Given the curve $y = \sqrt{x}+ 2$. Find a point on the curve where the tangent line is parallel to $y = \frac{1}{3}x − 10$.


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The gradient of the other line is clearly $m=\frac{1}{3}$.

Evaluate the derivative with respect to $x$ of the curve $y=\sqrt{x}+2$ and let $\frac{dy}{dx}=\frac{1}{3}$.


The slope of tangent of the curve $y = \sqrt{x} + 2$ is given by the derivative of $y$ with respect to $x$, viz.

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}}. $$

The other curve, $y = \frac13 x - 10$, is a straight line given in the form $y=mx+c$, where $m$ is the slope of the line. For the tangent of the first curve to be parallel to the straight line,

$$ m = \frac{dy}{dx} \\ \implies x = \frac94. $$

Hence, the corresponding point on the curve is $(9/4, 5)$.


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