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Question
Find the slop of a line tangent to the curve y=2x^2 at the point P(2.5,12.5) by finding the limit of the slopes of the scant lines PQ where Q has X-values 2,2.4,249 and2.499.
The y- coordinate for Q1 is ,so coordinates of Q1 are (1, 2)
y2-y1=2-4.5=-2.5
The x-coordinate P (1.5, 4.5)
x2-x1=1-1.5=-0.5
y2-y1/x2-x1=-2.5/-0.5=5
Looking at, "Find the slop of a line tangent to the curve y = 2x^2 at the point P(2.5,12.5) by finding the limit of the slopes of the scant lines PQ where Q has X-values 2, 2.4, 249 and 2.499."
gives me the following.
First of all, when x = 2.5, we're given y = 12.5.
When x = 2, y = 2*4 = 8.
When x = 2.4, 2.4^2 = 5.76, so y = 2*5.76 = 11.52.
When x = 2.49, 2.49^2 = 6.2001, so y = 2*6.2001 = 12.4002.
When x = 2.499, 2.499^2 = 6.245001, so y = 2*6.245001 = 12.490002.
Taking (y2-y1)/(x2-x1) for each case gives us
(12.5-8)/(2.5-2) = 4.5/0.5 = 9;
(12.5-11.52)/(2.5-2.4) = 0.98/0.1 = 9.8;
(12.5-12.4002)/(2.5-2.49) = 0.0998/0.01 = 9.98; and
(12.5-12.490002)/(2.4-2.499) = 0.009998/0.001 = 9.998.
From this, the value at 2.5 is guessed to be 10.
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| Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |
| Comment | scotto thankso, My text book is no help on this.Because of your help I can now move on .My textbook gives the answers ,but I could not figure out how to calculate them.This calculus word problem is the root of most of what I have to do.Thank you^2.Is their any way I could consult you personally further on other calculus type things. kenlarrington@yahoo.com | ||
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