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Derivatives |
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| Mar23-08, 05:16 PM | #1 |
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Derivatives
1. The problem statement, all variables and given/known data
Find the point on the parabola y= 4x^2 + 2x - 5 where the tangent line is perpendicular to the line 3x + 2y = 7. 2. Relevant equations 3. The attempt at a solution I don't know what to do since I was away the last 3 classes since I was away. Help me please. |
| Mar23-08, 05:25 PM | #2 |
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You want to find the slope of the line you're given and use the definition of a derivative as well as perpendicularity to solve for the x value you want.
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| Mar23-08, 05:27 PM | #3 |
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Two perpendicular lines have slopes that are negative reciprocals of each other, eg: a line with a slope 2 is perpendicular to a line with a slope -1/2.
Find the slope of the line, find the negative reciprocal of that slope. The derivative of a function is the slope of that graph at any point on the graph, so find the derivative of the parabola and see at what value of x it will equal the negative reciprocal of the slope you found earlier. |
| Mar23-08, 05:39 PM | #4 |
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Derivativesso..... 2y = -3x + 7 y= -3/2x + 7/2 slope = -3/2 so if it is perpendicular the slope is 2/3 is that my right slope? I now I have to do more but it that right so far? |
| Mar23-08, 05:41 PM | #5 |
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Yes.
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| Mar24-08, 01:51 PM | #6 |
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Correct.
dy/dx = 8x+2 You want the value of x when dy/dx is (2/3), as you said from above. Solving for x gets (-1/6). Plug this value into your original equation y=4x^2 etc. |
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