The vectors `vecA=sin(alphat) hati-cos(alphat) hatj`and `vecB=cos(alphat^2/4) hati +sin(alphat^2/4) hatj` are orthogonal to each other the value of t would be(where `alpha` is a positive constant)

Options

A. t=2
B. t=4
C. t=6
D. t=8

Answer

To solve the problem, we need to determine the value of \( t \) such that the vectors \( \vec{A} \) and \( \vec{B} \) are orthogonal. This means that their dot product must equal zero.

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \vec{A} = \sin(\alpha t) \hat{i} - \cos(\alpha t) \hat{j} \]
\[ \vec{B} = \cos\left(\frac{\alpha t^2}{4}\right) \hat{i} + \sin\left(\frac{\alpha t^2}{4}\right) \hat{j} \]

2. **Calculate the Dot Product**:
The dot product \( \vec{A} \cdot \vec{B} \) is given by:
\[ \vec{A} \cdot \vec{B} = \left(\sin(\alpha t) \cdot \cos\left(\frac{\alpha t^2}{4}\right)\right) + \left(-\cos(\alpha t) \cdot \sin\left(\frac{\alpha t^2}{4}\right)\right) \]
This simplifies to:
\[ \vec{A} \cdot \vec{B} = \sin(\alpha t) \cos\left(\frac{\alpha t^2}{4}\right) - \cos(\alpha t) \sin\left(\frac{\alpha t^2}{4}\right) \]

3. **Set the Dot Product to Zero**:
Since the vectors are orthogonal, we set the dot product to zero:
\[ \sin(\alpha t) \cos\left(\frac{\alpha t^2}{4}\right) - \cos(\alpha t) \sin\left(\frac{\alpha t^2}{4}\right) = 0 \]

4. **Use the Sine Difference Identity**:
We can recognize that the expression can be rewritten using the sine difference identity:
\[ \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \]
Let \( A = \alpha t \) and \( B = \frac{\alpha t^2}{4} \). Thus, we have:
\[ \sin\left(\alpha t - \frac{\alpha t^2}{4}\right) = 0 \]

5. **Solve for \( t \)**:
The sine function equals zero at integer multiples of \( \pi \):
\[ \alpha t - \frac{\alpha t^2}{4} = n\pi \quad (n \in \mathbb{Z}) \]
Rearranging gives:
\[ \alpha t - n\pi = \frac{\alpha t^2}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ 4\alpha t - 4n\pi = \alpha t^2 \]
Rearranging gives a quadratic equation:
\[ \alpha t^2 - 4\alpha t + 4n\pi = 0 \]

6. **Use the Quadratic Formula**:
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{4\alpha \pm \sqrt{(4\alpha)^2 - 4\alpha(4n\pi)}}{2\alpha} \]
Simplifying:
\[ t = \frac{4 \pm \sqrt{16 - 16n\pi}}{2} = 2 \pm 2\sqrt{1 - n\pi} \]

7. **Determine Valid Values of \( t \)**:
Since \( n \) must be chosen such that \( 1 - n\pi \geq 0 \), we can choose \( n = 0 \) to find:
\[ t = 2 \pm 2\sqrt{1} = 2 \pm 2 \]
This gives us \( t = 4 \) or \( t = 0 \). Since \( t \) must be positive, we have:
\[ t = 4 \]

### Final Answer:
The value of \( t \) is \( 4 \).