Question Number 147713 by Odhiambojr last updated on 22/Jul/21
$${Find}\:{a}\:{point}\:{on}\:{the}\:{curve}\:{y}=\sqrt{{x}} \\ $$$${where}\:{the}\:{tangent}\:{makes}\:{an}\:{angle}\: \\ $$$$\mathrm{45}°\:{with}\:{the}\:{positive}\:{x}-{axis} \\ $$
Answered by Olaf_Thorendsen last updated on 22/Jul/21
$$\frac{{dy}}{{dx}}\:=\:\mathrm{tan}\left(\theta\left({x}\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\mathrm{If}\:\theta\left({x}\right)\:=\:\mathrm{45}°,\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\mathrm{1}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Odhiambojr last updated on 22/Jul/21
$${thaks}\:{Mr}. \\ $$