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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #382 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=xln(y) , x=sin(t) and y=cos(t) .

Possible Answers:

$ln[sin(t)]sin(t)-\frac{cos^2(t)}{sin(t)}$

$ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

$ln[cos(t)]sin(t)-\frac{sin^2(t)}{cos(2t)}$

$ln[sin(t)]cos(t)-\frac{cos^2(t)}{sin(t)}$

Correct answer:

$ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

Explanation:

Find $\frac{dz}{dt}$ if z=xln(y) , x=sin(t) and y=cos(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(ln(y))(cos(t))+(x\frac{1}{y})(-sin(t))$

$\frac{dz}{dt}=ln(y)cos(t)-\frac{x}{y}sin(t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=sin(t) and y=cos(t) .

$\frac{dz}{dt}=ln(cos(t))cos(t)-\frac{sin(t)}{cos(t)}sin(t)$

$\frac{dz}{dt}=ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

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Example Question #381 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=x^2y-y^2x , x=t^2-5t and y=3t+4 .

Possible Answers:

25t^4 -150t^3 +268t^2 +508t +83

5t^4 +240t^3 -178t^2 -48t +80

15t^4 -140t^3 +168t^2 +408t +80

15t^4 -10t^3 +68t^2 +48t +8

Correct answer:

15t^4 -140t^3 +168t^2 +408t +80

Explanation:

Find $\frac{dz}{dt}$ if z=x^2y-y^2x , x=t^2-5t and y=3t+4 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xy-y^2)(2t-5)+(x^2-2yx)(3)$

$\frac{dz}{dt}=(2xy-y^2)(2t-5)+3(x^2-2xy)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^2-5t and y=3t+4 .

$\frac{dz}{dt}=(2(t^2-5t)(3t+4)-(3t+4)^2)(2t-5)+3((t^2-5t)^2-2(t^2-5t)(3t+4))$

$\frac{dz}{dt}=[2(3t^3-15t^2+4t^2-20t)-(9t^2+24t+16)](2t-5) +3[(t^4-10t^3+25t^2)-2(3t^3-15t^2+4t^2-20t)]$

$\frac{dz}{dt}=[6t^3-30t^2+8t^2-40t-9t^2-24t-16](2t-5) +3[t^4-10t^3+25t^2-6t^3+30t^2-8t^2+40t]$

$\frac{dz}{dt}=[6t^3-31t^2-64t-16](2t-5) +3[t^4-16t^3+47t^2+40t]$

$\frac{dz}{dt}=12t^4-62t^3-128t^2-32t -30t^3+155t^2+320t+80 +3t^4-48t^3+141t^2+120t$

$\frac{dz}{dt}=15t^4 -140t^3 +168t^2 +408t +80$

Report an Error

Example Question #384 : Partial Derivatives

Find $\frac{dz}{dt}$ if $z=e^{5xy}$ , $x=\sqrt{t}$ and y=t^2 .

Possible Answers:

$\frac{27}{4}t^{5/2}e^{5t^{3/2}}$

$\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

$\frac{25}{4}t^{3/4}e^{5t^{5/4}}$

$\frac{35}{9}t^{3}e^{5t^{5}}$

Correct answer:

$\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

Explanation:

Find $\frac{dz}{dt}$ if $z=e^{5xy}$ , $x=\sqrt{t}$ and y=t^2 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(e^{5xy}*5y)(1/2t^{-1/2})+(e^{5xy}*5x)(2t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, $x=\sqrt{t}$ and y=t^2 .

