Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1 : Applications Of Partial Derivatives

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

Example Question #1 : Applications Of Partial Derivatives

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

Possible Answers:

\(\displaystyle x+y=1\)

\(\displaystyle -x+z=1\)

\(\displaystyle x-z=2\)

\(\displaystyle y-z=-1\)

\(\displaystyle y-z=0\)

Correct answer:

\(\displaystyle -x+z=1\)

Explanation:

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

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

Taking partial derivatives, we have

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

\(\displaystyle f_y(x,y) = 2x\)

Substituting our \(\displaystyle x, y\) values into these, we get

\(\displaystyle f_x(0,1) =1\)

\(\displaystyle f_y(0,1) = 0\)

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

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

\(\displaystyle z - 1 = x\)

\(\displaystyle -x+z =1\).

 

Example Question #3 : Tangent Planes And Linear Approximations

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

Possible Answers:

None of the Above

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

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

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

Correct answer:

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

Explanation:

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

Step 1: Find \(\displaystyle f(x,y)\)

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

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

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

Step 3: Evaluate the partial derivative of x at \(\displaystyle -3\)

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

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

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

Step 5: Evaluate the partial derivative at \(\displaystyle y=2\)

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

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

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

Step 7: Write the equation of the tangent line:

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

Example Question #3 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to \(\displaystyle z=2x^2+4y^3\) at the point \(\displaystyle (1,5)\).

Possible Answers:

\(\displaystyle z=4x+300y-1002\)

\(\displaystyle z=x+30y-10\)

\(\displaystyle z=2x+300y-100\)

\(\displaystyle z=4x^2+3y^3-1002\)

Correct answer:

\(\displaystyle z=4x+300y-1002\)

Explanation:

To find the equation of the tangent plane, we find: \(\displaystyle f_x,f_y,z_0\) and evaluate \(\displaystyle f_x, f_y\) at the point given. \(\displaystyle f_x=\frac{\partial }{\partial x}(2x^2+4y^3)=4x\)\(\displaystyle f_y=\frac{\partial }{\partial y}(2x^2+4y^3)=12y^2\), and \(\displaystyle z_0=2(1^2)+4(5^3)=502\). Evaluating \(\displaystyle f_x, f_y\) at the point \(\displaystyle (1,5)\) gets us \(\displaystyle f_x=4, f_y=300\). We then plug these values into the formula for the tangent plane: \(\displaystyle z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\). We then get \(\displaystyle z-502=4(x-1)+300(y-5)\). The equation of the plane then becomes, through algebra, \(\displaystyle z=4x+300y-1002\)

Example Question #5 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to \(\displaystyle z=3\sin(x)+2\cos(y)\) at the point \(\displaystyle (\frac{\pi}{2},\frac{\pi}{6})\)

Possible Answers:

\(\displaystyle z=y-\frac{\pi}{6}\)

\(\displaystyle z=-y+\frac{\pi}{6}+3+\sqrt3\)

\(\displaystyle z=x-\frac{\pi}{3}+3+\sqrt3\)

\(\displaystyle z=y-\frac{\pi}{3}+2+\sqrt3\)

Correct answer:

\(\displaystyle z=-y+\frac{\pi}{6}+3+\sqrt3\)

Explanation:

To find the equation of the tangent plane, we find: \(\displaystyle f_x,f_y,z_0\) and evaluate \(\displaystyle f_x, f_y\) at the point given. \(\displaystyle f_x=\frac{\partial }{\partial x}(3\sin(x)+2(\cos(y))=3\cos(x)\)\(\displaystyle f_y=\frac{\partial }{\partial y}(3\sin(x)+2\cos(y))=-2\sin(y)\), and \(\displaystyle z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3\). Evaluating \(\displaystyle f_x, f_y\) at the point \(\displaystyle (\frac{\pi}{2},\frac{\pi}{6})\) gets us \(\displaystyle f_x=0, f_y=-1\). We then plug these values into the formula for the tangent plane: \(\displaystyle z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\). We then get \(\displaystyle z-(3+\sqrt3)=0(x-\frac{\pi}{2})+-1(y-\frac{\pi}{6})\). The equation of the plane then becomes, through algebra, \(\displaystyle z=-y+\frac{\pi}{6}+3+\sqrt3\)

