Find Tangent Vector & Vector Equation for Curve r(t)

In summary, we use the formula $\mathbf{r}(t)=\langle 2t,2t^2,-2t^3\rangle$ to find the tangent vector $\mathbf{r}^{\prime}(2)$ at the point P(4,8,-16) by plugging in t=2. We then use this vector to find the vector equation of the tangent line through P, which has the equation $L(t)=\mathbf{r}^{\prime}(2)t+(4,8,-16)$.
  • #1
Kris1
29
0
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. I've 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 I am unsure if this is right??
 
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  • #2
Kris said:
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. I've 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 I am 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!
 
  • #3
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??
 
  • #4
Kris said:
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?
 
  • #5
Actually no :/ I need to ask how I am going to find the vector?? Do I just solve the equation for L and then provide that as the answer?
 
  • #6
Kris said:
Actually no :/ I need to ask how I am 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!
 
  • #7
yes that clears things thankyou :) problem has been solved now
 

Related to Find Tangent Vector & Vector Equation for Curve r(t)

1. What is a tangent vector?

A tangent vector is a vector that describes the direction and rate of change of a curve at a specific point. It is a vector that is tangent to the curve at that point, meaning it touches but does not cross the curve.

2. How do you find the tangent vector for a curve?

To find the tangent vector for a curve, you can use the derivative of the curve's equation. The derivative will give you the slope of the curve at a specific point, which can then be used to calculate the tangent vector. Alternatively, you can use the parametric equations of the curve to find the tangent vector.

3. What is a vector equation for a curve?

A vector equation for a curve is an equation that describes the position of a point on the curve in terms of a vector. It is typically written as r(t) = , where x(t), y(t), and z(t) are the components of the position vector at time t. This equation allows us to easily find the coordinates of any point on the curve by plugging in a value for t.

4. How do you find the vector equation for a curve?

To find the vector equation for a curve, you can use the parametric equations of the curve. These equations will give you the x, y, and z coordinates of a point on the curve at any given time. You can then combine these equations into one vector equation, as described in the previous question.

5. Why is it important to find the tangent vector and vector equation for a curve?

Finding the tangent vector and vector equation for a curve is important because it allows us to understand the direction and rate of change of the curve at any given point. This information is useful in a variety of applications, such as physics, engineering, and computer graphics. It also allows us to easily find the coordinates of any point on the curve, which can be helpful in solving problems and making predictions.

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