$\frac{dz}{dt}=(e^{5t^{1/2}(t^2)}*5t^2)(1/2t^{-1/2})+(e^{5t^{1/2}(t^2)}*5t^{1/2})(2t)$

$\frac{dz}{dt}=(e^{5t^{5/2}}*5t^2)(1/2t^{-1/2})+(e^{5t^{5/2}}*5t^{1/2})(2t)$

$\frac{dz}{dt}=(e^{5t^{5/2}}*5/2t^{3/2})+(e^{5t^{5/2}}*10t^{3/2})$

$\frac{dz}{dt}=e^{5t^{5/2}}(5/2t^{3/2}+10t^{3/2})$

$\frac{dz}{dt}=\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

Report an Error

Example Question #31 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^2+7xy^2 , x=1/t and y=5t^2-t .

Possible Answers:

55t^2-10t+8-2t^3

525t^2-140t+7-2/t^3

-52t^5-40t^4+7t^3-2t^2

425t^2-240t-7+1/t^3

Correct answer:

525t^2-140t+7-2/t^3

Explanation:

Find $\frac{dz}{dt}$ if z=x^2+7xy^2 , x=1/t and y=5t^2-t .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2x+7y^2)(-1/t^2)+(14xy)(10t-1)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=1/t and y=5t^2-t .

$\frac{dz}{dt}=[2(1/t)+7(5t^2-t)^2](-1/t^2)+[14(1/t)(5t^2-t)](10t-1)$

$\frac{dz}{dt}=[2/t+7(25t^4-10t^3+t^2)](-1/t^2)+[(14/t)(5t^2-t)](10t-1)$

$\frac{dz}{dt}=[2/t+175t^4-70t^3+7t^2](-1/t^2)+(70t-14)(10t-1)$

$\frac{dz}{dt}=-2/t^3-175t^2+70t-7+700t^2-140t-70t+14$

$\frac{dz}{dt}= -175t^2+700t^2 +70t-140t-70t +14-7-2/t^3$

$\frac{dz}{dt}= 525t^2-140t+7-2/t^3$

Report an Error

Example Question #32 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=sin(x)cos(y) , x=t^4 and y=7-t .

Possible Answers:

4t^3cos(t^4)cos(7-t)+sin(t^4)sin(7-t)

-t^2cos(5t^4)sin(7-t)+sin(t^5)sin(7-t)

-t^3cos(5t^5)cos(7-t)+sin(t^5)sin(7-t)

14t^3cos(2t^5)sin(7-t)+cos(t^3)sin(7-t)

Correct answer:

4t^3cos(t^4)cos(7-t)+sin(t^4)sin(7-t)

Explanation:

Find $\frac{dz}{dt}$ if z=sin(x)cos(y) , x=t^4 and y=7-t .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=[cos(y)cos(x)](4t^3)+[sin(x)(-sin(y))](-1)$

$\frac{dz}{dt}=4t^3cos(x)cos(y)+sin(x)sin(y)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^4 and y=7-t .

$\frac{dz}{dt}=4t^3cos(t^4)cos(7-t)+sin(t^4)sin(7-t)$

Report an Error

Example Question #33 : Multi Variable Chain Rule

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables x, y , and are functions of the of the parameter :

$x = t\: \: \:\: \: \: \:y=\ln(t)\: \: \: \: \: \: \: \:z=t^2$

Possible Answers:

$t(e^{t^2}-1)$

$4t^2e^{t^2}+1$

$6te^{t^2}(1+t^2)$

$12t^2e^{t^2}(1+2t)$

$e^{t^2}(1-t)$

Correct answer:

$6te^{t^2}(1+t^2)$

Explanation:

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables x, y , and are functions of the of the parameter :

$x = t\: \: \:\: \: \: \:y=\ln(t)\: \: \: \: \: \: \: \:z=t^2$

Because each function is given in terms of the parameter we can conveniently write the function as a function of alone:

f(x(t), y(t), z(t))=f(t)

Solution 1

For the function described in this problem, the function can be written strictly in terms of the parameter by simply substituting the definitions given for , or .

$f(x,y,z)=f(t, \: \ln t,\: \:t^2 ) )=3te^{t^2+\ln t}=3t^2e^{t^2}$

$f(t)=3t^2e^{t^2}$

So now we have the function written in the form of a single variable function of We can use the usual chain-rule.

