Question

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!

Answer 1

The 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 re-write 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 \[x = 1\] ?

Answer 2

Sorry, 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.

Answer 3

\[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k\]

Answer 4

Grrrrr!

\[\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k)\]

Answer 5

let value of dy1/dx at x=1 be c.
c x dy2/dx =-1
2c=-1
solve the equation

Answer 6

dy1/dx at x=1 means substitute value of x as 1 in dy1/dx

Answer 7

as the two curves are perpendicular at the givn point the product of their slopes should be -1

Answer 8

Thank you.