In the equation `int(dt)/(sqrt(2at-t^(2)))=a^(x) sin^(-1)[t/a-1]`. The value of `x` is

Options

A. 1
B. `-1`
C. `0`
D. `2`

Answer

To solve the equation

\[ \int \frac{dt}{\sqrt{2at - t^2}} = a^x \sin^{-1}\left(\frac{t}{a} - 1\right), \]

we need to find the value of \( x \).

### Step 1: Analyze the left-hand side of the equation

We start with the integral on the left-hand side:

\[ \int \frac{dt}{\sqrt{2at - t^2}}. \]

To simplify this, we can factor out \( t \) from the expression under the square root. Notice that:

\[ 2at - t^2 = t(2a - t). \]

Thus, we can rewrite the integral as:

\[ \int \frac{dt}{\sqrt{t(2a - t)}}. \]

### Step 2: Change of variables

Let's make a substitution to simplify the integral. We can let \( t = 2a \sin^2(\theta) \). Then, \( dt = 4a \sin(\theta) \cos(\theta) d\theta \).

Substituting \( t \) into the integral gives us:

\[ \int \frac{4a \sin(\theta) \cos(\theta) d\theta}{\sqrt{2a(2a \sin^2(\theta)) - (2a \sin^2(\theta))^2}}. \]

### Step 3: Simplifying the integral

Now, we simplify the expression under the square root:

\[ 2a(2a \sin^2(\theta)) - (2a \sin^2(\theta))^2 = 2a(2a \sin^2(\theta)) - 4a^2 \sin^4(\theta). \]

This can be simplified further, but we can already see that the integral will yield a result that has a dependence on \( a \).

### Step 4: Analyze the right-hand side

Now, let's look at the right-hand side:

\[ a^x \sin^{-1}\left(\frac{t}{a} - 1\right). \]

The term \( \sin^{-1}\left(\frac{t}{a} - 1\right) \) is dimensionless, as it is an inverse sine function. Therefore, the only dimension that matters on the right-hand side is \( a^x \).

### Step 5: Equating dimensions

Now, we need to equate the dimensions from both sides. The left-hand side has dimensions of time \( [T] \) since the integral will yield a result in terms of time.

The right-hand side has dimensions \( [A]^x \) where \( [A] \) is the dimension of \( a \).

From the context, we can assume that \( a \) has dimensions of acceleration \( [L][T^{-2}] \). Thus, the dimensions of \( a^x \) will be:

\[ [L^{x}][T^{-2x}]. \]

### Step 6: Setting up the equation

Now, we can equate the dimensions from both sides:

\[ [T] = [L^{x}][T^{-2x}]. \]

This gives us:

\[ 1 = x \quad \text{(for length)} \quad \text{and} \quad 1 = -2x \quad \text{(for time)}. \]

### Step 7: Solving for \( x \)

From \( 1 = -2x \):

\[ x = -\frac{1}{2}. \]

However, since we are looking for the exponent of \( a \) that balances the equation, we find that \( x = 0 \) satisfies the dimensional analysis.

### Final Answer

Thus, the value of \( x \) is:

\[ \boxed{0}. \]