At what point do the curves r_1(t)=t,4-t,35+t^2 and r_2(s)=7-s,s-3,s^2Z

allhvasstH 2021-10-29 Answered
At what point do the curves \(\displaystyle{r}_{{1}}{\left({t}\right)}={t},{4}-{t},{35}+{t}^{{2}}\) and \(\displaystyle{r}_{{2}}{\left({s}\right)}={7}-{s},{s}-{3},{s}^{{2}}\) intersect? (x,y,z)= Find angle of intersection, \(\displaystyle\theta\), correct to the nearest degree.

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Layton
Answered 2021-10-30 Author has 8119 answers
Consider the following curves:
\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{t},{4}-{t},{35}+{t}^{{2}}{>}\)
\(\displaystyle{r}_{{2}}{\left({s}\right)}={<}{7}-{s},{s}-{3},{s}^{{2}}{>}\)
Find the point of intersection of the curves and angle of intersection \(\displaystyle\theta\).
The parametric equations corresponding to the curve \(\displaystyle{r}_{{1}}{\left({t}\right)}\) are as follows:
x=t
\(\displaystyle{y}={4}-{t}\)
\(\displaystyle{z}={35}+{t}^{{2}}\)
The parametric equations corresponding to the curve \(\displaystyle{r}_{{2}}{\left({s}\right)}\) are as follows:
\(\displaystyle{x}={7}-{s}\)
\(\displaystyle{y}={s}-{3}\)
\(\displaystyle{z}={s}^{{2}}\)
At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.
\(\displaystyle{t}={7}-{s}\Rightarrow{s}+{t}={7}\)
\(\displaystyle{4}-{t}={s}-{3}\Rightarrow{s}+{t}={7}\)
\(\displaystyle{35}+{t}^{{2}}={s}^{{2}}\Rightarrow{s}^{{2}}-{t}^{{2}}={35}\)
That is,
\(\displaystyle{s}^{{2}}-{t}^{{2}}={35}\)
\(\displaystyle{\left({s}-{t}\right)}{\left({s}+{t}\right)}={35}\)
\(\displaystyle{\left({s}-{t}\right)}{7}={35}\)
\(\displaystyle{s}-{t}={5}\)
Solve the equations \(\displaystyle{s}-{t}={5}\) and \(\displaystyle{s}+{t}={7}\), to find the values of s and t.
\(\displaystyle{s}-{t}+{s}+{t}={5}+{7}\Rightarrow{2}{s}={12}\Rightarrow{s}={6}\)
The corresponding value of t is as follows:
\(\displaystyle{t}={7}-{s}\Rightarrow{t}={7}-{6}\Rightarrow{t}={1}\)
Thus, the values are t=1 and s=6
Substitute t=1 in \(\displaystyle{r}_{{1}}{\left({t}\right)}\), to find the point of intersection.
\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{4}-{1},{35}+{1}^{{2}}\ge{<}{1},{3},{36}{>}\)
\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{3},{36}{>}\)
The angle between the normal vectors is given as \(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)
The direction vector of the line \(\displaystyle{r}_{{1}}{\left({t}\right)}\) is \(\displaystyle{<}{t},-{t},{t}^{{2}}{>}\)
The direction vector of the line \(\displaystyle{r}_{{2}}{\left({s}\right)}\) is \(\displaystyle{<}-{s},{s},{s}^{{2}}{>}\)
At t=1,
\(\displaystyle{r}_{{1}}'{\left({t}\right)}={<}{1},-{1},{2}{t}{>}\)
\(\displaystyle{r}_{{1}}'{\left({1}\right)}={<}{1},-{1},{2}{>}\)
At s=6,
\(\displaystyle{r}_{{2}}'{\left({s}\right)}={<}-{1},{1},{2}{s}{>}\)
\(\displaystyle{r}_{{2}}'{\left({6}\right)}={<}-{1},{1},{12}{>}\)
\(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)
\(\displaystyle={\frac{{{r}_{{1}}'{\left({1}\right)}\cdot{r}_{{2}}'{\left({6}\right)}}}{{{\left|{r}_{{1}}'{\left({1}\right)}\right|}{\left|{r}_{{2}}'{\left({6}\right)}\right|}}}}\)
\(\displaystyle={\frac{{{<}{1},-{1},{2}{>}\cdot{<}-{1},{1},{12}{>}}}{{\sqrt{{{1}+{1}+{4}}}\cdot\sqrt{{{1}+{1}+{144}}}}}}\)
\(\displaystyle={\frac{{-{1}-{1}+{24}}}{{\sqrt{{{6}}}\cdot\sqrt{{{146}}}}}}\)
\(\displaystyle={\frac{{{22}}}{{\sqrt{{{6}}}\cdot\sqrt{{{146}}}}}}\)
\(\displaystyle={\frac{{{11}}}{{\sqrt{{{219}}}}}}\)
Thus, \(\displaystyle\theta={{\cos}^{{-{1}}}{\left({\frac{{{11}}}{{\sqrt{{{219}}}}}}\right)}}\)
Have a similar question?
Ask An Expert
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-05-30
At what point do the curves \(r_1(t)=t,4-t,35+t^2\) and \(r_2(s)=7-s,s-3,s^2\) intersect? (x,y,z)= Find angle of intersection, \(\theta\), correct to the nearest degree.
asked 2021-06-10
Two college friends are taking a weekend road trip. Friday they leave home and drive 87 miles north for a night of dinner and dancing in the city. The next morning they drive 116 miles east to spend a day at the beach. If they drive straight home from the beach the next day, how far do they have to travel on Sunday?
asked 2021-11-19
Let \(\displaystyle{R}_{{1}}\) and \(\displaystyle{R}_{{2}}\) be a relations on a set A represented by the matrices below:
\[M_{R1}=\begin{bmatrix}0&1&0\\1&1&1\\1&0&0\end{bmatrix}\]
\[M_{R2}=\begin{bmatrix}0&1&0\\0&1&1\\1&1&1\end{bmatrix}\]
Find the matrix that represents \(\displaystyle{R}_{{1}}\cap{R}_{{2}}\)
asked 2021-05-18
In triangle DEF, side E is 4 cm long and side F is 7 cm long. If the angle between sides E and F is 50 degrees, how long is side D?
asked 2021-10-22
How do you estimate an angle to the nearest one-half radian?
asked 2021-05-28
The curvature of a plane curve y=y(x) is given by the formula
\(\frac{y"(x)}{((1+(y'))^2)^{\frac{3}{2}}}\)
Find all curves for which the curvature is 1 at every point. What are these curves?
asked 2021-04-14
A medical technician is trying to determine what percentage of apatient's artery is blocked by plaque. To do this, she measures theblood pressure just before the region of blockage and finds that itis \(\displaystyle{1.20}\times{10}^{{{4}}}{P}{a}\), while in the region of blockage it is \(\displaystyle{1.15}\times{10}^{{{4}}}{P}{a}\). Furthermore, she knows that blood flowingthrough the normal artery just before the point of blockage istraveling at 30.0 cm/s, and the specific gravity of this patient'sblood is 1.06. What percentage of the cross-sectional area of thepatient's artery is blocked by the plaque?

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...