Activity for Bennyshinichiâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #293104 |
Thank you for pointing this out, I was unaware that the use of incorrect terminology could lead to such significant misunderstandings. Thank you for your explanation, I'll keep that in mind in the future! (more) |
— | 24 days ago |
| Edit | Post #293011 | Initial revision | — | about 2 months ago |
| Question | — |
How to validate if the horizontal and vertical tangent lines exist for implicit functions? Given an implicit function $x^2 + y^3 - 15xy = 0$, I needed to calculate the horizontal tangent line and the vertical tangent line. When we calculate the derivative from this function we get:$$\frac{dy}{dx}= \frac{15y-2x}{3y^2-15x}$$ So mine approach of calculating the vertical tangent line was to s... (more) |
— | about 2 months ago |
| Comment | Post #291919 |
Thank you so much for your help! As I could follow the most parts, I will definitely look back on this once I have learned more about these topics. (more) |
— | 7 months ago |
| Comment | Post #291919 |
Thank you for your answer. May I ask how you derived this expression for $Q(t)$? Because I'm not very familiar with matrices. (more) |
— | 7 months ago |
| Edit | Post #291917 | Initial revision | — | 7 months ago |
| Question | — |
How can we prove that a point that follows another point has the same trajectory given a contant angle and ratio? So suppose that we have a point called $P$ that moves along the circle $O1$. The parametric equations of $P$ are given by:$$\begin{aligned}x(t) &= \cos(t) + 5 \\y(t) &= \sin(t) + 3 \\&\text{Or}\\\overrightarrow{P(t)} &= \begin{Bmatrix}\cos(t) + 5 \\\sin(t) + 3\end{Bmatrix}\end{aligned}$$ This point ... (more) |
— | 7 months ago |
| Comment | Post #291500 |
But substituting wouldn't result in the substituted value (in this case $\beta$) to come back in the equation, right? (more) |
— | 8 months ago |
| Comment | Post #291500 |
Aha, that makes a lot more sense now. I have one last question: how did we derive $\alpha^2 \mathbf{u}^2 - \beta\alpha \mathbf{v} \cdot \mathbf{u}$? Did you use $(\alpha \mathbf{u} - \beta \mathbf{v})^2$? But this expansion results in $\alpha^2 \mathbf{u}^2 - 2\beta\alpha \mathbf{v} \cdot \mathbf{u} ... (more) |
— | 8 months ago |
| Edit | Post #291497 |
Post edited: |
— | 8 months ago |
| Comment | Post #291500 |
Thank you for correcting me. I'll update my post using the correct equations. I understand your explanation up to the part where you wrote:
$$\begin{align}
\mathbf s \cdot (\alpha \mathbf u - \beta \mathbf v)
& = \mathbf s \cdot \alpha \mathbf u \\\\
& = \alpha^2 \mathbf u^2 - \beta\alpha\mathbf ... (more) |
— | 8 months ago |
| Edit | Post #291497 | Initial revision | — | 8 months ago |
| Question | — |
How to find a point on a vector equation with another vector equation and perpendicular distance? Given two lines, $l: 3x-y=9$ and $m: x-3y=3$, and an angle bisector $x-y=3$. Point $A\left(3,0\right)$ is where these three lines intersect. The problem asks us to find a point on the angle bisector that is at a perpendicular distance of $\sqrt{10}$ from line $m$. I know that we can use the distance ... (more) |
— | 8 months ago |