Given the curve $y = \sqrt{x}+ 2$. Find a point on the curve where the tangent line is parallel to $y = \frac{1}{3}x − 10$.
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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calculus
derivatives
tangent-line
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0Iv'e edited your post. Please ensure it is written as supposed to be. – 2017-02-08
2 Answers
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Hint:
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 $\frac{dy}{dx}=\frac{1}{3}$.
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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 $(9/4, 5)$.