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Calculus 3 : Calculus 3

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Example Questions

Example Question #748 : Vectors And Vector Operations

Find the determinant of the 3x3 matrix given:

$\begin{bmatrix} 1&-3 &1 \\ 3&-2 &1 \\ 4& 0& 5 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

To find the determinant of a 3x3 matrix $m=\begin{bmatrix} x&y &z\\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{bmatrix}$ . determinant=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)

Using the vectors from the problem statement, we get

determinant=1(-10-0)+3(15-4)+1(0+8)=31

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Example Question #745 : Vectors And Vector Operations

Find the determinant of the 3x3 matrix $\begin{bmatrix} -4&7 &1 \\ 3&0 &2 \\ -2&2 & 5 \end{bmatrix}$

Possible Answers:

155

121

Correct answer:

Explanation:

To find the determinant of a 3x3 matrix $M=\begin{bmatrix} x&y &z \\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{bmatrix}$ , you use the formula

determinant(M)=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)

Using the matrix from the problem statement, we get

-4(0-4)-7(15+4)+1(6-0)=-111

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Example Question #2751 : Calculus 3

Find the determinant of the 3x3 matrix $\begin{bmatrix} 1&0 &1 \\ 12&2 &4 \\ 5&10 &3 \end{bmatrix}$

Possible Answers:

100

Correct answer:

Explanation:

To find the determinant of a 3x3 matrix $M=\begin{bmatrix} x&y &z \\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{bmatrix}$ , you use the formula

determinant(M)=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)

Using the matrix from the problem statement, we get

1(6-40)-0(36-20)+1(120-10)=76

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Example Question #2752 : Calculus 3

Find the determinant of the matrix $\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1&-5 &3 \\ 2& 7& 0 \end{bmatrix}$

Possible Answers:

$\left \langle -21,-6,17\right \rangle$

$\left \langle -21,9,17\right \rangle$

$\left \langle -21,6,17\right \rangle$

$\left \langle -20,6,14\right \rangle$

Correct answer:

$\left \langle -21,6,17\right \rangle$

Explanation:

To find the determinant of a 3x3 matrix $m=\begin{bmatrix} x& y & z\\ a_1&a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}$ , we use the formula

determinant(m)=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)

Using the matrix from the problem statement, we get

$\hat{i}(0-21)-\hat{j}(0-6)+\hat{k}(7+10)=\left \langle -21,6,17\right \rangle$

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Example Question #2753 : Calculus 3

Calculate the determinant of Matrix .

$A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$

Possible Answers:

$\det(A)=6$

$\det(A)=-2$

$\det(A)=1$

$\det(A)=0$

$\det(A)=-6$

Correct answer:

$\det(A)=-6$

Explanation:

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract A_u from A_d .

$\det(A)=A_d-A_u=16-22=-6$

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Example Question #2751 : Calculus 3

Calculate the determinant of Matrix .

$A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$

Possible Answers:

$\det(A)=-2$

$\det(A)=1$

$\det(A)=6$

$\det(A)=-6$

$\det(A)=0$

Correct answer:

$\det(A)=-6$

Explanation:

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract A_u from A_d .

$\det(A)=A_d-A_u=16-22=-6$

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Example Question #382 : Partial Derivatives

Find $\frac{\partial f}{dz}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10e^z$

Correct answer:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

Explanation:

In order to find $\frac{\partial f}{dz}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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Example Question #383 : Partial Derivatives

Find $\frac{\partial f}{dy}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dy}=x+\frac{z}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+ye^{xy}$

$\frac{\partial f}{dy}=z^{10}+\frac{1}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+e^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

$\frac{\partial f}{dy}=\frac{x}{\ln(yz)}$

$\frac{\partial f}{dy}=xz^{10}$

Correct answer:

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Explanation:

In order to find $\frac{\partial f}{dy}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

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Example Question #2752 : Calculus 3

Find $\frac{\partial f}{dx}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$

$\frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Correct answer:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Explanation:

In order to find $\frac{\partial f}{dx}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

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Example Question #392 : Partial Derivatives

True or False

$\frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$

Possible Answers:

False

True

Correct answer:

True

Explanation:

True:

Since there are no y's in the equation, the derivative of a constant is .

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #748 : Vectors And Vector Operations

Example Question #745 : Vectors And Vector Operations

Example Question #2751 : Calculus 3

Example Question #2752 : Calculus 3

Example Question #2753 : Calculus 3

Example Question #2751 : Calculus 3

Example Question #382 : Partial Derivatives

Example Question #383 : Partial Derivatives

Example Question #2752 : Calculus 3

Example Question #392 : Partial Derivatives

All Calculus 3 Resources

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