course Mth 271 ??\????~???????assignment #012012. `query 12
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19:33:09 Class Notes #13 Explain how we obtain algebraically, starting from the difference quotient, the expression for the derivative of the y = x^2 function.
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RESPONSE --> We must start with taking the limit of {(y(x +dx)) - y(x)} / dx. By doing this, and going through the steps, we find that the derivative is 2x confidence assessment: 2
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19:33:25 The difference quotient is [ f(x + `dx ) - f(x) ] / `dx. In this case we get [ (x+`dx)^2 - x^2 ] / `dx = [ x^2 + 2 x `dx + `dx^2 - x^2 ] / `dx = [ 2 x `dx - `dx^2 ] / `dx = 2 x - `dx. Taking the limit as `dx -> 0 this gives us just 2 x. y ' = 2 x is the derivative of y = x^2. **
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RESPONSE --> self critique assessment:
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19:35:57 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> To begin this derivative we must also take the limit as dx approaches zero, starting out just like before, but the Binomial formula allows us to appropriatly factor out x^n. This becomes a long and semi-messy formula, but the long end can be stopped because it contains dx. As dx approaches zero, all of these terms cancel out leaving us with nx^n-1. confidence assessment: 2
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19:36:02 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> confidence assessment:
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19:36:13 The key is that (x + `dx)^n = x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. When we form the difference quotient the numerator is therefore f(x+`dx) - f(x) = (x + `dx)^n - x^n = (x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) - x^n = n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. The difference quotient therefore becomes (n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) / `dx = n x^(n-1) + n (n-1) / 2 * x^(n-2) `dx + ... +`dx^(n-1). After the first term n x^(n-1) every term has some positive power of `dx as a factor. Therefore as `dx -> 0 these terms all disappear and the limiting result is n x^(n-1). **
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RESPONSE --> self critique assessment: 2
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19:39:44 **** Explain how the derivative of y = x^3 is used in finding the equation of a tangent line to that graph.
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RESPONSE --> By drawing the graph, we can find a point on the graph and draw the tangent line. We see that this point is (-2,-8). Now that we found a point on the graph, our next step is to find the slope. We can find the slope by entering our x value of -2 into the derivative of the equation. We know the derivative is 3x^2. We find that our slope is 12. We now enter this into our equation for a line (y + 8) / (x+2) = 12 y + 8 = 12x + 24 y = 12x + 16 confidence assessment: 3
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19:39:51 The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. **
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RESPONSE --> self critique assessment: 3
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19:40:34 2.1.9 estimate slope of graph.................................................
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RESPONSE --> I estimated the slope to be -1/3 confidence assessment: 2
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19:41:01 You can use any two points on the graph to estimate the slope. The slope of a straight line is the same no matter what two points you use. Of course estimates can vary; the common approach of moving over 1 unit and seeing how many units you go up is a good method when the scale of the graph makes it possible to accurately estimate the distances involved. The rise and the run should be big enough that you can obtain good estimates. One person's estimate: my estimate is -1/3. I obtained this by seeing that for every 3 units of run, the tangent line fell by one unit. Therefore rise/run = -1/3. **
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RESPONSE --> This is how I estimated the answer to be, just didn't include in my answer, sorry. self critique assessment: 3
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19:43:50 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> lim dx approaches 0 [f(x + dx) - f(x)] / dx lim dx approaches 0 [(x = dx)^2 -4] -(x^2 - 4)] / dx lim dx approaches 0 [x^2 + 2xdx +dx^2 - 4 - (x^2 - 4)] dx lim dx approaches 0 2xdx +dx^2] / dx lim dx approaches 0 2x confidence assessment: 2
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19:44:38 f(t+`dt) = (t+'dt)^3+(t+'dt)^2. f(t) = t^3 + t^2. So [f(t+`dt) - f(t) ] / `dt = [ (t+'dt)^3+(t+'dt)^2 - (t^3 + t^2) ] / `dt. Expanding the square and the cube we get [t^3+3t^2'dt+3t('dt)^2+'dt^3]+[t^2+2t'dt+'dt^2] - (t^3 - t^2) } / `dt. } We have t^3 - t^3 and t^2 - t^2 in the numerator, so these terms subtract out, leaving [3t^2'dt+3t('dt)^2+'dt^3+2t'dt+'dt^2] / `dt. Dividing thru by `dt you are left with 3t^2+3t('dt)+'dt^2+2t+'dt. As `dt -> 0 you are left with just 3 t^2 + 2 t. **
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RESPONSE --> I copied the wrong problem in my answer. I did have this answer for 24, my problem I worked out was for 19. I'm sorry.
