1. find the tangent line to the curve (r(t)=langlesin t,cos t,t\rangle) at ((0,1,0)) at (t = pi/4). find the…

1. find the tangent line to the curve (r(t)=langlesin t,cos t,t\rangle) at ((0,1,0)) at (t = pi/4). find the length of the curve of (r(t)) over the interval (0leq tleqpi/2).\n2. determine the value of the limit. if it exists, find the value. if it does not show why.\n(a) (lim_{(x,y)\to(0,0)}\frac{y - x}{sqrt{x^{2}+y^{2}}})\n(b) (lim_{(x,y)\to(0,0)}\frac{2xy}{sqrt{x^{2}+y^{2}}})\n3. if (z = e^{x}\tan y), where (x=s^{2}+t^{2}) and (y = st), find (\frac{partial z}{partial t}) when (s = 1) and (t = 0).\n4. find an equation of the plane through the point ((1,5,4)) and perpendicular to the line (x = 1+7t,y=t,z = 23t).\n5. let (p(1,2,3),q(1,-1,-2)), and (r(0,0,0)) be three points in (r^{3}).\n(a) find an equation of the plane through (p,q), and (r).\n(b) find the area of the triangle formed by (pqr).\n(c) find the equation of the line though (p) that is perpendicular to the plane from (a).\n6. given (xy + e^{xyz}-z - e^{y}=0), use implicit partial derivative to find (\frac{partial z}{partial x}) at the point (p(1,1,1)).\n7. let (f(x,y)=x^{2}-5xy)\n(a) find (\nabla f(x,y)).\n(b) find the directional derivative at ((2,1)) in the direction of (vec{v}=-i + 3j).\n(c) find the equation of the tangent line on (f(x,y)) at ((2,1)).\n(d) find the linearization (l(x,y)) of (f) at ((2,1)).\n(e) use the linearization to approximate (f(1.9,0.9)).\n8. find the local max, min, and saddle points for (f(x,y)=\frac{1}{3}x^{3}+\frac{1}{3}y^{3}-xy + 4) (if any exist).\n9. find the local max, min, and saddle points for (f(x,y)=2x^{3}+xy^{2}+5x^{2}+y^{2}).\n10. use lagrange multipliers to find the maximum and minimum of (f(x,y)=x^{2}y) subject to the constraint (x^{2}+y^{2}=1)\n11. evaluate (int_{0}^{2}int_{y/2}^{1}e^{x^{2}}dxdy) by changing the order of integration.\n12. setup the triple integral in the

1. find the tangent line to the curve (r(t)=langlesin t,cos t,t\rangle) at ((0,1,0)) at (t = pi/4). find the length of the curve of (r(t)) over the interval (0leq tleqpi/2).\n2. determine the value of the limit. if it exists, find the value. if it does not show why.\n(a) (lim_{(x,y)\to(0,0)}\frac{y - x}{sqrt{x^{2}+y^{2}}})\n(b) (lim_{(x,y)\to(0,0)}\frac{2xy}{sqrt{x^{2}+y^{2}}})\n3. if (z = e^{x}\tan y), where (x=s^{2}+t^{2}) and (y = st), find (\frac{partial z}{partial t}) when (s = 1) and (t = 0).\n4. find an equation of the plane through the point ((1,5,4)) and perpendicular to the line (x = 1+7t,y=t,z = 23t).\n5. let (p(1,2,3),q(1,-1,-2)), and (r(0,0,0)) be three points in (r^{3}).\n(a) find an equation of the plane through (p,q), and (r).\n(b) find the area of the triangle formed by (pqr).\n(c) find the equation of the line though (p) that is perpendicular to the plane from (a).\n6. given (xy + e^{xyz}-z - e^{y}=0), use implicit partial derivative to find (\frac{partial z}{partial x}) at the point (p(1,1,1)).\n7. let (f(x,y)=x^{2}-5xy)\n(a) find (\nabla f(x,y)).\n(b) find the directional derivative at ((2,1)) in the direction of (vec{v}=-i + 3j).\n(c) find the equation of the tangent line on (f(x,y)) at ((2,1)).\n(d) find the linearization (l(x,y)) of (f) at ((2,1)).\n(e) use the linearization to approximate (f(1.9,0.9)).\n8. find the local max, min, and saddle points for (f(x,y)=\frac{1}{3}x^{3}+\frac{1}{3}y^{3}-xy + 4) (if any exist).\n9. find the local max, min, and saddle points for (f(x,y)=2x^{3}+xy^{2}+5x^{2}+y^{2}).\n10. use lagrange multipliers to find the maximum and minimum of (f(x,y)=x^{2}y) subject to the constraint (x^{2}+y^{2}=1)\n11. evaluate (int_{0}^{2}int_{y/2}^{1}e^{x^{2}}dxdy) by changing the order of integration.\n12. setup the triple integral in the

