CALCU-The figure below shows a point P(x_0, y_0) o

Solution by Steps

step 1

Given the parabola equation

y^2 = 4px

, differentiate both sides with respect to

x

to find

\frac{dy}{dx}

at the point

(x_0, y_0)

step 2

Differentiating

y^2 = 4px

gives

2y\frac{dy}{dx} = 4p

step 3

Solve for

\frac{dy}{dx}

to get

\frac{dy}{dx} = \frac{4p}{2y} = \frac{2p}{y}

step 4

At the point

(x_0, y_0)

,

\frac{dy}{dx}

becomes

\frac{2p}{y_0}

step 5

Since

\tan(\beta) = \frac{dy}{dx}

at

(x_0, y_0)

, we have

\tan(\beta) = \frac{2p}{y_0}

[question a] Answer

\tan(\beta) = \frac{2p}{y_0}

step 1

To find

\tan(\phi)

, use the coordinates of point

P(x_0, y_0)

and the focus

F(p, 0)

step 2

The slope of the line connecting

F

and

P

is

\frac{y_0 - 0}{x_0 - p} = \frac{y_0}{x_0 - p}

step 3

Since

\tan(\phi)

is the slope of the line connecting

F

and

P

, we have

\tan(\phi) = \frac{y_0}{x_0 - p}

[question b] Answer

\tan(\phi) = \frac{y_0}{x_0 - p}

step 1

Use the identity

\tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)}

to find

\tan(\alpha)

step 2

Substitute

\tan(\beta) = \frac{2p}{y_0}

and

\tan(\phi) = \frac{y_0}{x_0 - p}

into the identity

step 3

This gives

\tan(\alpha - \phi) = \frac{\frac{2p}{y_0} - \frac{y_0}{x_0 - p}}{1 + \frac{2p}{y_0} \cdot \frac{y_0}{x_0 - p}}

step 4

Simplify the expression to find

\tan(\alpha - \phi)

step 5

Since

\phi

is the angle between the line connecting

F

and

P

and the x-axis, and

\alpha

is the angle between the x-axis and line

L'

, which is parallel to the x-axis,

\phi = 0

and

\tan(\phi) = 0

step 6

Therefore,

\tan(\alpha) = \tan(\alpha - 0) = \tan(\alpha - \phi) = \frac{2p}{y_0}

step 7

Since

\tan(\alpha) = \tan(\beta)

, we conclude that

\alpha = \beta

by the uniqueness of the tangent function in the range

(-\frac{\pi}{2}, \frac{\pi}{2})

[question c] Answer

\tan(\alpha) = \frac{2p}{y_0}

and hence

\alpha = \beta

Key Concept

Tangent and slope in the context of a parabola

Explanation

The tangent to a curve at a point is represented by the derivative at that point. For a parabola, the slope of the tangent line can be found by differentiating the equation of the parabola. The reflective property of parabolas is shown by proving that the angles of incidence and reflection are equal, which is done by equating their tangents using the tangent subtraction formula.

Solution by Steps

step 1

To sketch the graph of

z = f(x, y)

, consider the two cases given by the function

step 2

For

xy \neq 0

,

f(x, y) = 0

. This means that for all points except along the x-axis and y-axis, the value of

z

is 0

step 3

For

xy = 0

,

f(x, y) = 1

. This means that along the x-axis and y-axis, the value of

z

is 1

step 4

The graph will show a flat plane at

z = 0

with lines at

z = 1

along the x-axis and y-axis

[question a] Answer

The graph is a flat plane at

z = 0

with spikes to

z = 1

along the x-axis and y-axis.

step 5

To determine if

f

is continuous at

(0,0)

, consider the limit of

f(x, y)

as

(x, y)

approaches

(0,0)

step 6

If the limit exists and equals

f(0,0)

