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xvdavis
 5 years ago
Hello, I am preparing for my Fundamentals of Engineering Exam. Here is math problem that I have below which I am having difficulty with:
Given: dy(1)/dx = 2/13 (1 + 5/2x  3/2  3/4k)
What is the value of k such that y(1) is perpendicular to the curve y(2)=2x at x=1?
I have the solution to this problem if you need to see it (Which I don't understand.)
Thanks!
xvdavis
 5 years ago
Hello, I am preparing for my Fundamentals of Engineering Exam. Here is math problem that I have below which I am having difficulty with: Given: dy(1)/dx = 2/13 (1 + 5/2x  3/2  3/4k) What is the value of k such that y(1) is perpendicular to the curve y(2)=2x at x=1? I have the solution to this problem if you need to see it (Which I don't understand.) Thanks!

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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0The notation you're using is a bit difficult to read. There is an "equation" button on the lower left hand corner of the text box which might make things easier. Here's my attempt to rewrite your problem: \[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x\frac{3}{2}\frac{3}{4}k\] Find \[k\] such that \[y_1\] is perpendicular to \[y_2=2x\] when \[x = 1\] ?

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, should be \frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x\frac{3}{2}\frac{3}{4}k) I forgot to close the parenthesis.

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x\frac{3}{2}\frac{3}{4}k\]

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Grrrrr! \[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x\frac{3}{2}\frac{3}{4}k)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let value of dy1/dx at x=1 be c. c x dy2/dx =1 2c=1 solve the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy1/dx at x=1 means substitute value of x as 1 in dy1/dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as the two curves are perpendicular at the givn point the product of their slopes should be 1
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