Find the decomposition of a(t) into tangential and r(t)=⟨t,12t2,16t3,t=1⟩

Step 1 Let r→(t) be a vector-valued function that denotes the position of an object as a function of time. Then a→(t)=r→ is the accelearion vector. The tangential and normal components of acceleration at→ and an→ are given by the formulas at→=v→.a→∣∣v∣∣→ and at→=∥v→×a→∥∥v→∥ We have given r→(t)=&lt;t,12t2,16t3&gt; y→(t)=&lt;1,t,12t2&gt; a→(t)=&lt;0,1,t&gt; Step 2 Then, at→=y→.a→∥y→∥ =&lt;1,t12t2&gt;.&lt;0,1,t&gt;1+t2+t44 at→(1)=1+121+1+14 3∠2=9∠4 =3∠2=3∠2 =1 Step 3 t→=∥y→×a→∥∥y→∥ a_ a=∥&lt;t22,−t,1&gt;∥1+t2+t44 =1+t2+t441+t2+t44 =1 at→(1)=1

Question: "Find the decomposition of a(t) into tangential and..."

varaderiyw

Answered question

2021-11-29

Find the decomposition of a(t) into tangential and

r (t) = ⟨ t, \frac{1}{2} t^{2}, \frac{1}{6} t^{3}, t = 1 ⟩

Answer & Explanation

Onlaceing

Beginner2021-11-30Added 15 answers

Step 1
Let $\vec{r} (t)$ be a vector-valued function that denotes the position of an object as a function of time. Then
$\vec{a} (t) = \vec{r}$ is the
accelearion vector.
The tangential and normal components of acceleration $a_{\vec{t}}$ and $a_{\vec{n}}$ are given by the formulas
$a_{\vec{t}} = \frac{\vec{v} . \vec{a}}{∣∣ \vec{v ∣∣}}$ and
$a_{\vec{t}} = \frac{∥ \vec{v} \times \vec{a} ∥}{∥ \vec{v} ∥}$
We have given
$\vec{r} (t) = < t, \frac{1}{2} t^{2}, \frac{1}{6} t^{3} >$
$\vec{y} (t) = < 1, t, \frac{1}{2} t^{2} >$
$\vec{a} (t) = < 0, 1, t >$
Step 2
Then,
$a_{\vec{t}} = \frac{\vec{y} . \vec{a}}{∥ \vec{y} ∥}$
$= \frac{< 1, t \frac{1}{2} t^{2} > . < 0, 1, t >}{\sqrt{1 + t^{2} + \frac{t^{4}}{4}}}$
$a_{\vec{t}} (1) = \frac{1 + \frac{1}{2}}{\sqrt{1 + 1 + \frac{1}{4}}}$
$^{3} ∠ 2 = \sqrt{^{9} ∠ 4}$
$=^{3} ∠ 2 =^{3} ∠ 2$
$= 1$
Step 3
$\vec{t} = \frac{∥ \vec{y} \times \vec{a} ∥}{∥ \vec{y} ∥}$
a_ $a = \frac{∥< \frac{t^{2}}{2}, - t, 1 >∥}{\sqrt{1 + t^{2} + \frac{t^{4}}{4}}}$
$= \frac{\sqrt{1 + t^{2} + \frac{t^{4}}{4}}}{\sqrt{1 + t^{2} + \frac{t^{4}}{4}}}$
$= 1$
$a_{\vec{t}} (1) = 1$

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