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The path r(t) = (t) + (21²-5) j describes motion on the parabola y = 2x2-5. Find the particle's velocity and acceleration vectors at t= 1, and sketch them as vectors on the curve. The velocity vector at t= 1 is v(1)=i+j (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

The path r(t) = (t) + (21²-5) j describes motion on the parabola y = 2x2-5. Find the particle's velocity and acceleration vectors at t= 1, and sketch them as vectors on the curve. The velocity vector at t= 1 is v(1)=i+j (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

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**Vector Functions and Motion Along a Curve** The path **r(t) = ⟨t, (2t² - 5)⟩** describes motion on the parabola **y = 2x² - 5**. Find the particle's velocity and acceleration vectors at **t = 1**, and sketch them as vectors on the curve. --- **The velocity vector at t = 1 is v(1) = ⟨[ ], [ ]⟩** *(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)* --- Explanation: 1. **Velocity Vector**: To find the velocity vector, differentiate the position vector **r(t)** with respect to time **t**. If **r(t) = ⟨t, (2t² - 5)⟩**, then: - **v(t)** = d**r(t)**/dt = ⟨1, 4t⟩ So, at t = 1: - **v(1)** = ⟨1, 4⟩ 2. **Acceleration Vector**: To find the acceleration vector, differentiate the velocity vector **v(t)** with respect to time **t**. If **v(t) = ⟨1, 4t⟩**, then: - **a(t)** = d**v(t)**/dt = ⟨0, 4⟩ So, at t = 1, the acceleration vector is: - **a(1)** = ⟨0, 4⟩ 3. **Sketching Vectors**: When sketching the vectors on the curve, make sure to: - Draw the parabola **y = 2x² - 5**. - Plot the point corresponding to **t = 1**, which is **(1, -3)**. - Draw the velocity vector **v(1) = ⟨1, 4⟩** originating from the point **(1, -3)**. - Draw the acceleration vector **a(1) = ⟨0, 4⟩** also originating from the point **(1, -3)**. --- By following these steps, you will be able to find and sketch the velocity
Transcribed Image Text:**Vector Functions and Motion Along a Curve** The path **r(t) = ⟨t, (2t² - 5)⟩** describes motion on the parabola **y = 2x² - 5**. Find the particle's velocity and acceleration vectors at **t = 1**, and sketch them as vectors on the curve. --- **The velocity vector at t = 1 is v(1) = ⟨[ ], [ ]⟩** *(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)* --- Explanation: 1. **Velocity Vector**: To find the velocity vector, differentiate the position vector **r(t)** with respect to time **t**. If **r(t) = ⟨t, (2t² - 5)⟩**, then: - **v(t)** = d**r(t)**/dt = ⟨1, 4t⟩ So, at t = 1: - **v(1)** = ⟨1, 4⟩ 2. **Acceleration Vector**: To find the acceleration vector, differentiate the velocity vector **v(t)** with respect to time **t**. If **v(t) = ⟨1, 4t⟩**, then: - **a(t)** = d**v(t)**/dt = ⟨0, 4⟩ So, at t = 1, the acceleration vector is: - **a(1)** = ⟨0, 4⟩ 3. **Sketching Vectors**: When sketching the vectors on the curve, make sure to: - Draw the parabola **y = 2x² - 5**. - Plot the point corresponding to **t = 1**, which is **(1, -3)**. - Draw the velocity vector **v(1) = ⟨1, 4⟩** originating from the point **(1, -3)**. - Draw the acceleration vector **a(1) = ⟨0, 4⟩** also originating from the point **(1, -3)**. --- By following these steps, you will be able to find and sketch the velocity
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