Solve the initial value problem for 𝕣r as a vector function of t. Differential equation: d2rdt2=−32k Initial conditions: r(0)=100k and drdt∣t=0=8i+8j

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Solve the initial value problem for  𝕣r  as a vector function of t. Differential equation:  d2rdt2=−32k  Initial conditions:  r(0)=100k and   drdt∣t=0=8i+8j

Onlaceing · Accepted Answer

Step 1 Since d2rdt=−32k=0i+0j−32k, so drdt=∫(0i+0j−32k)dt =[∫0dt]i+[∫0dt]j−[∫32dt]k =a1i+b1j−(32t+c1)k for some real constants a1,b1andc1. Using the initial value condition for drdt, we can find the values of a1,b1andc1 and hence drdt. Since drdt∣t=0=8i+8j,we have: drdt∣t=0=a1i+b1j−(32(0)+c1)k 8i+8j=a1i+b1j−c1k Therefore,8=a1,8=b1and0=−c1⇒c1=0 and hence drdt=8i+8j−32tk Step 2 Since drdt=8i+8j−32tk, so r(t)=∫[8i+8j−32tk]dt =[∫8dt]i+[∫8dt]j−[∫32tdt]k =(8t+a2)i+(8t+b2)j−(16t2+c2)k for some real constants a2,

Donald Proulx · Accepted Answer

Step 2    Since drdt=8i+8j−32tk, so    r(t)=∫[8i+8j−32tk]dt    =[∫8dt]i+[∫8dt]j−[∫32tdt]k    =(8t+a2)i+(8t+b2)j−(16t2+c2)k    for some real constants a2,b2andc2.Using the initial value condition for r,we can find the values of a2,b2andc2and hence the vector function r.    Since r(0)=100k, we have:    r(0)=(8(0)+a2)i+(8(0)+b2)j−(16(0)2+c2)k    100k=a2i+b2j−c2k    Therefore,0=a2,0=b2and100=−c2⇒c2=−100 and hence    r(t)=8ti+8tj−(16t2−100)k    Result    r(t)=8ti+8tj−(16t2−100)k

Solve the initial value problem for \mathbb {r} as a vector func

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