Find the equation of the line that is tangent to $\displaystyle y = x^2sin(2x)$ at $\displaystyle x = -\pi.$

Shown work would be appreciated, thank you.

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- Apr 13th 2010, 03:50 PMEixoDifferentiation of Sinusoidal Functions
Find the equation of the line that is tangent to $\displaystyle y = x^2sin(2x)$ at $\displaystyle x = -\pi.$

Shown work would be appreciated, thank you. - Apr 13th 2010, 04:04 PMskeeter
- Apr 13th 2010, 04:06 PMEixo
- Apr 13th 2010, 04:10 PMskeeter
- Apr 13th 2010, 04:30 PMEixo
Thank you for the help, it was the mental roadblock that I was struggling to get past.

$\displaystyle \frac{dy}{dx}=x^2 \cdot 2cos(2x)+2x \cdot 2sin(2x)$

I then subbed in $\displaystyle -\pi$ (typo from original post, will edit).

$\displaystyle =((-\pi)^2(2(1)))+(2(-\pi)(0))$

$\displaystyle =2\pi^2$

$\displaystyle y $@$\displaystyle -\pi = (-\pi)^2 \cdot sin(2(-\pi))

= 0$

Slope = Slope

$\displaystyle 2\pi^2=\frac{y-0}{x+\pi}$ (Cross multiply)

$\displaystyle 2\pi^2x+2\pi^3 = y$ - Apr 13th 2010, 04:37 PMskeeter
- Apr 13th 2010, 04:40 PMEixo
Sorry, will correct. My answer remains unchanged though.