Home Practice Questions AP Calculus AB Contextual Applications of Differentiation Question No. 25

Answer is: option4

II and III only

Solution:

"The particle is at a positive position on the \( x \)-axis at time \( t = \frac{a+b}{2} \)."

• We substitute \( t = \frac{a+b}{2} \) into \( x(t) \):

\[ x \left( \frac{a+b}{2} \right) = \left( \frac{a+b}{2} - a \right)^3 \left( \frac{a+b}{2} - b \right) \] \[ = \left( \frac{b-a}{2} \right)^3 \left( \frac{a-b}{2} \right) \]

Since \( b > a \), the term \( \frac{b-a}{2} \) is positive, so \( \left( \frac{b-a}{2} \right)^3 \) is also positive. However, \( \frac{a-b}{2} \) is negative. Thus, the product is negative.

Since \( x \left( \frac{a+b}{2} \right) < 0 \), statement I is false.

"The particle is at rest at time \( t = a \)."

• The velocity is given by:

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( (t-a)^3 (t-b) \right) \]

Using the product rule:

\[ v(t) = 3(t-a)^2 (t-b) + (t-a)^3 (1) \]
Factor out \( (t-a)^2 \):
\[ v(t) = (t-a)^2 [3(t-b) + (t-a)] \]

Substituting \( t = a \):

\[ v(a) = (b-a)^2 [3(a-b) + (a-a)] \] \[ = 0 \]

Since \( v(a) = 0 \), the particle is at rest at \( t = a \), so Statement II is true.

"The particle is moving to the right at \( t = b \)."

• The particle moves to the right if \( v(b) > 0 \).

Substituting \( t = b \) into the velocity expression:

\[ v(b) = (b-a)^2 [3(b-b) + (b-a)] \] \[ = (b-a)^2 [0 + (b-a)] \] \[ = (b-a)^3 \]

Since \( b > a \), we have \( (b-a)^3 > 0 \), meaning \( v(b) > 0 \), so the particle is moving to the right.

Statement III is true.

Final Answer:

Since Statement II and Statement III are true, the correct answer is: II and III only.