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!

1. mathteacher1729

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 $x = 1$ ?

2. mathteacher1729

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.

3. mathteacher1729

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

4. mathteacher1729

Grrrrr! $\frac{dy_1}{dx}=\frac{2}{13}(1+\frac{5}{2}x-\frac{3}{2}-\frac{3}{4}k)$

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. xvdavis

Thank you.