Question

Calc III: Change of Parameter
Please help me calculate dr/d(tao) by the chain rule.
t=(tao)^2

Answer 1

do i just plug-in (tao)^2 in place of t?

Answer 2

Are you telling us \(\vec{r}(t)=e^t\mathbf{\hat{i}}+4e^{-t}\mathbf{\hat{j}}\)?

Answer 3

and then take the derivative?

Answer 4

yes that is correct

Answer 5

Well, consider:$$\frac{d\vec{r}}{d\tau}=\frac{d\vec{r}}{dt}\cdot\frac{dt}{d\tau}$$

Answer 6

@Babyslapmafro you could, or you could do the above if you only want the derivative... it's just the chain rule. Both give the same result.

Answer 7

We can see clearly that $$\frac{d\vec{r}}{dt}=e^t\mathbf{\hat{i}}-4e^{-t}\mathbf{\hat{j}}$$

Answer 8

ok well the problem asks me to calculate dr/d(tao) by the chain rule and then check my result by expressing r in terms of tao and differentiating.

Answer 9

... and given \(t=\tau^2\) we see \(\dfrac{dt}{d\tau}=2\tau\)

Answer 10

So by the chain rule we have:$$\frac{d\vec{r}}{dt}=(e^t\mathbf{\hat{i}}-4e^{-t}\mathbf{\hat{j}})\cdot2\tau=2\tau e^t\mathbf{\hat{i}}-8\tau e^{-t}\mathbf{\hat{j}}$$

Answer 11

err that should be \(\dfrac{d\vec{r}}{d\tau}\)

Answer 12

aren't you supposed to replace t with tao as well

Answer 13

Anyways, observe that we're not done yet since we're mixing \(t,\tau\); substitute \(t=\tau^2\) to get our final form:$$\frac{d\vec{r}}{d\tau}=2\tau e^{\tau^2}\mathbf{\hat{i}}-8\tau e^{-\tau^2}\mathbf{\hat{j}}$$

Answer 14

bingo @Babyslapmafro :-)

Answer 15

Now consider substituting \(t=\tau^2\) directly:$$\vec{r}(t)=\vec{r}(\tau^2)=e^{\tau^2}\mathbf{\hat i}+4e^{-\tau^2}\mathbf{\hat j}$$

Answer 16

Calculate our derivative as usual:$$\frac{d\vec{r}}{d\tau}=2\tau e^{\tau^2}\mathbf{\hat i}-8\tau e^{-\tau^2}\mathbf{\hat j}$$which is exactly identical!

Answer 17

ok thanks for the help