, then

f

is continuous at

(0,0)

step 7

Approaching

(0,0)

along any line

y = mx

where

m

is a nonzero constant, the limit is

0

since

xy \neq 0

step 8

However, approaching

(0,0)

along the x-axis or y-axis, the function value is always

1

step 9

Since the limit depends on the path taken to approach

(0,0)

, the limit does not exist

step 10

Therefore,

f

is not continuous at

(0,0)

[question b] Answer

The function

f

is not continuous at

(0,0)

because the limit as

(x, y)

approaches

(0,0)

does not exist.

step 11

To "guess"

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y}

at

(0,0)

, look at the graph from part (a)

step 12

Along the x-axis,

f(x, 0) = 1

for all

x

, so

\frac{\partial f}{\partial x}

appears to be

0

at

(0,0)

step 13

Along the y-axis,

f(0, y) = 1

for all

y

, so

\frac{\partial f}{\partial y}

appears to be

0

at

(0,0)

step 14

To verify the guess for

\frac{\partial f}{\partial x}

, use the definition of the partial derivative

step 15

The partial derivative with respect to

x

is the limit of

\frac{f(x,0) - f(0,0)}{x - 0}

as

x

approaches

0

step 16

Since

f(x,0) = 1

and

f(0,0) = 1

, the limit is

\frac{1 - 1}{x} = 0

as

x

approaches

0

step 17

Therefore, the partial derivative

\frac{\partial f}{\partial x}

at

(0,0)

is

0

[question c] Answer

The estimated partial derivatives

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y}

at

(0,0)

are both

0

, and the estimation for

\frac{\partial f}{\partial x}

is verified to be correct.

step 18

To draw conclusions about the existence of partial derivatives and continuity, consider the results from parts (a), (b), and (c)

step 19

The partial derivatives at

(0,0)

exist and are both

0

step 20

However, the function is not continuous at

(0,0)

step 21

The existence of partial derivatives at a point does not imply continuity at that point

[question d] Answer

Partial derivatives may exist at a point where a function is not continuous. The existence of partial derivatives does not guarantee continuity.

Key Concept

Continuity and Partial Derivatives

Explanation

A function can have partial derivatives at a point even if it is not continuous at that point. The existence of partial derivatives alone does not ensure continuity.

Solution by Steps

step 1

To show that

(0,0,1)

is on the surface, we substitute

x=0

,

y=0

, and

z=1

into the given surface equation

x z^{2}-y z+\cos (x y)=1

step 2

Substituting the values, we get

0 \cdot 1^{2} - 0 \cdot 1 + \cos (0 \cdot 0) = 1

step 3

Simplifying the equation, we have

0 + 0 + \cos (0) = 1

. Since

\cos (0) = 1

, the equation simplifies to

1 = 1

[question a] Answer

The point

(0,0,1)

is indeed on the surface as the equation holds true when the values are substituted.

Key Concept

Substitution of coordinates into the surface equation to verify if a point lies on the surface.

Explanation

By substituting the coordinates of the point into the surface equation and simplifying, we can confirm whether the point satisfies the equation and thus lies on the surface.

step 1

To construct the equation of the platform tangent to the surface at

(0,0,1)

, we first need to find the gradient vector

\nabla f(x, y, z)

of the surface at that point

step 2

The gradient vector is given by

\nabla f(x, y, z) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)

, where

f(x, y, z) = x z^{2}-y z+\cos (x y) - 1

step 3

We calculate the partial derivatives at

(0,0,1)

:

\frac{\partial f}{\partial x} = z^{2} - y \sin(xy)

,

\frac{\partial f}{\partial y} = -z - x \sin(xy)

,

\frac{\partial f}{\partial z} = 2xz - y

step 4

Substituting

(0,0,1)

into the partial derivatives, we get

\frac{\partial f}{\partial x} = 1^{2} - 0 \cdot \sin(0 \cdot 0) = 1

,

\frac{\partial f}{\partial y} = -1 - 0 \cdot \sin(0 \cdot 0) = -1

, and

\frac{\partial f}{\partial z} = 2 \cdot 0 \cdot 1 - 0 = 0

step 5

The gradient vector at

(0,0,1)

is

\nabla f(0, 0, 1) = (1, -1, 0)

step 6

The equation of the tangent plane at

(0,0,1)

is given by

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

step 7

Simplifying the equation of the tangent plane, we get

x - y = 0

[question b] Answer

The equation of the platform tangent to the surface at

(0,0,1)

is

x - y = 0

.