Example Question #6 : Tangent Planes And Linear Approximations

Find the equation of the tangent plane to \(\displaystyle z=3x^2y\) at the point \(\displaystyle (2,4)\)

Possible Answers:

\(\displaystyle z=x+36y\)

\(\displaystyle z=48x+12y-96\)

\(\displaystyle z=48x-12\)

\(\displaystyle z=5x+50y\)

Correct answer:

\(\displaystyle z=48x+12y-96\)

Explanation:

To find the equation of the tangent plane, we need 5 things:\(\displaystyle f_x,f_y,z(x,y),f_x(x,y),f_y(x,y)\)

\(\displaystyle f_x=\frac{\partial }{\partial x}(3x^2y)=6xy\)

\(\displaystyle f_x(2,4)=6*2*4=48\)

\(\displaystyle f_y=\frac{\partial }{\partial y}(3x^2y)=3x^2\)

\(\displaystyle f_y(2,4)=3(2)^2=12\)

\(\displaystyle z(2,4)=3(2)^2(4)=48\)

Using the equation of the tangent plane

\(\displaystyle z-z(x,y)=f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0)\), we get

\(\displaystyle z-48=48(x-2)+12(y-4)\)

Through algebraic manipulation to get z by itself, we get

\(\displaystyle z=48x+12y-96\)

Example Question #1 : Applications Of Partial Derivatives

Find the absolute minimums and maximums of \(\displaystyle f(x,y)=10x^2-3y^2+10y\) on the disk of radius \(\displaystyle 4\)\(\displaystyle x^2+y^2\leq16\).

Possible Answers:

Absolute Minimum: \(\displaystyle (0,-4)\)

Absolute Maximum:  \(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Absolute Minimum: \(\displaystyle (0,4)\)

Absolute Maximum:\(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Absolute Minimum: \(\displaystyle (0,-4)\)

Absolute Maximum: \(\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Absolute Minimum: \(\displaystyle (0,-4)\)

Absolute Maximum: \(\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)\(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Absolute Minimum: \(\displaystyle (0,4)\)

Absolute Maximum: \(\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)\(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Correct answer:

Absolute Minimum: \(\displaystyle (0,-4)\)

Absolute Maximum: \(\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)\(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\)

Explanation:

The first thing we need to do is find the partial derivative in respect to \(\displaystyle x\), and \(\displaystyle y\).

\(\displaystyle \frac{\partial}{\partial x}=20x\)\(\displaystyle \frac{\partial}{\partial y}=-6y+10\)

 

We need to find the critical points, so we set each of the partials equal to \(\displaystyle 0\).

\(\displaystyle 0=20x\rightarrow x=0\)

\(\displaystyle 0=-6y+10 \rightarrow y=\frac{5}{3}\)

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

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

This is within our disk.

 

We now need to take a look at the boundary, \(\displaystyle x^2+y^2=16\). We can solve for \(\displaystyle x^2\), and plug it into \(\displaystyle f\).

\(\displaystyle x^2+y^2=16\)

\(\displaystyle x^2=16-y^2\)

\(\displaystyle g(y)=10(16-y^2)-3y^2+10y=-13y^2+10y+160\)

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

\(\displaystyle g'(y)=-26y+10\)

\(\displaystyle 0=-26y+10 \rightarrow y=\frac{5}{13}\)

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

\(\displaystyle g(-4)=-13(-4)^2+10(-4)+160=-88\)

\(\displaystyle g(4)=-13(4)^2+10(4)+160=-8\)

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

Now we need to figure out the values of \(\displaystyle x\) these correspond to.

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

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

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

Now lets summarize our results as follows:

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

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

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

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

 

From this we can conclude that there is an absolute minimum at \(\displaystyle (0,-4)\), and two absolute maximums at \(\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\) and \(\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)\).

Example Question #1 : Lagrange Multipliers

Find the minimum and maximum of \(\displaystyle f(x,y)=2x-5y\), subject to the constraint \(\displaystyle x^2+y^2=144\).

