What is the slope of the line normal to the tangent line of $f (x) = 2 x^{2} - x \sqrt{x^{2} - x}$ $f(x) = 2x^2-xsqrt(x^2-x)$ at $x = 2$ $x= 2$ ?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T11:42:39

$The Reqd. Slope= \frac{16 + 5 \sqrt{2}}{103}$ $"The Reqd. Slope="(16+5sqrt2)/103$ .

Recall that, $f'(2)$ gives the slope of the Tangent Line to the

Curve C : $y=f(x)=2x^2-xsqrt(x^2-x)$

at the pt. $P(2,f(2))=P(2,8-2sqrt2)$ .

As Normal at pt. $P$ is $bot$ to the tgt., its slope is given by,

$1/(f'(2)), if f'(2)!=0$ .

Now, $f(x)=2x^2-xsqrt(x^2-x) = 2x^2-sqrt(x^4-x^3).$

$:. f'(x)=2(2x)-1/(2sqrt(x^4-x^3))*(x^4-x^3)'$

$=4x-1/(2sqrt(x^4-x^3))*(4x^3-3x^2)$

$=4x-(x^2(4x-3))/(2xsqrt(x^2-x))$

$:. f'(x)=4x-(x(4x-3))/(2sqrt(x^2-x))$

$rArr f'(2)=8-(2(8-3))/(2sqrt(4-2))=8-5/sqrt2=(8sqrt2-5)/sqrt2 !=0.$

Hence, the Reqd. Slope $=sqrt2/(8sqrt2-5)=(sqrt2(8sqrt2+5))/{(8sqrt2)^2-5^2}$

$=(16+5sqrt2)/103$ .

What is the slope of the line normal to the tangent line of f(x) = 2x^2-xsqrt(x^2-x) f(x)=2x2−x√x2−xf(x) = 2x^2-xsqrt(x^2-x) at x= 2 x=2 x= 2 ?