Apply the product rule and and the chain-rule:

$\frac{df}{dt}=3t^2\left(\frac{d}{dt}e^{t^2} \right )+e^{t^2}\left(\frac{d}{dt}3t^2 \right )=6te^{t^2}+3t^2e^{t^2}(2t)$

$=6te^{t^2}+6t^3e^{t^2}$

$=6te^{t^2}(1+t^2)$

Solution 2

The derivative of with respect to will be the sum of the derivatives of each variable , , and with respect to

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$ (1)

In Equation (1) we use the partial derivative notation for a derivative of with respect to the variables , , and since it is a multi-variable function, we have to specify which variable we are differentiating unless we are differentiating with respect to

We return to the standard derivative notation when computing the derivatives with respect to since each function , , , and can be written as a single variable function of Now simply apply (3) to the function term-by-term:

$\frac{\partial f}{\partial x}\frac{dx}{dt}=3e^{z+y}$

$\frac{\partial f}{\partial y}\frac{dy}{dt}=3xe^{z+y}\frac{1}{t}$

$\frac{\partial f}{\partial z}\frac{dz}{dt}=3xe^{z+y}(2t)=6xte^{z+y}$

Now ad the terms to get an expression for $\frac{df}{dt}$ and then rewrite in terms of

$\frac{df}{dt}=3xe^{z+y}\frac{1}{t}+6xte^{z+y}+3e^{z+y}$

$=3e^{z+y}(1+\frac{x}{t}+2xt)$

$=3e^{t^2+\ln t}\left(1+\frac{t}{t}+2t^2\right)$

$=6e^{t^2+\ln t}\left(1+t^2\right)$

$=6te^{t^2}\left(1+t^2\right)$

Therefore:

$\frac{df}{dt}=6te^{t^2}\left(1+t^2\right)$

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Example Question #34 : Multi Variable Chain Rule

Find the two variable function f(x,y) that satisfies the following:

$\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )=2$

$\frac{\partial f}{\partial x}(0,1)=1$

f(1,0)=2

Possible Answers:

f(x,y)=2x-3y(4-x)

f(x,y)=y-2x+4

f(x,y)=(2y-1)x+3

f(x,y)=(2x-1)y-1

f(x,y)=2xy-3(4-x)

Correct answer:

f(x,y)=(2y-1)x+3

Explanation:

$\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )=2$

$\frac{\partial f}{\partial x}(0,1)=1$

f(1,0)=2

The first step is to integrate the function with respect to since that is the outer most derivative in this problem.

$\frac{\partial f}{\partial x}= \int \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )dy=\int 2dy=2y+C_1$

$f(x,y)=\int \frac{\partial f}{\partial x}dx=\int (2y+C_1)dx=(2y+C_1)x+C_2$

Now apply the constraints to compute the two constants of integration.

$\frac{\partial f}{\partial x}(0,1)=2+C_1 = 1 \: \: \: \: \: \: \:\Rightarrow \: \: \: \: \: \: \:C_1=-1$

So far we have:

f(x,y)=(2y-1)x+C_2

Now apply the second constraint given:

$f(1,0)=2\: \: \: \: \: \:\ \Rightarrow \: \: \: \: \: -1+C_2 =2\: \: \: \: \: \Rightarrow \: \: \: \:C_2 =3$

f(x,y)=(2y-1)x+3

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Example Question #1 : Applications Of Partial Derivatives

Find the linear approximation to $z=6+\frac{x^2}{4}+\frac{y^2}{16}$ at (-2,4) .

Possible Answers:

$L(x,y)\approx 4+x-\frac{1}{2}y$

$L(x,y)\approx -x+\frac{1}{2}y$

$L(x,y)\approx 4-x+\frac{1}{2}y$

$L(x,y)\approx 4+\frac{1}{2}y$

$L(x,y)\approx 4-x-\frac{1}{2}y$

Correct answer:

$L(x,y)\approx 4-x+\frac{1}{2}y$

Explanation:

The question is really asking for a tangent plane, so lets first find partial derivatives and then plug in the point.