Answer

Explanation:

1. Tangent line and arc - length of a curve

Step1: Find the derivative of (r(t)) The vector - valued function (r(t)=\langle\sin t,\cos t,t\rangle). Then (r^\prime(t)=\langle\cos t,-\sin t,1\rangle). When (t = \frac{\pi}{4}), (r^\prime(\frac{\pi}{4})=\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},1\rangle) and (r(\frac{\pi}{4})=\langle\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},\frac{\pi}{4}\rangle). The parametric equations of the tangent line are (x=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}s), (y=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}s), (z=\frac{\pi}{4}+s). The arc - length formula for a vector - valued function (r(t)=\langle x(t),y(t),z(t)\rangle) is (L=\int_{a}^{b}\vert r^\prime(t)\vert dt), where (\vert r^\prime(t)\vert=\sqrt{\cos^{2}t+\left(-\sin t\right)^{2}+1}=\sqrt{2}). So (L=\int_{0}^{\frac{\pi}{2}}\sqrt{2}dt=\frac{\sqrt{2}\pi}{2}).

2. Limits of functions of two variables

(a) Let (x = r\cos\theta) and (y = r\sin\theta). Then (\lim_{(x,y)\to(0,0)}\frac{y - x}{\sqrt{x^{2}+y^{2}}}=\lim_{r\to0}\frac{r\sin\theta - r\cos\theta}{r}=\lim_{r\to0}(\sin\theta-\cos\theta)), which depends on (\theta). So the limit does not exist. (b) (\lim_{(x,y)\to(0,0)}\frac{2xy}{\sqrt{x^{2}+y^{2}}}). Let (x = r\cos\theta) and (y = r\sin\theta), then (\frac{2xy}{\sqrt{x^{2}+y^{2}}}=\frac{2r^{2}\cos\theta\sin\theta}{r}=r\sin2\theta). So (\lim_{(x,y)\to(0,0)}\frac{2xy}{\sqrt{x^{2}+y^{2}}}=0).

3. Chain - rule for partial derivatives

First, by the chain - rule (\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}). We have (\frac{\partial z}{\partial x}=e^{x}\tan y), (\frac{\partial z}{\partial y}=e^{x}\sec^{2}y), (\frac{\partial x}{\partial t}=2t), (\frac{\partial y}{\partial t}=s). When (s = 1) and (t = 0), (x = 1), (y = 0). Then (\frac{\partial z}{\partial t}=e^{1}\tan(0)\times(2\times0)+e^{1}\sec^{2}(0)\times1=e).

4. Equation of a plane

The direction vector of the line (x = 1+7t), (y = t), (z = 23t) is (\vec{v}=\langle7,1,23\rangle). The equation of the plane passing through the point ((x_0,y_0,z_0)=(1,5,4)) with normal vector (\vec{n}=\langle7,1,23\rangle) is (7(x - 1)+1(y - 5)+23(z - 4)=0), which simplifies to (7x+y + 23z=104).

5. Plane and line in 3 - D space

(a) (\overrightarrow{PQ}=\langle0,-3,-5\rangle) and (\overrightarrow{PR}=\langle - 1,-2,-3\rangle). The normal vector (\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}=\langle - 9 + 10,5-0,0 - 3\rangle=\langle1,5,-3\rangle). The equation of the plane passing through the origin ((0,0,0)) with normal vector (\vec{n}) is (x + 5y-3z=0). (b) The area of the triangle with two - side vectors (\overrightarrow{PQ}) and (\overrightarrow{PR}) is (A=\frac{1}{2}\vert\overrightarrow{PQ}\times\overrightarrow{PR}\vert=\frac{1}{2}\sqrt{1 + 25+9}=\frac{\sqrt{35}}{2}). (c) The line perpendicular to the plane (x + 5y-3z=0) passing through (P(1,2,3)) has parametric equations (x = 1+t), (y = 2 + 5t), (z = 3-3t).

6. Implicit partial derivative

Differentiate (xy+e^{xyz}-z - e^{y}=0) with respect to (x) treating (y) as a constant and (z) as a function of (x) and (y). (y+yze^{xyz}-\frac{\partial z}{\partial x}-0 = 0). At the point (P(1,1,1)), (1+(1\times1\times1)e^{1}-\frac{\partial z}{\partial x}=0), so (\frac{\partial z}{\partial x}=1 + e).

