Jump to content


  • Content Count

  • Joined

  • Last visited

Community Reputation

0 Neutral

About KFS

  • Rank

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

  1. In Example 2, everything was fine, what I was doing matched perfectly with the book until the part "Setting this equal to zero gives *equation*". How is this? The expression on the left has no p, how can it have a p on the right? If the derivative dS/dx is set to 0 how is an x on the right, at the other side of =? I understand the rest. It's just that I don't see how one expression results in the other. That's all. Thank you. P.S: Sorry if images are annoying, english is not my first language, so this way it's easier for me to explain and for you to understand.
  2. Thanks for the help. Yes, I am familiar with composition of functions and the chain rule. It's just that word problems are hard for me.
  3. In Example 1: I don't understand the procedure of this exercise. I do know why the derivative of the angle with respect to time is 2π/30. What I don't understand is the rest. It says x=10tanθ, how is this possible? It gives me, and the computer, by tanθ=opposite/adjacent, x=10/tanθ, not 10tanθ. What I don't understand either is why both derivatives (dx/dθ) and (dθ/dt) are multiplying. Why is x=8 choosen? Why not any other number? Thanks for the help.
  4. How do I find the sum of 1/(1+k^2) from k=1 to 1000? The book didn`t provide any information of how to solve rational sums, nor formulas for greater powers than 1. I have no idea how to do this. Thank you.
  5. I just don't understand this. What's the process to get these answers? I tried the same process as with closed interval but I was wrong. Why none, none, none, none? I'm stuck. Show me how you would do it, or any other example you put to help me understand. Any help is welcome, thank you for your answers. (EXAMPLE 2, (b)).
  6. Hello. This the problem: The population of Booneville increases at a rate of r(t)= (3.62)(1+0.8t^2) people per year, where t is the time in years from 1970. The population in 1976 was 726. What was is it in 1984? I' ve been trying integrating and then evaluating the integral at t=8 (8 years, 1984-1976=8). The answer must be 3195. I want to learn this because I saw the next problems in my textbook are similar to this one. It might be silly, but any help is welcome. Thank you.
  7. Ok, thank you very much for the explanation. But 10 and 20 were not in the problem' statement, so where do they come from?
  8. Thanks for the answer DrP. But please explain it more clear. I know nothing about similar triangles. Thank you.
  9. Hello. This is not even a problem but an example my book gives, but I want to understand it in order to do the exercises. It is as follows: A light L is being raised up a pole. The light shines on the object Q, casting a shadow on the ground. At a certain moment the light is 40 meters off the ground, rising at 5 meters per minute. How fast is the shadow shrinking at that instant? Then the book gives the answer that I don't understand. It reads: Let the height be y at time t and the lenght of the shadow be x. By similiar triangles, x/10=(x+20)/y; i.e., xy=10(x+20).The explanation goes on but I only want to know about the similar triangles part. I don't know where the equation and the 10 and 20 come from. Just that, I have no problem with the rest. Thank you in advance.
  10. Hello. My problem is as follows: Suppose x^4+y^2+y-3=0. a) Compute dy/dx by implicit differentiation. b) What is dy/dx when x=1 and y=1? c) Solve for y in terms of x (by the quadratic formula) and compute dy/dx directly. Compare with your answer in part a).I solved a) and b). a)=-4x^3/(2y + 1), and b)=-4/3. I'm stuck at c). This is what I've been doing: Using the quadratic formula to solve for y in x^4+y^2+y-3=0 gives y=-1±√-4x^4+13/(2). Then applying the chain rule in the result of y gives -4x^3/(√-4x^4+13). But it must give the same as a)=-4x^3/(2y + 1) . Where am I failing at? How can I solve it?Thank you for your time.
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.