course Mth 271 ??????????????assignment #012012. `query 12
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17:26:28 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 --> y=x^2 y'(x)= lim as x approaches 0 is (y(x+deltax)-y(x))/delta x lim x-----0 ((x+deltax)^2-x^2)/deltax limx----0 (x^2+2xdeltax+deltax^2-x^2)/deltax limx----0 (2xdeltax+deltax^2)/deltax limx----0 2x+deltax limx----0 2x confidence assessment: 3
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17:26:47 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 --> yup self critique assessment: 3
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17:30:45 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> lim deltax---0 ((x^n+nx(n-1)deltax+ (n(n-1)/2)x(n-2)deltax^2+.....)-x^n)/deltax x^n and -x^n cancel, divide by deltax lim deltax--0 nx^(n-1)+(n(n-1)/2)x(n-2)deltax +.... lim deltax-0 nx^(n-1) confidence assessment: 3
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17:30:54 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 --> yup self critique assessment: 3
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17:31:40 **** 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 --> y'=3x2 using the rules confidence assessment: 3
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17:32:05 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 --> yup self critique assessment: 3
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17:32:30 2.1.9 estimate slope of graph.................................................
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RESPONSE --> m= -1/3 confidence assessment: 3
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17:32:44 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 --> yup self critique assessment: 3
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17:33:11 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> okay on this problem i dont know how to expand (x+deltax)^3 confidence assessment: 0
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19:37:50 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 --> okay, i see self critique assessment: 3
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19:39:03 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 --> y'= 2x+2 b/c of y=at^2+bt+c and y'=2at+b y'(-3)= -4 confidence assessment: 3
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19:41:19 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 --> okay i didnt see anything like this in the book or notes self critique assessment: 3
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19:44:41 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 --> x cannot equal zero b/c of a cusp confidence assessment: 3
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19:45:11 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 --> okay this is the answer to problem 51 and the question asked for the answer to problem 52 self critique assessment: 3
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19:45:45 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> x cannot equal -2 or 2 due to vertical asymptotes at those values confidence assessment: 3
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19:46:02 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 --> yup self critique assessment: 3
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19:46:40 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 --> b/c the graph approaches 2 from both sides confidence assessment: 3
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19:46:59 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 --> ok self critique assessment: 3
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19:47:34 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> no the limit does not exist at 2 b/c of asymptote so the function is not differentiable at 2 confidence assessment: 3
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19:47:42 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 --> yup self critique assessment: 3
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19:48:00 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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