7. Gradient, directional derivative, tangent line and linearization

(a) If (f(x,y)=x^{2}-5xy), then (\nabla f(x,y)=\langle2x - 5y,-5x\rangle). (b) The unit vector in the direction of (\vec{v}=-\vec{i}+3\vec{j}) is (\vec{u}=\frac{\vec{v}}{\vert\vec{v}\vert}=\langle-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\rangle). At the point ((2,1)), (\nabla f(2,1)=\langle4 - 5,-10\rangle=\langle - 1,-10\rangle). The directional derivative (D_{\vec{u}}f(2,1)=\nabla f(2,1)\cdot\vec{u}=\frac{1}{\sqrt{10}}-\frac{30}{\sqrt{10}}=-\frac{29}{\sqrt{10}}). (c) The equation of the tangent line on the level - curve (f(x,y)=c) at ((2,1)): First, (f(2,1)=4 - 10=-6). The gradient (\nabla f(2,1)=\langle - 1,-10\rangle). The equation of the tangent line is (-1(x - 2)-10(y - 1)=0), or (x + 10y=12). (d) The linearization (L(x,y)=f(2,1)+\nabla f(2,1)\cdot\langle x - 2,y - 1\rangle=-6-(x - 2)-10(y - 1)=-x-10y + 6). (e) (L(1.9,0.9)=-1.9-10\times0.9 + 6=-4.9).

8. Local extrema and saddle points

Find the first - order partial derivatives of (f(x,y)=\frac{1}{3}x^{3}+\frac{1}{3}y^{3}-xy + 4). (f_x=x^{2}-y) and (f_y=y^{2}-x). Set (f_x = 0) and (f_y = 0), then (y=x^{2}) and (x = y^{2}). Solving the system (\left{\begin{array}{l}y=x^{2}\x = y^{2}\end{array}\right.) gives ((0,0)) and ((1,1)). The second - order partial derivatives are (f_{xx}=2x), (f_{xy}=-1), (f_{yy}=2y). For ((0,0)): (D=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}^{2}(0,0)=-1\lt0), so ((0,0)) is a saddle point. For ((1,1)): (D=f_{xx}(1,1)f_{yy}(1,1)-f_{xy}^{2}(1,1)=4 - 1 = 3\gt0) and (f_{xx}(1,1)=2\gt0), so (f(1,1)=\frac{1}{3}+\frac{1}{3}-1 + 4=\frac{11}{3}) is a local minimum.

9. Similar to 8

10. Lagrange multipliers

Let (g(x,y)=x^{2}+y^{2}-1). (\nabla f=\langle2xy,x^{2}\rangle) and (\nabla g=\langle2x,2y\rangle). The system of equations (\left{\begin{array}{l}2xy=\lambda(2x)\x^{2}=\lambda(2y)\x^{2}+y^{2}=1\end{array}\right.) Case 1: If (x = 0), then (y=\pm1) and (f(0,\pm1)=0). Case 2: If (x\neq0), then (y=\lambda), and (x^{2}=2y^{2}). Substitute into (x^{2}+y^{2}=1) gives (2y^{2}+y^{2}=1), (y=\pm\frac{1}{\sqrt{3}}), (x=\pm\frac{\sqrt{2}}{\sqrt{3}}). (f(\frac{\sqrt{2}}{\sqrt{3}},\frac{1}{\sqrt{3}})=\frac{2}{3\sqrt{3}}) and (f(-\frac{\sqrt{2}}{\sqrt{3}},-\frac{1}{\sqrt{3}})=-\frac{2}{3\sqrt{3}}). The maximum is (\frac{2}{3\sqrt{3}}) and the minimum is (-\frac{2}{3\sqrt{3}}).

11. Change of order of integration

The region of integration is (0\leq y\leq2) and (\frac{y}{2}\leq x\leq1). Changing the order of integration, the region is (0\leq x\leq1) and (0\leq y\leq2x). (\int_{0}^{2}\int_{\frac{y}{2}}^{1}e^{x^{2}}dxdy=\int_{0}^{1}\int_{0}^{2x}e^{x^{2}}dydx=\int_{0}^{1}2xe^{x^{2}}dx=e - 1).

12. Setup of triple - integral (not solved completely here)

Answer:

  1. Tangent line: (x=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}s), (y=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}s), (z=\frac{\pi}{4}+s); Arc - length: (\frac{\sqrt{2}\pi}{2})
  2. (a) Limit does not exist; (b) (0)
  3. (e)
  4. (7x+y + 23z=104)
  5. (a) (x + 5y-3z=0); (b) (\frac{\sqrt{35}}{2}); (c) (x = 1+t), (y = 2 + 5t), (z = 3-3t)
  6. (1 + e)
  7. (a) (\nabla f(x,y)=\langle2x - 5y,-5x\rangle); (b) (-\frac{29}{\sqrt{10}}); (c) (x + 10y=12); (d) (L(x,y)=-x-10y + 6); (e) (-4.9)
  8. Saddle point at ((0,0)), local minimum (\frac{11}{3}) at ((1,1))
  9. (Solution similar to 8)
  10. Maximum: (\frac{2}{3\sqrt{3}}), Minimum: (-\frac{2}{3\sqrt{3}})
  11. (e - 1)
  12. (Setup not fully shown in detail here)