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MHF Helper

#### myngo9191

2. Find the equation of the line tangent to at $$\displaystyle y=\frac{5}{13}x+$$$$\displaystyle \frac{21}{13}$$
3. Find the point P on the graph of closest to the point 4. Calculate the surface area of revolution of about the x-axis over $$\displaystyle y=(4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}$$

$$\displaystyle (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}{x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{3}}$$

$$\displaystyle u=4-x^\frac{2}{3}$$
$$\displaystyle du=-\frac{2}{3}x^{-\frac{1}{3}}dx$$
$$\displaystyle -3du=2x^{-\frac{1}{3}}$$

$$\displaystyle 2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}$$

#### skeeter

MHF Helper
2. Find the equation of the line tangent to at $$\displaystyle y=\frac{5}{13}x+$$$$\displaystyle \frac{21}{13}$$

ok
3. Find the point P on the graph of closest to the point 4. Calculate the surface area of revolution of about the x-axis over $$\displaystyle y=(4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}$$

$$\displaystyle (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}{x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{3}}$$

$$\displaystyle u=4-x^\frac{2}{3}$$
$$\displaystyle du=-\frac{2}{3}x^{-\frac{1}{3}}dx$$
$$\displaystyle -3du=2x^{-\frac{1}{3}}$$

$$\displaystyle 2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}$$
$$\displaystyle A = 2\pi \int_2^3 y\sqrt{1+(y')^2} \, dx$$

$$\displaystyle y = (4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y' = -\frac{(4-x^\frac{2}{3})^\frac{1}{2}}{x^\frac{1}{3}}$$

$$\displaystyle (y')^2 = \frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}$$

$$\displaystyle 1 + (y')^2 = \frac{4}{x^\frac{2}{3}}$$

$$\displaystyle A = 2\pi \int_2^3 (4-x^\frac{2}{3})^\frac{3}{2} \cdot \frac{2}{x^\frac{1}{3}} \, dx$$

$$\displaystyle A = -\frac{12\pi}{5} \left[(4-x^\frac{2}{3})^\frac{5}{2}\right]_2^3 \approx 29.658$$

#### myngo9191

I didn't even know it's an optimization problem and no I haven't done any. I went look in the book and it says I should find the critical point by solving for x. I'm not sure if that's the way to go but here it is...

$$\displaystyle 2(x-10)+1\rightarrow x=9\frac{1}{2}$$

#### Defunkt

MHF Hall of Honor
The point closest to the graph is the point where the minimum of the distance function is achieved.

Minimum is achieved at an extremal point, so...

#### myngo9191

I'm sorry but I'm really lost and I have no idea what to do with that problem.