Gradient of a curve

**Tim82** · May 10th 2011, 08:47 PM

Q:
The gradient of a curve with equation y=f(x) is given by f1 (x) = x + sin2x.

If f(0) = 1, find f(\pi )

Any help would be appreciated

**Prove It** · May 10th 2011, 09:00 PM

If you're finding f(pi), just substitute x = pi and simplify...

**islam** · May 10th 2011, 09:49 PM

Originally Posted by Prove It

If you're finding f(pi), just substitute x = pi and simplify...

i think he is asking about f(pi) and not f1(pi) so u have to find f(x) by integrating the gradient and then just substitute f(0)=1 to find the value of the constant

**HallsofIvy** · May 11th 2011, 04:15 AM

"f1" is not a standard notation for the ordinary derivative but you may indeed be right.
Tim82, how about clarifying?

**Sambit** · May 11th 2011, 04:27 AM

Originally Posted by islam

i think he is asking about f(pi) and not f1(pi) so u have to find f(x) by integrating the gradient and then just substitute f(0)=1 to find the value of the constant

Exactly. So $f(x)=\int x dx+\int \sin 2x dx=\frac{x^2}{2}-\frac{\cos 2x}{2}+c$

Putting $f(0)=1$ we have $c=\frac{3}{2}$

So $f(x)=\frac{x^2}{2}-\frac{\cos 2x}{2}+\frac{3}{2}$

So $f(\pi)=\frac{{\pi}^2}{2}-\frac{\cos 2\pi}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}-\frac{1}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}+1$

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