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Integration of a function that is the extreme value of a function having three variables
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Consider $$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ where $a,b,t ∈ R$.Given that $f(t)$ and $g(b)$ are the minimum values of $phi(a,b,t)$.
Based on this statement there are two questions:
$int fracdxf(x) $ is
$$
(a) tan^-1(x-1)+c \
(b) frac12tan^-1left(fracx-12right) + c \
(c) ln left[x-1+sqrtx^2-2x+2right]+c \
(d) frac12lnleft|fracx-1x+1right| + c
$$If $inte^xg(x)dx = e^x(Ax^2 + Bx + C) + D$, where $D$ is constant of integration then $(A+B+C)$ is equal to
$$
(a)2 \
(b) frac145 \
(c) 5 \
(d) frac195
$$
My attempt : Since the function $phi$ is a function of three variables, I didn't know how to proceed with the question. Any hint will be very helpful.
integration indefinite-integrals maxima-minima
asked Jul 20 at 5:21
MathsLearner
657213
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Consider $$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ where $a,b,t ∈ R$.Given that $f(t)$ and $g(b)$ are the minimum values of $phi(a,b,t)$.
Based on this statement there are two questions:
$int fracdxf(x) $ is
$$
(a) tan^-1(x-1)+c \
(b) frac12tan^-1left(fracx-12right) + c \
(c) ln left[x-1+sqrtx^2-2x+2right]+c \
(d) frac12lnleft|fracx-1x+1right| + c
$$If $inte^xg(x)dx = e^x(Ax^2 + Bx + C) + D$, where $D$ is constant of integration then $(A+B+C)$ is equal to
$$
(a)2 \
(b) frac145 \
(c) 5 \
(d) frac195
$$
My attempt : Since the function $phi$ is a function of three variables, I didn't know how to proceed with the question. Any hint will be very helpful.
integration indefinite-integrals maxima-minima
asked Jul 20 at 5:21
MathsLearner
657213
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider $$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ where $a,b,t ∈ R$.Given that $f(t)$ and $g(b)$ are the minimum values of $phi(a,b,t)$.
Based on this statement there are two questions:
$int fracdxf(x) $ is
$$
(a) tan^-1(x-1)+c \
(b) frac12tan^-1left(fracx-12right) + c \
(c) ln left[x-1+sqrtx^2-2x+2right]+c \
(d) frac12lnleft|fracx-1x+1right| + c
$$If $inte^xg(x)dx = e^x(Ax^2 + Bx + C) + D$, where $D$ is constant of integration then $(A+B+C)$ is equal to
$$
(a)2 \
(b) frac145 \
(c) 5 \
(d) frac195
$$
My attempt : Since the function $phi$ is a function of three variables, I didn't know how to proceed with the question. Any hint will be very helpful.
integration indefinite-integrals maxima-minima
asked Jul 20 at 5:21
MathsLearner
657213
Consider $$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ where $a,b,t ∈ R$.Given that $f(t)$ and $g(b)$ are the minimum values of $phi(a,b,t)$.
Based on this statement there are two questions:
$int fracdxf(x) $ is
$$
(a) tan^-1(x-1)+c \
(b) frac12tan^-1left(fracx-12right) + c \
(c) ln left[x-1+sqrtx^2-2x+2right]+c \
(d) frac12lnleft|fracx-1x+1right| + c
$$If $inte^xg(x)dx = e^x(Ax^2 + Bx + C) + D$, where $D$ is constant of integration then $(A+B+C)$ is equal to
$$
(a)2 \
(b) frac145 \
(c) 5 \
(d) frac195
$$
My attempt : Since the function $phi$ is a function of three variables, I didn't know how to proceed with the question. Any hint will be very helpful.
integration indefinite-integrals maxima-minima
asked Jul 20 at 5:21
MathsLearner
657213
edited Jul 20 at 5:49
asked Jul 20 at 5:21
MathsLearner
657213
asked Jul 20 at 5:21
MathsLearner
657213
asked Jul 20 at 5:21
MathsLearner
657213
657213
add a comment |Â
add a comment |Â
1 Answer
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To help you with the first, consider
$$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ from which
$$fracpartialpartial aphi(a,b,t) =4 a^3-10 a$$
$$fracpartialpartial bphi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+frac334$$ Complete the square and go on.
I am sure that you can take it from here.
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
To help you with the first, consider
$$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ from which
$$fracpartialpartial aphi(a,b,t) =4 a^3-10 a$$
$$fracpartialpartial bphi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+frac334$$ Complete the square and go on.
I am sure that you can take it from here.
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
To help you with the first, consider
$$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ from which
$$fracpartialpartial aphi(a,b,t) =4 a^3-10 a$$
$$fracpartialpartial bphi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+frac334$$ Complete the square and go on.
I am sure that you can take it from here.
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To help you with the first, consider
$$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ from which
$$fracpartialpartial aphi(a,b,t) =4 a^3-10 a$$
$$fracpartialpartial bphi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+frac334$$ Complete the square and go on.
I am sure that you can take it from here.
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
To help you with the first, consider
$$phi(a,b,t) = a^4 - 5a^2 +b^2 + 5t^2 -4bt -2t + frac334$$ from which
$$fracpartialpartial aphi(a,b,t) =4 a^3-10 a$$
$$fracpartialpartial bphi(a,b,t) =2 b-4 t$$ Since you look for an extremum, set these partial derivatives equal to $0$ and solve these two equations for $(a,b)$. Because of the first, you will find three solutions for $a$ and, for each of them, $b=2t$. You need to find which solution of $a$ (let us call it $a_*$ gives the minimum value of $phi(a,b,t)$.
So $$f(t)=t^2-2 t+a_*^4-5 a_*^2+frac334$$ Complete the square and go on.
I am sure that you can take it from here.
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
answered Jul 20 at 7:33
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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