:

The question is referring to the following series:
More than a calculus problem, this is an algebra (advanced topics) problem. F(x+1) should not be confused with f(x), because any algebraic work has to be operated on x+1. To get x²-5x it is necessary first to square the quantity x+1. So we are left with x²+2x+1. Now to get -5x, 7x has to be subtracted. But REMEMBER that we only work with x+1. So we add -7(x+1). Now 1-7 is -6, so we add +6 to cancel that. Therefore, we get:
f(x+1)=(x+1)²-7(x+1)+6. More generally,
f(x)=x²-7x+6.
Now in the calculus part, we set its derivative equal to zero and we get 2x-7=0. So the answer is x=7/2. Sweet mathematics, right?

Topic:

Advanced Topics

______________________________________

01/21/02) The opposite of the square of a number is chosen. It is then added to its half, again added to the result's half and so on. What is the input that makes the result of this process maximized?

Solution:

The question is referring to the following series:
-x² + (-x²/2) + (-x²/4) + ...
According to geometric series formulas, if r, ratio of multiplication is less than 1, the sum is t₁/(1-r). In this particular case, it is
-x²/(1-1/2), which is -2x².
Now, to maximize this result, we have to set its derivative equal to zero. So -4x=0. Therefore, the required input is 0. Incredible, isn't it?

Topic:

Series and Derivatives

______________________________________

01/14/02) What is the largest number that exceeds its square? Its cube? Its n^th power?

Solution:

By setting up an equation for this problem, we obtain x-x², and taking its derivative, we get 1-2x and solving it for x when the derivative is zero, the answer is 1/2. The cubic one is also the same process, so x is square root of 1/3 or root3/3. The n^th power can be solved in the same way too. x-xⁿ is the function and 1-nx^n-1 is the derivative. So the largest number that exceeds its n^th power, is (n-1)^th root of 1/n. Very interesting!

Topic:

Optimization

______________________________________

01/07/02) A rectangular cube is to be made in such a way that one side is x, the other is 1m less than x and the last side is 3m bigger than x. The cube has to have the lowest volume possible. What should the sides be to fulfill this requirement?

Solution:

Ironically, there is no solution for this question. We first have to put the problem in mathematical terms. So volume is x(x-1)(x+3), and V'(x), which is the rate of change of the volume, is 3x²+4x-2. By setting this equal to zero, we get two values for x, which are approximately 0.535 and -1.87. However, x cannot be less than 1, because that would make the volume negative, something that does NOT make sense. Therefore, there is no solution to this problem.

Topic:

Optimization

______________________________________

12/31/01) Find the result of dividing the 65^th derivative of 2x⁶⁶ by 65 factorial.

Solution:

When taking the derivative of an exponential function, we bring down the exponent as a coefficient and subtract one from the exponent for the new one. Therefore, the 65^th derivative of 2x⁶⁶ is simply 2*66!x, which turns into 2*66x or 132x when divided by 65!. A little thinking would be sufficient, right?

Topic:

Derivatives

______________________________________

12/24/01) Compute:

Solution:

12/24/01) First let us simplify the expression in the limit. Let's call it A. Now wouldn't you agree that A=(b+A)1/2, since it is the same expression A in the square root? Therefore, by squaring both sides and solving for A, we would get A= 1±(1-4b)1/2/2, of which the negative part can be eliminated. When b approaches zero, the expression is just (1+1)/2, which is 1. Isn't it nice?

Topic:

Limits and Algebra

______________________________________

12/17/01) Where does the function y = |3x²| - |x⁴| -9 change its concavity?
Where does y = | |3x²| - |x⁴| | change from decreasing to increasing or vice versa?

Solution:

The first part is very simple, since the absolute values are meaningless. 3x² and x⁴ are always positive. Therefore, by taking the second derivative of the function, we would find the inflection points to be -root2/2 and +root2/2.
The second part is more complex. Firstly, the inner absolute values can be eliminated again. So we only work with |x²-x⁴|. Since x² and x⁴ meet at points x=1 and x=-1, those are the places the function is not differentiable at, but should be analyzed. This function can be rewritten as x² - x⁴ when x is between -1 and 1; and x⁴ - x² when x is anywhere else. Therefore the first derivative of the function is 2x-4x³ between -1 and 1. Hence, it is 4x³-2x anywhere else. Setting these equal to zero, we get x=0, x=-root2/2, x=root2/2. However, the points -1 and 1, at which the function is not differentiable, are also extremas, since the function is flipped at the points. So the answer set is {-1,-root2/2,0,root2/2,1}. Although tedious, it was lovely, wasn't it?

Topic:

Extremas and Absolute Values

______________________________________

12/10/01) How many local max/mins are there in the function y = sin (1/x), between -2/and 2/. Find a formula for the local max/mins. Is 2/2001one of the local extremas? What about 1/1000?

Solution:

The function sin(1/x) oscillates infinitely, when x approaches zero. When differentiated, the derivative function is -cos(1/x)/x². The derivative function has infinite zeros. Therefore, the number of local max/mins are also infinite. Wherever the cos(1/x) equals zero, the original function has local extremas. The formula for the extremas is 2/(n) where n can be any nonzero odd integer. Therefore, 2/2001 is an extrema, whereas 1/1000 is not. Fascinating, huh?

Topic:

Extremas

______________________________________

12/03/01)

Part 1:

Find the 127^th derivative of the following expression:

Part 2:

Find the 110^th derivative of y = x¹⁰⁰ +10cos(x) - 8sin(x).

Solution:

The sin and cos function are cyclical in their derivatives. In every four interval, they start to repeat their pattern. So the answer to part 1 is simply the third derivative of the given function, which is sin x - cos x. By the way, the csc and sec can be easily simplified to sines and cosines.

For the second part, you have to be a little cleverer! x¹⁰⁰ becomes a coefficient times x in the 99^th derivative. Therefore, in the hundreth derivative, it is just zero and for the solution of this problem, it can be ignored!! The rest is just like the first part, since it is cyclical. 110^th derivative is just like the second one. So the answer is -10cos(x)+8sin(x). GREAT, huh?

Topic:

Cyclical Derivatives

______________________________________

11/26/01) Find the following limit and show work.

Solution:

The key is to multiply the numinator and the denominator by (x^2/3 + x^1/3.h^1/3 + h^2/3). By the algebraic formula for difference of cubes, the numinator becomes x-h and the denominator is (x-h) times the multiplied expression. So the (x-h) cancels and as h approaches x, the denominator becomes 3x^2/3. So the limit equals 1/3x^2/3. Lovely, huh?

Topic:

Limits

______________________________________

Find dy/dx:
11/19/01: (y-x)^x2=cos(x-y)

Solution:

x²(y-x)^x2-1(dy/dx-1)={-sin(x-y)}(1-dy/dx)
by algebra you would get: dy/dx=1, surprising, huh?

Topic:

Implicit Differentiation

______________________________________

11/12/2001) Water is being poured into a conical container of height of 12 in. and radius of 5 in., with a speed of 2 in³/sec. How fast is the surface area of the water changing at the height of 6 in.?

Solution:

By the volume formula, we find dh/dt to be 0.10186 in/s. Using the ratio of h to r, we find the dr/dt to be 0.04244 in/s and by Mr. Pythagoras, we calculate s, the diagonal of h and r, to be 0.11035 in/s.

Now, with all these information, we use the surface area equation. Using the multiplication rule and all the fun of calculus we find dA/dt to be 2.59 in²/s. Interesting, huh?

Topic:

Related Rates of Change

Contact Me:

Email: omid1986@yahoo.com

Back to Home Page!