Question 1204791: Let f'''(t) = -10t - 10 sqrt(t). I found f''(t) = -5t^2 - 20/3t^(3/2) + C but I am struggling to find f'(t) and f(t). I had found an equation for f'(t) = -5/3t^3 - 8/3t^(5/2) + D but the computer says it's not correct.
Found 3 solutions by MathLover1, ikleyn, math_tutor2020:
Answer by MathLover1(20849)   (Show Source):
Answer by ikleyn(52767)  (Show Source):
You can put this solution on YOUR website! .
As I see from the error, made by you in your post, you just know the basic rules
of taking antiderivatives, but are not uniformly attentive.
When you worked on f'(t), you missed the term C*t in this antiderivative.
Taking antiderivatives is a part of Calculus, which you need to learn from your textbook,
if you want to know the subject.
Taking antiderivative (indefinite integrals) is a key part of Calculus.
To start, look into this introduction
https://www.sparknotes.com/math/calcab/introductiontointegrals/section1/
Normal Calculus student must know the rules of taking antiderivatives as good
as he (or she) knows the multiplication table.
Antiderivative f''(t) in the post by @MathLover1 is INCORRECT.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
The tickmark notation is very limited, so I'll use exponents to represent the derivative level.
f^3 = 3rd derivative
f^2 = 2nd derivative
f^1 = 1st derivative
f = original function
f^3 = -10t - 10*sqrt(t)
f^3 = -10t - 10*t^(1/2)
f^2 = integral[ f^3 ]
f^2 = integral[ -10t - 10*t^(1/2) ]
f^2 = -10*(1/(1+1))*t^(1+1) - 10*(1/(0.5+1))*t^(1/2+1) + C
f^2 = -5t^2 - (20/3)t^(3/2) + C
As a check,
f^3 = derivative[ f^2 ]
f^3 = derivative[ -5t^2 - (20/3)t^(3/2) + C ]
f^3 = derivative[ -5t^2 ] - derivative[ (20/3)t^(3/2) ] + derivative[ C ]
f^3 = 2*(-5)t^(2-1) - (3/2)*(20/3)*t^(3/2-1) + 0
f^3 = -10t - 10*t^(1/2)
f^3 = -10t - 10*sqrt(t)
The CAS (computer algebra system) feature of GeoGebra can be used to verify this.
WolframAlpha is another good option.
Feel free to explore other calculators.
Then,
f^1 = integral[ f^2 ]
f^1 = integral[ -5t^2 - (20/3)t^(3/2) + C ]
f^1 = (-5/3)t^3 - (20/3)*(1/(3/2+1))*t^(3/2+1) + C*t + D
f^1 = (-5/3)t^3 - (8/3)*t^(5/2) + C*t + D
As a check,
f^2 = derivative[ f^1 ]
f^2 = derivative[ (-5/3)t^3 - (8/3)*t^(5/2) + C*t + D ]
f^2 = (-5/3)*3*t^(3-1) - (8/3)*(5/2)*t^(5/2-1) + C
f^2 = -5*t^2 - (20/3)*t^(3/2) + C
Lastly,
f = integral[ f^1 ]
f = integral[ (-5/3)t^3 - (8/3)*t^(5/2) + C*t + D ]
f = (-5/3)*(1/(3+1))*t^(3+1) - (8/3)*(1/(5/2+1))*t^(5/2+1) + (C/2)*t^2 + D*t + E
f = (-5/12)*t^4 - (16/21)*t^(7/2) + (C/2)*t^2 + D*t + E
I'll leave the derivative check for the student to carry out.
Summary- original function = f = (-5/12)*t^4 - (16/21)*t^(7/2) + (C/2)*t^2 + D*t + E
- 1st derivative = f^1 = (-5/3)t^3 - (8/3)*t^(5/2) + C*t + D
- 2nd derivative = f^2 = -5*t^2 - (20/3)*t^(3/2) + C
where C, D, and E are constants. Each answer has been verified.
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