#$&* course Mth 271 10/24 1159p Question: `qClass Notes #13
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Given Solution: `a The difference quotient is [ f(x + `dx ) - f(x) ] / `dx. In this case we get [ (x+`dx)^2 - x^2 ] / `dx = [ x^2 + 2 x `dx + `dx^2 - x^2 ] / `dx = [ 2 x `dx + `dx^2 ] / `dx = 2 x + `dx. Taking the limit as `dx -> 0 this gives us just 2 x. y ' = 2 x is the derivative of y = x^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q **** Explain how the binomial formula is used to obtain the derivative of y = x^n. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: According to the Binomial Formula, to form the derivative, the original exponent becomes a coefficient, and the new exponent is (n-1), so y’ = n x^(n-1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was too lazy to type the whole thing and I only explained it for the terms that were there. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q **** Explain how the derivative of y = x^3 is used in finding the equation of a tangent line to that graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative gives you the slope of the graph at the given point of tangency. y’ =3x so the slope is 3. A point would be needed to plug into point-slope form of y-y1=m(x-x1) for the equation of the tangent line. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. ** Self-critique (if necessary):n/a self-critique Rating:3 ********************************************* Question: `q2.1.9 estimate slope of graph............................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. Draw a line through a point on the graph. 2.Find the slope of that line. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Apparently there was supposed to be a graph here that I did not see? I thought it meant how to estimate the slope of a graph in general. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q 2.1.24 limit def to get y' for y = t^3+t^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Derivative of the sum of 2 functions = sum of the derivatives of the 2 functions: y’ for t^3 = 3t^2 y’ for t^2 = 2t y’ = 3t^2+ 2t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a f(t+`dt) = (t+'dt)^3+(t+'dt)^2. f(t) = t^3 + t^2. So [f(t+`dt) - f(t) ] / `dt = [ (t+'dt)^3+(t+'dt)^2 - (t^3 + t^2) ] / `dt. Expanding the square and the cube we get [t^3+3t^2'dt+3t('dt)^2+'dt^3]+[t^2+2t'dt+'dt^2] - (t^3 - t^2) } / `dt. } We have t^3 - t^3 and t^2 - t^2 in the numerator, so these terms subtract out, leaving [3t^2'dt+3t('dt)^2+'dt^3+2t'dt+'dt^2] / `dt. Dividing thru by `dt you are left with 3t^2+3t('dt)+'dt^2+2t+'dt. As `dt -> 0 you are left with just 3 t^2 + 2 t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not understand the directions. Did not realize I was supposed to expand it. ------------------------------------------------ Self-critique Rating:3
#$&* course Mth 271 10/24 1159p Question: `qClass Notes #13
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q **** Explain how the binomial formula is used to obtain the derivative of y = x^n. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: According to the Binomial Formula, to form the derivative, the original exponent becomes a coefficient, and the new exponent is (n-1), so y’ = n x^(n-1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The key is that (x + `dx)^n = x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. When we form the difference quotient the numerator is therefore f(x+`dx) - f(x) = (x + `dx)^n - x^n = (x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) - x^n = n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. The difference quotient therefore becomes (n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) / `dx = n x^(n-1) + n (n-1) / 2 * x^(n-2) `dx + ... +`dx^(n-1). After the first term n x^(n-1) every term has some positive power of `dx as a factor. Therefore as `dx -> 0 these terms all disappear and the limiting result is n x^(n-1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was too lazy to type the whole thing and I only explained it for the terms that were there. ------------------------------------------------ Self-critique Rating:3
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Given Solution: `a The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. ** Self-critique (if necessary):n/a self-critique Rating:3 ********************************************* Question: `q2.1.9 estimate slope of graph............................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. Draw a line through a point on the graph. 2.Find the slope of that line. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Apparently there was supposed to be a graph here that I did not see? I thought it meant how to estimate the slope of a graph in general. ------------------------------------------------ Self-critique Rating:2
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not understand the directions. Did not realize I was supposed to expand it. ------------------------------------------------ Self-critique Rating:3
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Given Solution: `aSTUDENT 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If the slope is different at different points of the curve, it seems like my equation would be right at the point I chose. Don’t you always need 2 points to find the slope? ------------------------------------------------ Self-critique Rating:2
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I must have done something wrong - I’m guessing that the derivative of x^2-4 is not 2x-4 - because I can understand why x cannot equal 2 because that gives a 0 on the bottom, but -2 would not. S ritique Rating:2 ********************************************* Question: `q **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This appears to be the same question. I graphed it this time and it makes sense that the function isn’t defined at x=+-2 because the graph does not pass through x=+-2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative is defined on(-infinity,-2)u(-2,2)u(2,infinity). The reason the derivative doesn't exist at x = +-2 is that the function isn't even defined at x = +- 2. The derivative at 2, for example, is defined as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. If f(2) is not defined then this expression is not defined. The derivative therefore does not exist. At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I need more practice with problems where the derivative does not exist. This is not really clear, and I can’t really explain what I don’t understand about it. elf-critique Rating:1 ********************************************* Question: `qIf x is close to but not equal to 2, what makes you think that the function is differentiable at x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x is close to 2 it is still not 2 so it could be differentiable at any other point than +-2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):n/a elf-critique Rating:3 ********************************************* Question: `qIf x is equal to 2, is the function differentiable? Explain why or why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No because we just determined that f(2) does not exist so it cannot be differentiable at +-2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD ANSWER FROM STUDENT: if the function does not have limits at that point then it is not differentiableat at that point. Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!