Welcome to our community

Be a part of something great, join today!

Calculus Curves

Kris

New member
May 22, 2013
29
For the curve defined by
r(t) = 3*t*i + 2*t^2*j − 3*t^4*k


Find the tangent vector r′(t0) at the point P(4,8,−16), given that the position vector of P is r(t0).

and

Find the vector equation of the tangent line to the trajectory through P.

Im unsure as to how to go about solving this. Ive looked for help in my textbook but it doesn't have any examples.

Is there a formula that I could use?? I think I have to differentiate r(t) then plug in the points but im unsure if this is right??
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
For the curve defined by
r(t) = 3*t*i + 2*t^2*j − 3*t^4*k


Find the tangent vector r′(t0) at the point P(4,8,−16), given that the position vector of P is r(t0).

and

Find the vector equation of the tangent line to the trajectory through P.

Im unsure as to how to go about solving this. Ive looked for help in my textbook but it doesn't have any examples.

Is there a formula that I could use?? I think I have to differentiate r(t) then plug in the points but im unsure if this is right??
Are you sure that $\mathbf{r}(t)=\langle 3t,2t^2,-3t^4\rangle$ is the correct vector valued function and that you have the correct point? For problems like this, you typically first figure out what $t_0$ is by setting $\mathbf{r}(t_0)=P$; in this case, we'd like to find a $t_0$ such that $3t_0=4$, $2t_0^2=8$ and $-3t_0^4=-16$. If you tried to solve for $t_0$ in each case, you'd get different values (which is a bad thing -- it tells you that the point doesn't lie on the vector valued curve and hence finding a tangent line makes no sense); this is why I want to make sure you have the correct function/point.

Once we get that taken care of, then you can easily find $\mathbf{r}^{\prime}(t_0)$ and furthermore the equation of the tangent line which has the equation $L(t) = \mathbf{r}^{\prime}(t_0)t + P$.

I hope this makes sense!
 

Kris

New member
May 22, 2013
29
Apologies the curve is actually.

r(t) = 2*t*i + 2*t^2*j − 2*t^3*k

I provided the wrong equation from something else I was working on. I was wondering why it indeed looked odd :/ Does this look better??
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Apologies the curve is actually.

r(t) = 2*t*i + 2*t^2*j − 2*t^3*k

I provided the wrong equation from something else I was working on. I was wondering why it indeed looked odd :/ Does this look better??
Yes, it does look better now. With that said, when you set $\mathbf{r}(t_0)=P$, we get the equations $2t_0=4$, $2t_0^2=8$ and $-2t_0^3=-16$. When you solve each of them for $t_0$, we get $t_0=2$. Thus, we're supposed to find $\mathbf{r}^{\prime}(2)$ and then the tangent line $L(t) = \mathbf{r}^{\prime}(2)t+(4,8,-16)$.

Do you think you can take things from here?
 

Kris

New member
May 22, 2013
29
Actually no :/ I need to ask how Im going to find the vector?? Do I just solve the equation for L and then provide that as the answer?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Actually no :/ I need to ask how Im going to find the vector?? Do I just solve the equation for L and then provide that as the answer?
I'm trying not to work everything out for you, so please bear with me. :)

You first need to compute $\mathbf{r}^{\prime}(2)$. Since $\mathbf{r}(t)=\langle 2t,2t^2,-2t^3\rangle$, it follows that $\mathbf{r}^{\prime}(t)=\ldots$ and thus $\mathbf{r}^{\prime}(2)=\ldots$. Once you have this, you then plug this into the line equation I provided to get $L(t)$ and simplify.

I hope this clarifies things!
 

Kris

New member
May 22, 2013
29
yes that clears things thankyou :) problem has been solved now