Last Week's Solution:
01/28/02) If f(x+1)=x2-5x, what is the local extrema of f(x)?
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
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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?
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 nth power?
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 nth power can be solved in the same way too. x-xn is the function and 1-nxn-1 is the derivative. So the largest number that exceeds its nth 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:
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?
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 65th derivative of 2x66 by 65 factorial. Solution:
12/31/01) Find the result of dividing the 65th derivative of 2x66 by 65 factorial.
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 65th 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) Compute:
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:
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?
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?
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)
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 99th 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. 110th 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.
The key is to multiply the numinator and the denominator by (x2/3 + x1/3.h1/3 + h2/3). 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 3x2/3. So the limit equals 1/3x2/3. Lovely, huh? Topic: Limits
Find dy/dx: 11/19/01: (y-x)x2=cos(x-y)
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.?
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
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