Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

varsity tutors app store varsity tutors android store

Example Questions

Example Question #382 : Partial Derivatives

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

 

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Keep in mind, when taking the derivative with respect to \(\displaystyle x\)\(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\)\(\displaystyle x\) is treated as a constant.

 

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

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

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

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

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

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

 

Example Question #381 : Partial Derivatives

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

Possible Answers:

\(\displaystyle 25t^4 -150t^3 +268t^2 +508t +83\)

\(\displaystyle 5t^4 +240t^3 -178t^2 -48t +80\)

\(\displaystyle 15t^4 -140t^3 +168t^2 +408t +80\)

\(\displaystyle 15t^4 -10t^3 +68t^2 +48t +8\)

Correct answer:

\(\displaystyle 15t^4 -140t^3 +168t^2 +408t +80\)

Explanation:

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

Keep in mind, when taking the derivative with respect to \(\displaystyle x\)\(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\)\(\displaystyle x\) is treated as a constant.

 

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

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

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

 

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

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

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

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

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

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

 

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

Example Question #384 : Partial Derivatives

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

 

Possible Answers:

\(\displaystyle \frac{27}{4}t^{5/2}e^{5t^{3/2}}\)

\(\displaystyle \frac{25}{2}t^{3/2}e^{5t^{5/2}}\)

\(\displaystyle \frac{25}{4}t^{3/4}e^{5t^{5/4}}\)

\(\displaystyle \frac{35}{9}t^{3}e^{5t^{5}}\)

Correct answer:

\(\displaystyle \frac{25}{2}t^{3/2}e^{5t^{5/2}}\)

Explanation:

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

Keep in mind, when taking the derivative with respect to \(\displaystyle x\)\(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\)\(\displaystyle x\) is treated as a constant.

 

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

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

 

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

 

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

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

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

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

\(\displaystyle \frac{dz}{dt}=\frac{25}{2}t^{3/2}e^{5t^{5/2}}\)

Example Question #31 : Multi Variable Chain Rule

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

 

Possible Answers:

\(\displaystyle 55t^2-10t+8-2t^3\)

\(\displaystyle 525t^2-140t+7-2/t^3\)

\(\displaystyle -52t^5-40t^4+7t^3-2t^2\)

\(\displaystyle 425t^2-240t-7+1/t^3\)

Correct answer:

\(\displaystyle 525t^2-140t+7-2/t^3\)

Explanation:

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

Keep in mind, when taking the derivative with respect to \(\displaystyle x\)\(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\)\(\displaystyle x\) is treated as a constant.

 

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

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

 

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

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

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

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

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

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

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

Example Question #32 : Multi Variable Chain Rule

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

 

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

 

Keep in mind, when taking the derivative with respect to \(\displaystyle x\)\(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\)\(\displaystyle x\) is treated as a constant.

 

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

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

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

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

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

 

Example Question #33 : Multi Variable Chain Rule

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

 

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

Possible Answers:

\(\displaystyle t(e^{t^2}-1)\)

\(\displaystyle 4t^2e^{t^2}+1\)

\(\displaystyle 6te^{t^2}(1+t^2)\)

\(\displaystyle 12t^2e^{t^2}(1+2t)\)

\(\displaystyle e^{t^2}(1-t)\)

Correct answer:

\(\displaystyle 6te^{t^2}(1+t^2)\)

Explanation:

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

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

Because each function is given in terms of the parameter \(\displaystyle t\) we can conveniently write the function as a function of \(\displaystyle t\) alone: 

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

 

Solution 1

For the function described in this problem, the function can be written strictly in terms of the parameter \(\displaystyle t\) by simply substituting the definitions given for \(\displaystyle x\)\(\displaystyle y,\) or \(\displaystyle z\)

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

 \(\displaystyle f(t)=3t^2e^{t^2}\)                                                         

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

 

Apply the product rule and and the chain-rule:  

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

 \(\displaystyle =6te^{t^2}+6t^3e^{t^2}\)                                                        

 \(\displaystyle =6te^{t^2}(1+t^2)\)

 

Solution 2

The derivative of \(\displaystyle f\) with respect to \(\displaystyle t\) will be the sum of the derivatives of each variable \(\displaystyle x\)\(\displaystyle y\), and \(\displaystyle z\) with respect to \(\displaystyle t.\)

 

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

 

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

 

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

 

 

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

 

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

 

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

 

Now ad the terms to get an expression for \(\displaystyle \frac{df}{dt}\) and then rewrite in terms of \(\displaystyle t.\)

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

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

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

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

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

 

Therefore: 

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

 

 

 

Example Question #34 : Multi Variable Chain Rule

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

 

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

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

\(\displaystyle f(1,0)=2\)

 

 

 

 

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

\(\displaystyle f(1,0)=2\)

 

The first step is to integrate the function with respect to \(\displaystyle y\) since that is the outer most derivative in this problem. 

 

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

 

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

 

 

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

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

So far we have: 

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

Now apply the second constraint given: 

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

 

 

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

 

Example Question #1 : Tangent Planes And Linear Approximations

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

Possible Answers:

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

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

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

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

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

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 : Tangent Planes And Linear Approximations

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

Possible Answers:

\(\displaystyle x-z=2\)

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

\(\displaystyle y-z=0\)

\(\displaystyle x+y=1\)

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

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 : Applications Of Partial Derivatives

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

Possible Answers:

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

None of the Above

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

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

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:

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

Learning Tools by Varsity Tutors