MHB N27.09 Derivative of tan and phase shift

[karush]

Find the function with the given derivative
whose graph passes through point P.
$$r'\left(\theta\right) =6+\sec^2 \left({\theta}\right), P\left(\frac{\pi}{4},0\right)$$
[desmos="0,2pi,-10,10"]6+sec^2(x)[/desmos]

The phase shift appears to be 1 but not sure how to get that

How do add another equation to desmos?

 

[Bushy]

You have $$\sec\left({\theta}\right)$$ in your equation and $$\sec^2\left({\theta}\right)$$ in your graph. Which one is it?

 

[karush]

$$\sec^2 \left({\theta}\right)$$

It appears you have to use x in desmos

 

[Bushy]

How about $$ r\left(\theta\right) = \int r'(\theta) d\theta = \int 6+\sec ^2 \left({\theta}\right)d\theta = 6\theta +\tan(\theta) +C $$

 

[karush]

$$6\theta+\tan\left({\theta +1 }\right)$$
Seems to be the answer
But desmos doesn't like $\theta

 

[Bushy]

just use x instead of theta

 

[MarkFL]

What you are given is the IVP:

$$\d{r}{\theta}=\sec^2(\theta)+6$$ where $$r\left(\frac{\pi}{4}\right)=0$$

Integrating w.r.t $\theta$, we obtain:

$$\int_0^{r(\theta)}\,du=\int_{\frac{\pi}{4}}^{\theta}\sec^2(v)+6\,dv$$

Applying the FTOC, there results:

$$r(\theta)-0=\left[\tan(v)+6v\right]_{\frac{\pi}{4}}^{\theta}=\tan(\theta)+6\theta-\tan\left(\frac{\pi}{4}\right)-6\left(\frac{\pi}{4}\right)=\tan(\theta)+6\theta-\frac{3\pi+2}{2}$$

 

[karush]

Well that graphed to the answer but it seemed to be a vertical shift downward,
Why would you do this vs a phase shift?

 

[MarkFL]

Well that graphed to the answer but it seemed to be a vertical shift downward,
Why would you do this vs a phase shift?

We are given an IVP, and I simply used a standard method for solving such a problem. An IVP consists of an ODE and an initial condition...so we can use the initial and final values (boundaries) as the limits of integration, and then apply the FTOC to get the solution satisfying both the ODE and the initial values.