Solve the initial value problem for r as a vector function of t. Differential equation: drdt=−ti−tj−tk Initial condition: r(0)=i+2j+3k

Name: Solve the initial value problem for r as a vector function of t. Differential equation: drdt=−ti−tj−tk Initial condition: r(0)=i+2j+3k
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the initial value problem for r as a vector function of t. Differential equation: drdt=−ti−tj−tk Initial condition: r(0)=i+2j+3k

Question

Solve the initial value problem for  r  as a vector function of t. Differential equation:  drdt=−ti−tj−tk  Initial condition:  r(0)=i+2j+3k

Gennenzip · Accepted Answer

r=∫(−ti−tj−tk)dt=−t22i−t22j−t22k+C;  r(0)=0i−0j−0k+C=i+2j+3k⇒C=i+2j+3k⇒  Result:  r=(−t22+1)i+(−t22+2)j+(−t22+3)k

Andre BalkonE · Accepted Answer

Step 1: Integrate both sides of the differential equation with respect to t:∫drdtdt=∫(−ti−tj−tk)dtStep 2: Apply the integration rule for each component:r(t)=∫−tidt+∫−tjdt+∫−tkdtStep 3: Evaluate the integrals:r(t)=−12t2i−12t2j−12t2k+C where C is the constant of integration.Step 4: Use the initial condition r(0)=i+2j+3k to determine the value of C:i+2j+3k=−12(0)2i−12(0)2j−12(0)2k+CSimplifying the equation:C=i+2j+3kStep 5: Substitute the value of C back into the solution:r(t)=−12t2i−12t2j−12t2k+i+2j+3kSimplifying further:r(t)=(1−12t2)i+(2−12t2)j+(3−12t2)kTherefore, the solution to the initial value problem is:r(t)=(1−12t2)i+(2−12t2)j+(3−12t2)k

Jazz Frenia · Accepted Answer

To solve the initial value problem for r as a vector function of t, with the given differential equation and initial condition, we can proceed as follows.The differential equation is given by:d𝐫dt=−t𝐢−t𝐣−t𝐤where 𝐢, 𝐣, and 𝐤 are the standard unit vectors in the Cartesian coordinate system.To solve this differential equation, we&#039;ll integrate both sides with respect to t. Integrating the right-hand side:∫(−t𝐢−t𝐣−t𝐤)dt=∫−t𝐢dt−∫t𝐣dt−∫t𝐤dt⇒−12t2𝐢−12t2𝐣−12t2𝐤+𝐂where 𝐂 is a constant vector of integration.Now, we need to find the constant vector 𝐂 using the initial condition:𝐫(0)=𝐢+2𝐣+3𝐤Substituting t = 0 and 𝐫(0) into the solution expression, we get:𝐢+2𝐣+3𝐤=−12(0)2𝐢−12(0)2𝐣−12(0)2𝐤+𝐂Simplifying, we find:𝐢+2𝐣+3𝐤=𝐂Therefore, the constant vector 𝐂 is equal to 𝐢+2𝐣+3𝐤.Substituting this value back into the solution expression, we have:𝐫(t)=−12t2𝐢−12t2𝐣−12t2𝐤+𝐂𝐫(t)=−12t2𝐢−12t2𝐣−12t2𝐤+(𝐢+2𝐣+3𝐤)𝐫(t)=(𝐢−12t2𝐢)+(2𝐣−12t2𝐣)+(3𝐤−12t2𝐤)𝐫(t)=𝐢(1−12t2)+𝐣(2−12t2)+𝐤(3−12t2)Hence, the solution to the initial value problem is:𝐫(t)=(1−12t2)𝐢+(2−12t2)𝐣+(3−12t2)𝐤

fudzisako · Accepted Answer

Step 1:First, let&#039;s separate the variables and integrate both sides of the differential equation:∫d𝐫dtdt=∫(−t𝐢−𝐣−𝐤)dt∫d𝐫=∫(−t𝐢−𝐣−𝐤)dt𝐫=∫(−t𝐢−𝐣−𝐤)dtStep 2:Next, we can integrate each component separately. Since the constant of integration will be a vector, we can denote it as 𝐂:𝐫=∫(−t𝐢)dt−∫𝐣dt−∫𝐤dt+𝐂𝐫=(−12t2𝐢)−(t𝐣)−(t𝐤)+𝐂Step 3:Now, let&#039;s apply the initial condition 𝐫(0)=𝐢+2𝐣+3𝐤 to find the constant 𝐂:𝐫(0)=(−12(0)2𝐢)−(0𝐣)−(0𝐤)+𝐂𝐢+2𝐣+3𝐤=𝐂Therefore, the constant of integration 𝐂 is 𝐢+2𝐣+3𝐤.Hence, the solution to the initial value problem is:𝐫(t)=−12t2𝐢−t𝐣−t𝐤+(𝐢+2𝐣+3𝐤)We can simplify this further:𝐫(t)=(−12t2+1)𝐢+(2−t)𝐣+(3−t)𝐤

Solve the initial value problem for r as a vector function of t. Differe

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