course MTH 271 10/12, 1445 揵ٕӢsw|١assignment #012
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00:23:30 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 --> Difference quotient= [f(x + 'dx) - f(x)] / 'dx y=x^2 y'(x)=( y(x + 'dx) - y(x) ) / 'dx =( (x + 'dx)^2 - x^2 ) / 'dx =(x^2 + 2x'dx + 'dx^2) - x^2) / 'dx =(2x'dx + 'dx^2) / 'dx ='dx (2x + 'dx) / 'dx =2x + 'dx Limit as 'dx approaches 0: 2x confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:23:38
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RESPONSE --> confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:23:49 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 Rating:ent: 3
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00:30:42 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> y=x^n y'=( y(x + 'dx) - y(x) ) / 'dx =[(x^n + nx^(n-1)'dx + (n(n-1)/2)x^(n-2)'dx^2 + etc) - x^n ] / 'dx =(nx^(n-1)'dx + (n(n-1)/2)x^(n-2)'dx^2 + etc) / 'dx =(nx^(n-1) + (n(n-1)/2)x^(n-2) 'dx + etc As 'dx approaches 0: nx^(n-1) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:30:50 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 Rating:ent: 3
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00:38:43 **** 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 --> You can find the equation of any line given a single point and the slope of the line. Using the function y=x^3, we can select an x value on the graph, and by plugging it into the equation of the function, find the y value, thus giving us our point. (x1, y1). The slope of the tangent line is equal to the derivative of the function y=x^3. Therefore, the slope is equal to 3x^2. Using our x1 value, we can determine the slope. By using the point-slope equation of : y - y1=m(x - x1), we are able to determine the equatino of the tanget line. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:38:52 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 Rating:ent: 3
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00:41:28 2.1.9 estimate slope of graph.................................................
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RESPONSE --> Slope=change in y / change in x Slope= -1 / 3 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:41:39 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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00:57:07 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> y=t^3 + t^2 y'=( f(t + 'dt) - f(t) ) / 'dt y'=( ( (t + 'dt)^3 + ( t + 'dt)^2 ) - t^3 - t^2 ) / 'dt y'=( ( (t^3 +3t^2'dt + 3t'dt^2 + 'dt^3) + (t^2 + 2t'dt + 'dt^2) ) -t^3 -t^2) / 'dt =(3t^2'dt + 3t'dt^2 + 'dt^3 + 2t'dt + 'dt^2 ) / 'dt = 'dt (3t^2 + 3t'dt + 'dt^2 + 2t + 'dt ) / 'dt =(3t^2 + 3t'dt + 'dt^2 + 2t + 'dt ) As 'dt approaches 0: 3t^2 + 2t confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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00:57:17 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:00:30 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 --> Slope of tangent line is equal to derivative of function. y=x^2 + 2x + 1 y'=2x + 2 Substituting in known x value of -3: m=2(-3) + 2= -4 Using point-slope formula: y-4=-4(x--3) y-4 =-4(x+3) y-4=-4x -12 y=-4x - 8 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:00:46 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:08:24 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 +-2. At these two points, vertical asymptotes exist that make the function not continuous. Being differentiable implies being continuous. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:09:24 **** 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. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:09:36 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:10:56 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 --> The function will be differntiable at x b/c the value is approaching the asymptote but not touching it, meaning x is not really 2. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:11:12 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 Rating:ent: 3
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01:12:20 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> The function is not differentiable at x=2 because a vertical asymptote exists here that forms a discontinuity. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:12:44 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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