Partial Derivatives

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AP Calculus BC › Partial Derivatives

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$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow1-}\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}\end{align*}$

$\frac{(25\cdot \pi )}{21}$

CORRECT

$\infty$

$-\infty$

$-\frac{(25\cdot \pi )}{21}$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}\rightarrow\frac{0}{0}\&\frac{50\cdot \pi \cdot (tan(10\cdot \pi x)^{2} + 1)}{\frac{42}{x}}\rightarrow\frac{50\cdot \pi}{42}\&lim_{x\rightarrow1-}\frac{5tan(10\cdot \pi x)}{7ln(x^{6})}=\frac{(25\cdot \pi )}{21}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2cos(z^{3})tan(y) - 3^xsin(z^{2})e^{(y^{2})}\&\text{In the direction of }\overrightarrow{v}=(-4,-18,0).\end{align*}$

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

CORRECT

${36cos(z^{3})\cdot (tan(y)^{2} + 1) + 4.39\cdot 3^xsin(z^{2})e^{(y^{2})} + 36\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${-81\cdot 3^xyzcos(z^{2})e^{(y^{2})}}$

${111z^{2}sin(z^{3})tan(y) - 20.3\cdot 3^xsin(z^{2})e^{(y^{2})} - 36.9cos(z^{3})\cdot (tan(y)^{2} + 1) - 36.9\cdot 3^xzcos(z^{2})e^{(y^{2})} - 36.9\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-18)^2+(0)^2}=18.44\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{18.44}=-0.22\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.44}=-0.98\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.22)(-1.1\cdot 3^xsin(z^{2})e^{(y^{2})})+(-0.98)(- 2cos(z^{3})\cdot (tan(y)^{2} + 1) - 2\cdot 3^xysin(z^{2})e^{(y^{2})})+(0)(6z^{2}sin(z^{3})tan(y) - 2\cdot 3^xzcos(z^{2})e^{(y^{2})})\&D_{\overrightarrow{u}}(x,y,z)=1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}\end{align*}$

Evaluate the following limit:

$\lim_{x\rightarrow 2^{-}}f(x), f(x)=\left\{\begin{matrix} \ln(2-x), x\leq 2\ x-3, x> 2 \end{matrix}\right.$

$-\infty$

CORRECT

$\infty$

Explanation

To evaluate the limit, we must determine whether it is right or left sided. The negative sign "exponent" on the 2 indicates that we are approaching from the left, or numbers slightly less than 2. So, we choose the part of the piecewise function that is for x values less than or equal to 2. Now, evaluating the limit, as natural log approaches zero, we get $-\infty$ .

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{-{(cos{(x^3)}cos{(y^2)})}}{z}\&\text{In the direction of }\overrightarrow{v}=(0,1,15).\end{align*}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

CORRECT

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.0098ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(90.2x^2ysin{(x^3)}sin{(y^2)})}}{z^2}}$

${\frac{{(15cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(30.1ycos{(x^3)}sin{(y^2)})}}{z} +\frac{ {(45.1x^2cos{(y^2)}sin{(x^3)})}}{z}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(15)^2}=15.03\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{15.03}=0.07\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(3x^2cos{(y^2)}sin{(x^3)})}}{z})+(0.07)(\frac{{(2ycos{(x^3)}sin{(y^2)})}}{z})+(1)(\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2})\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}\end{align*}$

Evaluate the limit of the following function:

$\lim_{x\rightarrow 11^+}f(x), f(x)=\left\{\begin{matrix} x+3, x\leq 11\ \ln(x-11), x> 11 \end{matrix}\right$

$-\infty$

CORRECT

$\infty$

Explanation

To evaluate the limit, we must first determine whether the limit is right or left sided; the plus sign "exponent" on indicates that we are evaluating the limit from the right, or using values slightly larger than . The second function of the piecewise function corresponds to these values, and when we evaluate the limit using this function we get $-\infty$ , as when the natural log function approaches zero, it approaches $-\infty$ .