Learning Materials
Features
Discover
Chapter 5: Problem 73
For \(y=x(c-x)\) \(\frac{d y}{d x}=c-2 x\) For \(y=x^{2}+a x+b\) \(\frac{d y}{d x}=2 x+a\) At \((1,0)\) \(0=(C-1) \Rightarrow C=1\) \(0=1+a+b, 1-2=2+a\) \(\Rightarrow \quad a=-3\) \(\Rightarrow b=2\)
Over 30 million students worldwide already upgrade their learning with 91影视!
All the tools & learning materials you need for study success - in one app.
Get started for free
\(1^{2}=x^{2}+y^{2}\) \(1 \frac{d l}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t}\) \(y=x^{3 / 2}\) \(\frac{d y}{d t}=\frac{3}{2} x^{1 / 2} \frac{d x}{d t}\) Using (1) \& (2) $11 \sqrt{x^{2}+x^{3}}=x \frac{d x}{d t}+\frac{3}{2} x^{3 / 2} x^{1 / 2} \frac{d x}{d t}$ \(\frac{d x}{d t}=\frac{66}{3+\frac{27}{2}}=\frac{66 \times 2}{33}=4\)
\(y=\frac{x^{2}}{4}-3 x+10\) For \(y=2-\frac{x^{2}}{4}\) eqn of tangent \(y-2=m x+m^{2}\) Solving (II) \& (I) \(m x+2+m^{2}=\frac{x^{2}}{4}-3 x+10\) \(\Rightarrow \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0\) \(\mathrm{D}=0\) \((m+3)^{2}-\left(8-m^{2}\right)=0\) \(2 m^{2}+6 m+1=0\) \(\mathrm{m}_{1}+\mathrm{m}_{2}=-3\)
\(\mathrm{x}=\mathrm{t}^{2}+\mathrm{t}+1\) \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=2 \mathrm{t}+1\) \(\mathrm{y}=\mathrm{t}^{2}-\mathrm{t}+1\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1\) \(\frac{d y}{\mathrm{~d} \mathrm{x}}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\) $\Rightarrow \mathrm{y}-\left(\mathrm{t}^{2}-\mathrm{t}+1\right)=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\left(\mathrm{x}-\left(\mathrm{t}^{2}+\mathrm{t}+1\right)\right)$ $\Rightarrow\left(\mathrm{t}-\mathrm{t}^{2}\right)=\left(\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\right)\left(-\left(\mathrm{t}^{2}+\mathrm{t}\right)\right)$ $\Rightarrow 2 \mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-\mathrm{t}^{2}=\mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-2 \mathrm{t}^{2}$ \(\Rightarrow 2 \mathrm{t}^{2}=0\) \(\Rightarrow \mathrm{t}=0\)
Ellipse \(\rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) \(\Rightarrow \frac{x}{8} \frac{d x}{d t}+\frac{2 y}{9} \frac{d y}{d t}=1\) $\Rightarrow \frac{4 \sqrt{1-y^{2} / 9}}{8} \frac{d x}{d t}+\frac{2 y}{9} \times \frac{d y}{d x} \times \frac{d x}{d t}=1$ Now put \(\frac{d y}{d x}=1 \quad \& \quad y=1\) $\Rightarrow \frac{4}{6 \sqrt{2}} \frac{d x}{d t}+\frac{2}{9} \frac{d x}{d t}=1$ \(\Rightarrow \frac{d x}{d t}\left(\frac{\sqrt{2}}{3}+\frac{2}{9}\right)=1\) \(\Rightarrow \frac{d x}{d t}=\frac{9}{3 \sqrt{2}+2}\)
\(x y^{2}=1\) \(y^{2}+2 x y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-y}{2 x}\) \(-\frac{d x}{d y}=\frac{2 x}{y}=\frac{2}{y^{3}}\) \(y-y_{1}=\frac{2}{y_{1}^{3}}\left(x-x_{1}\right)\) \(+y_{1}^{4}=2 x_{1}\) \(y_{1}^{6}=2\) \(y_{1}=\pm 2^{1 / 6}\) \(x_{1}=\pm 2^{-\sqrt{3}}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine