Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #3581 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3cos{(z^2)}ln{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-4,0,-14).\end{align*}\)

Possible Answers:

\(\displaystyle {\frac{{(29.1y^3cos{(z^2)})}}{x} + 43.7y^2cos{(z^2)}ln{(x^2)} - 29.1y^3zln{(x^2)}sin{(z^2)}}\)

\(\displaystyle {28y^3zln{(x^2)}sin{(z^2)} -\frac{ {(8y^3cos{(z^2)})}}{x}}\)

\(\displaystyle {\frac{-{(175y^2zsin{(z^2)})}}{x}}\)

\(\displaystyle {1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}}\)

Correct answer:

\(\displaystyle {1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(\frac{{(2y^3cos{(z^2)})}}{x})+(0)(3y^2cos{(z^2)}ln{(x^2)})+(-0.96)(-2y^3zln{(x^2)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}\end{align*}\)

Example Question #3582 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^2)}cos{(z)})}}{y^2}\\&\text{In the direction of }\overrightarrow{v}=(0,6,-11).\end{align*}\)

Possible Answers:

\(\displaystyle {-\frac{ {(25.1cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(12.5cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(25.1xsin{(x^2)}cos{(z)})}}{y^2}}\)

\(\displaystyle {\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}}\)

\(\displaystyle {\frac{-{(50.1xsin{(x^2)}sin{(z)})}}{y^3}}\)

\(\displaystyle {-\frac{ {(0.461cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(0.774cos{(x^2)}sin{(z)})}}{y^2}}\)

Correct answer:

\(\displaystyle {\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(6)^2+(-11)^2}=12.53\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.53}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{12.53}=0.48\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{12.53}=-0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(2xsin{(x^2)}cos{(z)})}}{y^2})+(0.48)(\frac{-{(2cos{(x^2)}cos{(z)})}}{y^3})+(-0.88)(\frac{-{(1cos{(x^2)}sin{(z)})}}{y^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}\end{align*}\)

Example Question #3583 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^2)}x^3z^2\\&\text{In the direction of }\overrightarrow{v}=(15,17,0).\end{align*}\)

Possible Answers:

\(\displaystyle {45\cdot3^{(y^2)}x^2z^2 + 37.4\cdot3^{(y^2)}x^3yz^2}\)

\(\displaystyle {299\cdot3^{(y^2)}x^2yz}\)

\(\displaystyle {45.3\cdot3^{(y^2)}x^3z + 68\cdot3^{(y^2)}x^2z^2 + 49.8\cdot3^{(y^2)}x^3yz^2}\)

\(\displaystyle {1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2}\)

Correct answer:

\(\displaystyle {1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(17)^2+(0)^2}=22.67\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{22.67}=0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{22.67}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{22.67}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.66)(3\cdot3^{(y^2)}x^2z^2)+(0.75)(2.2\cdot3^{(y^2)}x^3yz^2)+(0)(2\cdot3^{(y^2)}x^3z)\\&D_{\overrightarrow{u}}(x,y,z)=1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2\end{align*}\)

Example Question #3584 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(z^4ln{(x^2)})}}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,15,12).\end{align*}\)

Possible Answers:

\(\displaystyle {\frac{{(48z^3ln{(x^2)})}}{y} -\frac{ {(15z^4ln{(x^2)})}}{y^2}}\)

\(\displaystyle {\frac{{(76.8z^3ln{(x^2)})}}{y} -\frac{ {(19.2z^4ln{(x^2)})}}{y^2} +\frac{ {(38.4z^4)}}{{(xy)}}}\)

\(\displaystyle {\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}}\)

\(\displaystyle {\frac{-{(154z^3)}}{{(xy^2)}}}\)

Correct answer:

\(\displaystyle {\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(15)^2+(12)^2}=19.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{19.21}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{19.21}=0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(2z^4)}}{{(xy)}})+(0.78)(\frac{-{(1z^4ln{(x^2)})}}{y^2})+(0.62)(\frac{{(4z^3ln{(x^2)})}}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}\end{align*}\)

Example Question #3585 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3ln{(z^2)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(6,-10,0).\end{align*}\)

Possible Answers:

\(\displaystyle {1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}}\)

\(\displaystyle {0.78x^2ln{(z^2)}sin{(y^3)} + 2.22x^3y^2cos{(y^3)}ln{(z^2)}}\)

\(\displaystyle {\frac{{(210x^2y^2cos{(y^3)})}}{z}}\)

\(\displaystyle {\frac{{(23.3x^3sin{(y^3)})}}{z} + 35x^2ln{(z^2)}sin{(y^3)} + 35x^3y^2cos{(y^3)}ln{(z^2)}}\)

Correct answer:

\(\displaystyle {1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-10)^2+(0)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{11.66}=0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{11.66}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.51)(3x^2ln{(z^2)}sin{(y^3)})+(-0.86)(3x^3y^2cos{(y^3)}ln{(z^2)})+(0)(\frac{{(2x^3sin{(y^3)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}\end{align*}\)

Example Question #3586 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^3ze^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-7,0,7).\end{align*}\)

Possible Answers:

\(\displaystyle {21x^2ze^{(y)} - 7x^3e^{(y)}}\)

\(\displaystyle {2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}\)

\(\displaystyle {-29.7x^2e^{(y)}}\)

\(\displaystyle {- 9.9x^3e^{(y)} - 29.7x^2ze^{(y)} - 9.9x^3ze^{(y)}}\)

Correct answer:

\(\displaystyle {2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(0)^2+(7)^2}=9.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{9.9}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{9.9}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{9.9}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-3x^2ze^{(y)})+(0)(-1x^3ze^{(y)})+(0.71)(-1x^3e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.13x^2ze^{(y)} - 0.71x^3e^{(y)}\end{align*}\)

Example Question #3587 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{cos{(x)}}{{(y^2z)}}\\&\text{In the direction of }\overrightarrow{v}=(0,9,15).\end{align*}\)

Possible Answers:

\(\displaystyle {-\frac{ {(17.5cos{(x)})}}{{(y^2z^2)}} -\frac{ {(35cos{(x)})}}{{(y^3z)}} -\frac{ {(17.5sin{(x)})}}{{(y^2z)}}}\)

\(\displaystyle {-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}\)

\(\displaystyle {\frac{-{(35sin{(x)})}}{{(y^3z^2)}}}\)

\(\displaystyle {-\frac{ {(0.74cos{(x)})}}{{(y^2z^2)}} -\frac{ {(0.52cos{(x)})}}{{(y^3z)}}}\)

Correct answer:

\(\displaystyle {-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(15)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{17.49}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(1sin{(x)})}}{{(y^2z)}})+(0.51)(\frac{-{(2cos{(x)})}}{{(y^3z)}})+(0.86)(\frac{-{(1cos{(x)})}}{{(y^2z^2)}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}\end{align*}\)

Example Question #3588 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3z^3sin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(20,12,0).\end{align*}\)

Possible Answers:

\(\displaystyle {20y^3z^3cos{(x)} + 36y^2z^3sin{(x)}}\)

\(\displaystyle {23.3y^3z^3cos{(x)} + 70y^2z^3sin{(x)} + 70y^3z^2sin{(x)}}\)

\(\displaystyle {210y^2z^2cos{(x)}}\)

\(\displaystyle {0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}\)

Correct answer:

\(\displaystyle {0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(12)^2+(0)^2}=23.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{23.32}=0.86\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{23.32}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.86)(y^3z^3cos{(x)})+(0.51)(3y^2z^3sin{(x)})+(0)(3y^3z^2sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}\end{align*}\)

Example Question #3589 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-zcos{(x^2)}ln{(y)}\\&\text{In the direction of }\overrightarrow{v}=(2,-13,0).\end{align*}\)

Possible Answers:

\(\displaystyle {26.3xzsin{(x^2)}ln{(y)} -\frac{ {(13.1zcos{(x^2)})}}{y} - 13.1cos{(x^2)}ln{(y)}}\)

\(\displaystyle {\frac{{(26.3xsin{(x^2)})}}{y}}\)

\(\displaystyle {\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}}\)

\(\displaystyle {0.045xzsin{(x^2)}ln{(y)} -\frac{ {(0.98zcos{(x^2)})}}{y}}\)

Correct answer:

\(\displaystyle {\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-13)^2+(0)^2}=13.15\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{13.15}=0.15\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{13.15}=-0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{13.15}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.15)(2xzsin{(x^2)}ln{(y)})+(-0.99)(\frac{-{(1zcos{(x^2)})}}{y})+(0)(-1cos{(x^2)}ln{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}\end{align*}\)

Example Question #3590 : Calculus 3

\(\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}ln{(z)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(5,0,10).\end{align*}\)

Possible Answers:

\(\displaystyle {1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}}\)

\(\displaystyle {0.607x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.792cos{(x^3)}sin{(y)})}}{z}}\)

\(\displaystyle {33.5x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(11.2cos{(x^3)}sin{(y)})}}{z} - 11.2cos{(x^3)}cos{(y)}ln{(z)}}\)

\(\displaystyle {\frac{{(33.5x^2sin{(x^3)}cos{(y)})}}{z}}\)

Correct answer:

\(\displaystyle {1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(0)^2+(10)^2}=11.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{11.18}=0.45\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.18}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{11.18}=0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.45)(3x^2sin{(x^3)}ln{(z)}sin{(y)})+(0)(-1cos{(x^3)}cos{(y)}ln{(z)})+(0.89)(\frac{-{(1cos{(x^3)}sin{(y)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}\end{align*}\)

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