Possible Answers:

\(\displaystyle f(4.46,11.14)\) is a maximum

\(\displaystyle f(-4.46,-11.14)\) is a minimum

\(\displaystyle f(4.46,11.14)\) is a maximum

\(\displaystyle f(4.46,-11.14)\) is a minimum

\(\displaystyle f(-4.46,11.14)\) is a maximum

\(\displaystyle f(4.46,-11.14)\) is a minimum

 

\(\displaystyle f(4.46,-11.14)\) is a maximum

\(\displaystyle f(-4.46,11.14)\) is a minimum

There are no maximums or minimums

Correct answer:

\(\displaystyle f(4.46,-11.14)\) is a maximum

\(\displaystyle f(-4.46,11.14)\) is a minimum

Explanation:

First we need to set up our system of equations.

\(\displaystyle 2=2\lambda x \rightarrow x=\frac{1}{\lambda}\)

\(\displaystyle -5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}\)

\(\displaystyle x^2+y^2=144\)

Now lets plug in these constraints.

\(\displaystyle (\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144\)

 \(\displaystyle \frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144\)

\(\displaystyle \frac{29}{4\lambda^2}=144\)

Now we solve for \(\displaystyle \lambda\)

\(\displaystyle \frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}\)

If

 \(\displaystyle \lambda=\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46\)\(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14\)

 

If

 \(\displaystyle \lambda=-\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46\)\(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14\)

 

Now lets plug in these values of \(\displaystyle x\), and \(\displaystyle y\) into the original equation.

\(\displaystyle f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62\)

\(\displaystyle f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62\)

 

We can conclude from this that \(\displaystyle f(4.46,-11.14)\) is a maximum, and \(\displaystyle f(-4.46,11.14)\) is a minimum.

Example Question #2 : Lagrange Multipliers

Find the absolute minimum value of the function \(\displaystyle f(x,y) = x^2+y^2\) subject to the constraint \(\displaystyle x^2 +2y^2 = 1\).

Possible Answers:

\(\displaystyle \sqrt2\)

\(\displaystyle 0\)

\(\displaystyle 2\sqrt2\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle \frac{1}{2}\)

Explanation:

Let \(\displaystyle g = x^2 +2y^2\)To find the absolute minimum value, we must solve the system of equations given by

\(\displaystyle \triangledown f = \lambda\triangledown g, g =1\).

So this system of equations is

\(\displaystyle f_x = \lambda g_x\), \(\displaystyle f_y = \lambda g_y\), \(\displaystyle g =1\).

Taking partial derivatives and substituting as indicated, this becomes

\(\displaystyle 2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1\).

From the left equation, we see either \(\displaystyle x=0\) or \(\displaystyle \lambda = 1\). If \(\displaystyle x=0\), then substituting this into the other equations, we can solve for \(\displaystyle y, \lambda\), and get \(\displaystyle y = \pm \sqrt{2}/2\), \(\displaystyle \lambda = 1/2\), giving two extreme candidate points at \(\displaystyle (0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})\).

On the other hand, if instead \(\displaystyle \lambda =1\), this forces \(\displaystyle y = 0\) from the 2nd equation, and \(\displaystyle x = \pm 1\) from the 3rd equation. This gives us two more extreme candidate points; \(\displaystyle (-1,0),(1,0)\).

 

Taking all four of our found points, and plugging them back into \(\displaystyle f\), we have

\(\displaystyle f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}\).

Hence the absolute minimum value is \(\displaystyle \frac{1}{2}\).

 

Example Question #3 : Lagrange Multipliers

Find the dimensions of a box with maximum volume such that the sum of its edges is \(\displaystyle 60\) cm.

Possible Answers:

\(\displaystyle 6 \times 5 \times 4\)

\(\displaystyle 6 \times 6 \times 6\)

\(\displaystyle 5 \times 4 \times 3\)

\(\displaystyle 5 \times 5 \times 5\)

Correct answer:

\(\displaystyle 5 \times 5 \times 5\)

Explanation:

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