Finding inverse of a Sin function (problem from Mooculus)

[Ryaners]

I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h^-1(20).

The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h^-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h^-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h^-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h⁻¹(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!

 

[Samy_A]

I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h^-1(20).

The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h^-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h^-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h^-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h⁻¹(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!

You forgot about the condition: "If h is restricted to the domain [3.5, 10.5]"
arcsin in a multivalued function, that makes calculating the inverse of h tricky.
(I mean $\sin(0)=0$, but also $\sin(\pi)=0$, etc...)

Maybe just set h(t) = 20, and see what value(s) in the domain [3.5, 10.5] satisfy this equation.

 

[Ray Vickson]

I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h^-1(20).

The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h^-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h^-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h^-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h⁻¹(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!

If h is restricted to [3.5,10.5] (as you stated) then findig where h(t) = 20 makes no sense. Did you mean t in [3.5,10.5]? If that is what you mean then I would avoid starting with the arcsin function and just look at the equation h(t) = 20 directly; later, after clarifying the rough locations of the relevant root or roots, I would increase the accuracy by then using the arcsiin function at that point. Remember: the standard arcsin function yields angles between -π/2 and +π/2.

 

[Ryaners]

If h is restricted to [3.5,10.5] (as you stated) then findig where h(t) = 20 makes no sense. Did you mean t in [3.5,10.5]? If that is what you mean then I would avoid starting with the arcsin function and just look at the equation h(t) = 20 directly; later, after clarifying the rough locations of the relevant root or roots, I would increase the accuracy by then using the arcsiin function at that point. Remember: the standard arcsin function yields angles between -π/2 and +π/2.

Maybe just set h(t) = 20, and see what value(s) in the domain [3.5, 10.5] satisfy this equation.

Thanks Ray - it's asking me to find h^-1(20), not h(20). Apologies if I've misinterpreted your reply or that was a typo. The question is exactly as stated in the textbook & I'm disinclined to think it's a mistake. It's pretty confusing, though.
If h(t) has an inverse h^-1(t) (on the interval h ∈ [3.5, 10.5]), the 'output' of h(20) will be the height h at t=20, right? Is h^-1(20) not then the time at h = 20, which is not a permitted value of h? Maybe it is a mistake, and they do mean t ∈ [3.5, 10.5]. (Those seem like more sensible as values for time than for height with a ferris wheel in mind, though maybe bringing the real world into textbook examples is a bad idea..!)

 

[Samy_A]

It says: "If h is restricted to the domain [3.5, 10.5]", so they quite clearly mean that you have to find a t ∈ [3.5, 10.5] satisfying h(t)=20.

 

[Ryaners]

It says: "If h is restricted to the domain [3.5, 10.5]", so they quite clearly mean that you have to find a t ∈ [3.5, 10.5] satisfying h(t)=20.

Ah yes ok, h being a function & not a variable. My bad. Thanks for pointing that out.

 

[Ray Vickson]

Thanks Ray - it's asking me to find h^-1(20), not h(20). Apologies if I've misinterpreted your reply or that was a typo. The question is exactly as stated in the textbook & I'm disinclined to think it's a mistake. It's pretty confusing, though.
If h(t) has an inverse h^-1(t) (on the interval h ∈ [3.5, 10.5]), the 'output' of h(20) will be the height h at t=20, right? Is h^-1(20) not then the time at h = 20, which is not a permitted value of h? Maybe it is a mistake, and they do mean t ∈ [3.5, 10.5]. (Those seem like more sensible as values for time than for height with a ferris wheel in mind, though maybe bringing the real world into textbook examples is a bad idea..!)

No, I knew perfectly well what you MEANT, which is different from what you SAID. Here it is exactly (via cut and paste): "If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h^-1(20)". As written, that statement makes no sense; $h^{-1}(20)$ refers to the solution of the equation $h(t) = 20$, but you said you were restricting $h$ to [3.5, 10.5], which most definitely excludes the value '20' for h. Beginners such as students and self-learners might have more trouble figuring what you actually meant, so that is why it is always good to avoid such errors if you can. In this case it seems that your textbook needs some editing, so the problem begins there.

I suppose one could say that the word "domain" clears it up, because the domain of h(t) is a set of "allowed" t-values. However, it is a somewhat weird way of saying it, and as I indicated, beginners might have trouble interpreting it. It would have been better if they said "the domain of h is [3.5,10.5]" or something similar.