Applications of Partial Derivatives - AP Calculus BC

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Question

Find the absolute minimums and maximums of on the disk of radius , $x^2+y^2\leq16$ .

Answer

The first thing we need to do is find the partial derivative in respect to , and .

$\frac{\partial}{\partial x}=20x$ , $\frac{\partial}{\partial y}=-6y+10$

We need to find the critical points, so we set each of the partials equal to .

$0=20x\rightarrow x=0$

$0=-6y+10 \rightarrow y=\frac{5}{3}$

We only have one critical point at $f\Big(0, \frac{5}{3}\Big)$ , now we need to find the function value in order to see if it is inside or outside the disk.

$f\Big(0, \frac{5}{3}\Big)=10(0)^2-3\Big(\frac{5}{3}\Big)^2+10\Big(\frac{5}{3}\Big)=\frac{25}{3}\approx8.33333$

This is within our disk.

We now need to take a look at the boundary, . We can solve for , and plug it into .

We will need to find the absolute extrema of this function on the range $-4\leq y \leq 4$ . We need to find the critical points of this function.

$0=-26y+10 \rightarrow y=\frac{5}{13}$

The function value at the critical points and end points are:

$g\Big(\frac{5}{13}\Big)=-13\Big(\frac{5}{13}\Big)^2+10\Big(\frac{5}{13}\Big)+160=\frac{2105}{13}\approx161.92$

Now we need to figure out the values of these correspond to.

$y=-4: x^2=(-4)^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=4: x^2=4^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=\frac{5}{13}: x^2=\Big(\frac{5}{13}\Big)^2-16 \rightarrow x^2=\frac{2679}{169}\rightarrow x\approx\pm\ 3.98$

Now lets summarize our results as follows:

$g(-4)=-88 \rightarrow f(0,-4)=-88$

$g(4)=-8 \rightarrow f(0,4)=-8$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(-\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

From this we can conclude that there is an absolute minimum at , and two absolute maximums at $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ and $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ .

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Question

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at .

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=2z=4$

y:

$\frac{\delta f}{\delta y}=6yz=60$

z:

$\frac{\delta f}{\delta z}=2x+3y^2=2+75=77$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=4$

y:

$\frac{\delta f}{\delta y}=5$

z:

$\frac{\delta f}{\delta z}=6$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=5yz=60$

y:

$\frac{\delta f}{\delta y}=5xz=40$

z:

$\frac{\delta f}{\delta z}=5xy=30$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=6yz+2y+z=3$

y:

$\frac{\delta f}{\delta y}=6xz+2x=60$

z:

$\frac{\delta f}{\delta z}=6xy+x=3$

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Question

Find the slope of the function at the point $(3,2,2\pi)$

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point $(3,2,2\pi)$

x:

$\frac{\delta f}{\delta x}=y^2sin(z)=0$

y:

$\frac{\delta f}{\delta y}=2xysin(z)=0$

z:

$\frac{\delta f}{\delta z}=xy^2cos(z)=12$

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Question

Find the slope of the function at the point $(\pi,\pi,\pi)$

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point $(\pi,\pi,\pi)$

x:

$\frac{\delta f}{\delta x}=sin(y)cos(y)=0$

y:

$\frac{\delta f}{\delta y}=xcos(y)cos(z)=\pi$

z:

$\frac{\delta f}{\delta z}=-xsin(y)cos(y)=0$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=3x^2y^2=108$

y:

$\frac{\delta f}{\delta y}=2x^3y=48$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=3y=6$

y:

$\frac{\delta f}{\delta y}=3x+6y^2=30$

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Question

Find the slope of the function at the point $(3,\pi)$

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point $(3,\pi)$

x:

$\frac{\delta f}{\delta x}=sin(y)=0$

y:

$\frac{\delta f}{\delta y}=xcos(y)=-3$

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Question

Find the slope of the function at the point

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point

x:

$\frac{\delta f}{\delta x}=5y^2-4xy=80-32=48$

y:

$\frac{\delta f}{\delta y}=10xy-2x^2=80-8=72$

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Question

Find the slope of the function at the point $(2,\pi)$

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point $(2,\pi)$

x:

$\frac{\delta f}{\delta x}=ycos(xy)=\pi$

y:

$\frac{\delta f}{\delta y}=xcos(xy)=2$

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Question

Find the slope of the function at the point $(\frac{1}{4},\pi)$

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at at the point $(\frac{1}{4},\pi)$

x:

$\frac{\delta f}{\delta x}=\frac{y}{cos^2(xy)}=2\pi$

y:

$\frac{\delta f}{\delta y}=\frac{x}{cos^2(xy)}=\frac{1}{2}$

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Question

Find the equation of the tangent plane to $z = \ln \left(x^2-3xy+y \right )$ at .

