Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 10

MATH100 A December 2023

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Question 10
Find the equation of the tangent line to the curve ${\textstyle y=x{\sqrt {x}}}$ at the point ${\textstyle (1,1)}$. Your answer must be written in the form ${\textstyle y=mx+b}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that the slope of the tangent line at a point ${\textstyle x=a}$ is ${\textstyle f'(a)}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Knowing that the slope of the tangent line is the derivative of the function, we start by calculting ${\textstyle {\frac {dy}{dx}}}$: $${\begin{aligned}{\frac {\mathrm {d} y}{\mathrm {d} x}}&={\sqrt {x}}+x({\frac {1}{2}}x^{-1/2})\\&={\frac {3}{2}}{\sqrt {x}}.\end{aligned}}$$ So, the slope of the tangent line at ${\textstyle (1,1)}$ is ${\textstyle m={\frac {dy}{dx}}|_{x=1}={\frac {3}{2}}}$. To determine the ${\textstyle y}$-intercept of the tangent line, we can either (1) sub in ${\textstyle y=1}$ and ${\textstyle x=1}$ into the equation of the line and solve for ${\textstyle b}$, OR (2) we could use the point-slope formula and rearrange it to be in the form ${\textstyle y=mx+b}$. Both methods will be shown below: Sub in ${\textstyle y=1}$ and ${\textstyle x=1}$: $${\begin{aligned}1&=({\frac {3}{2}})(1)+b\\b&=1-{\frac {3}{2}}\\b&=-{\frac {1}{2}}\end{aligned}}$$ Point-slope: $${\begin{aligned}(y-1)&={\frac {3}{2}}(x-1)\\y&={\frac {3}{2}}x+1-{\frac {3}{2}}\\y&={\frac {3}{2}}x-{\frac {1}{2}}\end{aligned}}$$ In either case, the equation of the tangent line at ${\textstyle (1,1)}$ is ${\textstyle y={\frac {3}{2}}x-{\frac {1}{2}}}$.