### Home > APCALC > Chapter 11 > Lesson 11.2.1 > Problem11-44

11-44.

A point travels along the curve $y = xe^{x}$ with a horizontal velocity of $\frac { d x } { d t } = 4$ units per second. How fast is the point’s distance from the origin changing when $x = -1$?

Let $d =$ distance from the origin. Then:

$d^2=x^2+y^2=x^2+(xe^x)^2$

Differentiate:

$2dd^\prime=(2x+2(xe^x)(1e^x+xe^x))x^\prime$

Evaluate when $x = -1$:

$d=\sqrt{(-1)^2+(-1e^{-1})^2}=\sqrt{1+\frac{1}{e^2}}$

$2\sqrt{1+\frac{1}{e^2}}d^\prime=(2(-1)+2(-1e^{-1})(1e^{-1}-1e^{-1}))(4)$

Solve for $d′$.