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19:52:13 2.1.32 tan line to y = x^2+2x+1 at (-3,4) What is the equation of your tangent line and how did you get it?
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RESPONSE --> The equation of my tangent line is 2x+2 I arrived at this by taking the derivative of x^2 + 2x + 1, then taking the limit as dx approached 0. f(x + dx) - f(x) {[ (x + dx)^2 + 2(x + dx) + 1] - ( x^2 + 2x + 1)} / dx (x^2 + 2xdx + dx^2 + 2x + 2dx + 1 - x^2 - 2x - 1) / dx (2xdx + dx^2 + 2dx) / dx 2x + dx + 2 When we take the limit as dx approaches 0, we are left with 2x + 2 confidence assessment: 2
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19:54:56 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: f ' (x)=2x+2. I got this using the method of finding the derivative that we learned in the modeling projects The equation of the tangent line is 2x+2. I obtained this equation by using the differential equation. the slope is -4...i got it by plugging the given x value into the equation of the tan line. INSTRUCTOR COMMENT: If slope is -4 then the tangent line can't be y = 2x + 2--that line has slope 2. y = 2x + 2 is the derivative function. You evaluate it to find the slope of the tangent line at the given point. You have correctly found that the derivative is -4. Now the graph point is (-3,4) and the slope is -4. You need to find the line with those properties--just use point-slope form. You get y - 4 = -4(x - -3) or y = -4 x - 8. **
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RESPONSE --> I made the same mistake as the student did in the problem. I just forgot to stick the x-coordinate and the slope -4 into the point slope equation, then I obtained the correct equation. self critique assessment: 2
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19:56:20 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown) At what points is the function differentiable, and why?
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RESPONSE --> Differentiable at all points except at x= 0. This is because if we plug zero in the equation it would not be continiuous. confidence assessment: 2
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19:59:17 At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist. The reason the derivative doesn't exist for this function this is that the function isn't even defined at x = +- 2. So there if x = +- 2 there is no f(2) to use when defining the derivative as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. f(2+`dx) is fine, but f(2) just does not exist as a real number. **
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RESPONSE --> Is this question 54? This is sort of the same answer I had for 54. Either way, I was a bit confused with this concept in the book, however, after reading this explanation, I am beginning to understand it a little better. self critique assessment: 1
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20:00:21 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> The graph is differentiable at all points except x =0. This is because the tangent line cannot be horizontal. If it is, the derivative. does not exist. confidence assessment: 1
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20:00:56 The derivative is defined on(-infinity,-2)u(-2,2)u(2,infinity). The reason the derivative doesn't exist at x = +-2 is that the function isn't even defined at x = +- 2. The derivative at 2, for example, is defined as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. If f(2) is not defined then this expression is not defined. The derivative therefore does not exist. At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist.**
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RESPONSE --> Ok, I dont' understand. Were both of these #52? self critique assessment: 1
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20:01:47 If x is close to but not equal to 2, what makes you think that the function is differentiable at x?
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RESPONSE --> Because AT 2, the derivative does not exist. Any point before and after 2, the derivative exists. confidence assessment: 1
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20:01:59 If x is close to 2 you have a nice smooth curve close to the corresponding point (x, f(x) ), so as long as `dx is small enough you can define the difference quotient and the limit will exist. **
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RESPONSE --> self critique assessment: 2
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20:02:27 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> No, it is not differentiable because the limit/derivative does not exist at this point. confidence assessment: 1
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20:02:35 GOOD ANSWER FROM STUDENT: if the function does not have limits at that point then it is not differentiableat at that point.
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RESPONSE --> self critique assessment: 2
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20:02:39 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment:
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