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Question Number 147713 by Odhiambojr last updated on 22/Jul/21

Find a point on the curve y=(√x)  where the tangent makes an angle   45° with the positive x-axis

$${Find}\:{a}\:{point}\:{on}\:{the}\:{curve}\:{y}=\sqrt{{x}} \\ $$$${where}\:{the}\:{tangent}\:{makes}\:{an}\:{angle}\: \\ $$$$\mathrm{45}°\:{with}\:{the}\:{positive}\:{x}-{axis} \\ $$

Answered by Olaf_Thorendsen last updated on 22/Jul/21

(dy/dx) = tan(θ(x)) = (1/(2(√x)))  If θ(x) = 45°, (1/(2(√x))) = 1 ⇒ x = (1/4)

$$\frac{{dy}}{{dx}}\:=\:\mathrm{tan}\left(\theta\left({x}\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\mathrm{If}\:\theta\left({x}\right)\:=\:\mathrm{45}°,\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\mathrm{1}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by Odhiambojr last updated on 22/Jul/21

thaks Mr.

$${thaks}\:{Mr}. \\ $$

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