$f(x,y)=6+\frac{x^2}{4}+\frac{y^2}{16}$ , $f(-2,4)=6+\frac{(-2)^2}{4}+\frac{(4)^2}{16}=6+1+1=8$

$f_x(x,y)=\frac{x}{2}$ , $f_x(-2,4)=\frac{-2}{2}=-1$

$f_y(x,y)=\frac{y}{8}$ , $f_y(-2,4)=\frac{4}{8}=\frac{1}{2}$

Remember that we need to build the linear approximation general equation which is as follows.

$L(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

$L(x,y)\approx 8-(x+2)+\frac{1}{2}(y-4)$

$L(x,y)\approx 8-x-2+\frac{1}{2}y-2$

$L(x,y)\approx 4-x+\frac{1}{2}y$

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Example Question #1 : Tangent Planes And Linear Approximations

Find the tangent plane to the function $f(x,y) = 2xy+e^{-x}$ at the point (0,1,1) .

Possible Answers:

y-z=-1

x+y=1

-x+z=1

y-z=0

x-z=2

Correct answer:

-x+z=1

Explanation:

To find the equation of the tangent plane, we use the formula

z- z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) .

Taking partial derivatives, we have

f_x(x,y) = $2y-e^{-x}$

f_y(x,y) = 2x

Substituting our x, y values into these, we get

f_x(0,1) =1

f_y(0,1) = 0

Substituting our point into x_0, y_0, z_0 , and partial derivative values in the formula we get

z - 1 = 1(x-0)+0(y-1)

z - 1 = x

-x+z =1 .

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Example Question #3 : Applications Of Partial Derivatives

Find the Linear Approximation to $z=2+\frac {x^2}{9}+\frac {y^2}{4}$ at (-3,2) .

Possible Answers:

$4-\frac {2}{3}(x+3)+(y+2)$

None of the Above

$-4+\frac {2}{3}(x+3)+(y-1)$

$4+\frac {2}{3}(x+3)+(y-2)$

Correct answer:

$4+\frac {2}{3}(x+3)+(y-2)$

Explanation:

We are just asking for the equation of the tangent plane:

Step 1: Find f(x,y)

$f(x,y)=2+\frac {(-3)^2}{9}+\frac {2^2}{4}=2+\frac {9}{9}+\frac {4}{4}=2+1+1=4$

Step 2: Take the partial derivative of $\frac {x^2}{9}$ with respect with (x,y):

$\partial(\frac {x^2}{9})=\frac {2x}{9}$

Step 3: Evaluate the partial derivative of x at

$\frac {2(3)}{9}=\frac {6}{9}=\frac {2}{3}$

Step 4: Take the partial derivative of $\frac {y^2}{4}$ with respect to (x,y) :

$\partial (\frac{y^2}{4})=\frac {2y}{4}$

Step 5: Evaluate the partial derivative at y=2

$\frac {2(2)}{4}=\frac {4}{4}=1$ .

Step 6: Convert (x,y) back into binomials:

$x\rightarrow (x+3)$
$y\rightarrow (y-2)$

Step 7: Write the equation of the tangent line:

$4+\frac {2}{3}(x+3)+(y-2)$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #382 : Partial Derivatives

Example Question #381 : Partial Derivatives

Example Question #384 : Partial Derivatives

Example Question #31 : Multi Variable Chain Rule

Example Question #32 : Multi Variable Chain Rule

Example Question #33 : Multi Variable Chain Rule

Example Question #34 : Multi Variable Chain Rule

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Tangent Planes And Linear Approximations

Example Question #3 : Applications Of Partial Derivatives

All Calculus 3 Resources

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