I found this question and couldn’t do it and its really bugging me.

Find the coordinates of the point at which the curve with the equation

y= 2x^2 + 5 sqr root 3x + 1 is inclined at 60 degrees to the x-axis.

I used tan theta = m to get 1.73 then tried to differentiate the equation but got stuck on the 5 sqr root 3x bit.

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# our old friend differentiation

Started by william, Apr 10 2007 07:39 AM

2 replies to this topic

### #1

Posted 10 April 2007 - 07:39 AM

"millis times ten to the power of minus 3 yeah?"

### #2

Posted 10 April 2007 - 01:02 PM

I found this question and couldn’t do it and its really bugging me.

Find the coordinates of the point at which the curve with the equation

y= 2x^2 + 5 sqr root 3x + 1 is inclined at 60 degrees to the x-axis.

I used tan theta = m to get 1.73 then tried to differentiate the equation but got stuck on the 5 sqr root 3x bit.

Clarification: is it or ?Find the coordinates of the point at which the curve with the equation

y= 2x^2 + 5 sqr root 3x + 1 is inclined at 60 degrees to the x-axis.

I used tan theta = m to get 1.73 then tried to differentiate the equation but got stuck on the 5 sqr root 3x bit.

I'm going to assume the first. Don't let the square root on the 3 put you off, it's still just a constant (the constant in front of a term such as x, x

^{2}, etc. is called it's coefficient). We could just as easily write (but we don't as this is less accurate), so can you see that the coefficient being more complicated doesn't actually change how we differentiate with respect to x. This should still give you the "constant" in your function dy/dx.

Incidentally, tan(60) is an exact value that you are supposed to learn. It's probably advisable to leave it as this exact value.

If it is the second equation from the first line, then things are considerably more complicated. Finding the value of x that satisfies dy/dx=root3 involves solving a cubic in root(x), which I'm almost certain you wouldn't be asked to do.

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