Answer

The definition of a tangent plane of a surface $z=f \left(x,y \right )$ is given by

$z_T = f \left(x_0, y_0 \right )+f_x\left(x_0, y_0 \right )\left(x-x_0\right )+f_y\left(x_0, y_0 \right )\left(y-y_0 \right )$ .

Therefore, we need to first find , given below as

$f_x=\frac{2x-3y}{x^2-3xy+y}\Rightarrow f_x\left(5, 1 \right )=\frac{7}{11}$

$f_y=\frac{-3x+1}{x^2-3xy+y}\Rightarrow f_y\left(5, 1 \right )=-\frac{14}{11}$

Noting that

$f \left(5,1 \right )= \ln (11)$ ,

we can write our complete expression for the tangent line as

$z_T=\ln (11)+\frac{7}{11}\left(x-5 \right )-\frac{14}{11}\left(y-1 \right )$ .

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Question

Calculate $\vec{R}$ if $\vec{\bigtriangledown} f = \vec{R}$ and

Answer

By definition $\vec{\bigtriangledown}=\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}+\hat{z}\frac{\partial}{\partial z}$ . Therefore,

$\vec{R}=\vec{\bigtriangledown}f=\hat{x}\frac{\partial f}{\partial x}+\hat{y}\frac{\partial f}{\partial y}+\hat{z}\frac{\partial f}{\partial z}$ , so we will need to find the partial derivatives of , shown below as

$\begin{align} \frac{\partial f}{\partial x}&= 2(xyz^3-y) \ \frac{\partial f}{\partial y}&= x^2z^3-2x+5z\ \frac{\partial f}{\partial z}&= 3x^2yz^2+5y \end{align}$

Therefore,

$\vec{R}= 2\left( xyz^3-y\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

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Question

Calculate $R = \vec{\bigtriangledown } \cdot \vec{f}$ given $\vec{f}= (x+3y)\hat{x}+(x^2+\sqrt{z})\hat{y}+(xy\sqrt{z})\hat{z}$

Answer

By definition,

$\vec{\bigtriangledown } \cdot \vec{f} = \frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$ , where are the respective components of $\vec{f}$ .

Therefore, we need to calculate the above terms, shown as

$\begin{align} \frac{\partial f_x}{\partial x}&=1 \ \frac{\partial f_y}{\partial y}&= 0\ \frac{\partial f_z}{\partial z}&= \frac{xy}{2\sqrt{z}} \end{align}$

Therefore,

$R = 1 +\frac{xy}{2\sqrt{z}}$ .

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Question

Find the equation of the plane that passes through the points , and

Answer

Step 1:

Let

Using these three points we will find two vectors and . \[You can find PQ and QR too\]

$PQ=(3,5,2)-(2,2,4)=(1,3,-2)\ PR=(1,2,1)-(2,2,4)=(-1,0,-3)$

Step 2:

We are required to find a perpendicular (normal) vector to both and . So we need to take their cross product

$\begin{pmatrix}i &j &k \ 1& 3& 2\ 1& 0& -3\end{pmatrix}=(-9-0)i-(-3-2)j+(0-3)k$

We have found our normal vector.

Step 3: We will use the following formula to find the final answer

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0\ -9(x-1)+5(y-2)-3(z-1)=0\ -9x+9+5y-10-3z+3=0\ -9x+5y-3z+2=0$

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Question

Find the gradient vector for

$\vec{F}=<e^xyz,ye^x,xyz>$

Answer

Suppose that

$\vec{F}=<P,Q,R>$

then

$abla\vec{F}=\left<\frac{\partial{F}}{\partial{x}},\frac{\partial{F}}{\partial{y}},\frac{\partial{F}}{\partial{z}}\right>$

taking the respective partial derivatives and putting them into order as stated in the formula above yields

$abla\vec{F}=<e^xyz,e^x,xy>$

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Question

Find , where $f(x,y,z)=xzy+\tan(xz)$

Answer

The gradient vector of f, , is equal to $\left \langle f_x, f_y, f_z\right \rangle$ .

So, we must find the partial derivatives of the function with respect to x, y, and z, keeping the other variables constant for each partial derivative:

$f_x=yz+z\sec^2(xz)$

$f_z=xy+x\sec^2(xz)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

Plugging this in to a vector, we get

$\left \langle yz+z\sec^2(xz), xz, xy+x\sec^2(xz)\right \rangle$

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