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AP Calculus BC 3.6 Calculating Higher- Order Derivatives Exam Style Questions – FRQ

AP Calculus BC 3.6 Calculating Higher- Order Derivatives Exam Style Questions - FRQ

Question

(a) Topic-7.3-Sketching Slope Fields

(b) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema

(c) Topic-7.7-Finding Particular Solutions Using Initial Conditions and Separation of Variables

3. The depth of seawater at a location can be modeled by the function H that satisfies the differential equation \(\frac{dH}{dt}=\frac{1}{2}(H-1)cos(\frac{t}{2})\), where H ( t) is measured in feet and t is measured in hours after noon (t = 0). It is known that H(0 ) = 4.

(a) A portion of the slope field for the differential equation is provided. Sketch the solution curve, y = H ( t),through the point ( 0, 4) .

(b) For 0< t< 5, it can be shown that H (t ) >  1. Find the value of t, for 0 < t < 5, at which H has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.

(c) Use separation of variables to find y = H(t), the particular solution to the differential equation \(\frac{dH}{dt}=\frac{1}{2}(H-1)cos(\frac{t}{2})\)with initial condition H(o ) = 4.

▶️Answer/Explanation

3(a) A portion of the slope field for the differential equation is provided. Sketch the solution curve, y = H(t), through the point (0, 4).

3(b) For 0 < t < 5, it can be shown that H(t) > 1. Find the value of t, for 0 <t < 5, at which H has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.

Because H(t) > 1, then \(\frac{dH}{dt}=0\) implies \(cos\frac{t}{2}=0\).

This implies that t = \(\pi \) is a critical point.

For \(o< t< \pi ,\frac{dH}{dt}> 0\) and for \(\pi < t< 5 ,\frac{dH}{dt}< 0\). Therefore,  t = \(\pi\) is the location of a relative maximum value of H

3(c) Use separation of variables to find y = H(t), the particular solution to the differential equation \(\frac{dH}{dt}=\frac{1}{2}(H-1)cos(\frac{t}{2})\) with initial condition H(0) = 4. 

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