Key Concept

Gradient vector and tangent plane equation.

Explanation

The gradient vector at a point on a surface is perpendicular to the tangent plane at that point. By finding the gradient vector, we can construct the equation of the tangent plane.

step 1

To show that the curve is tangent to the surface, we need to show that the velocity vector of the curve at the point of tangency is parallel to the gradient vector of the surface at that point

step 2

The velocity vector of the curve is given by

v(t) = \frac{d \mathbf{r}}{d t} = \frac{d}{d t} (\ln t \mathbf{i} + t \ln t \mathbf{j} + t \mathbf{k})

step 3

Differentiating, we get

v(t) = \frac{1}{t} \mathbf{i} + (\ln t + 1) \mathbf{j} + \mathbf{k}

step 4

To find the point of tangency, we set the position vector equal to the point

(0,0,1)

:

\ln t \mathbf{i} + t \ln t \mathbf{j} + t \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}

step 5

Solving for

t

, we find that

t = 1

since

\ln 1 = 0

step 6

Substituting

t = 1

into the velocity vector, we get

v(1) = \frac{1}{1} \mathbf{i} + (\ln 1 + 1) \mathbf{j} + \mathbf{k} = \mathbf{i} + \mathbf{j} + \mathbf{k}

step 7

The gradient vector at

(0,0,1)

is

\nabla f(0, 0, 1) = (1, -1, 0)

as found in the previous question

step 8

Since the velocity vector

v(1)

and the gradient vector

\nabla f(0, 0, 1)

are not parallel (they are not scalar multiples of each other), the curve is not tangent to the surface at

(0,0,1)

[question c] Answer

The curve given by

\mathbf{r}(t) = (\ln t) \mathbf{i} + (t \ln t) \mathbf{j} + t \mathbf{k}

is not tangent to the surface at the point

(0,0,1)

because the velocity vector at

t=1

is not parallel to the gradient vector of the surface at that point.

Key Concept

Tangency condition between a curve and a surface.

Explanation

For a curve to be tangent to a surface at a point, the velocity vector of the curve at that point must be parallel to the gradient vector of the surface at the same point. In this case, they are not parallel, so the curve is not tangent to the surface.

Solution by Steps

step 2

The triangle has a base of 3 and height of 2, so the radius of the disks as a function of

x

is

r(x) = \frac{2}{3}x

step 3

The volume of the solid formed by the triangle is

V_{\text{triangle}} = \int_0^3 \pi \left(\frac{2}{3}x\right)^2 dx

step 4

To find the volume of the solid formed by rotating the quarter-circle around the z-axis, we use the shell method. The volume of a cylindrical shell with radius

r

, height

h

, and thickness

dr

is

V = 2\pi rh dr

step 5

The quarter-circle has a radius of 2, so the height of the shells as a function of

r

is

h(r) = 2 - r

step 6

The volume of the solid formed by the quarter-circle is

V_{\text{circle}} = \int_0^2 2\pi r(2 - r) dr

step 7

The total volume of the trophy is

V_{\text{total}} = V_{\text{triangle}} + V_{\text{circle}}

step 8

For part (b), we generalize the volume calculation by replacing the dimensions 3 and 2 with

r

and

s

respectively

step 9

The volume of a similar solid with dimensions

r

and

s

is

V(r, s) = \int_0^r \pi \left(\frac{s}{r}x\right)^2 dx + \int_0^s 2\pi r(s - r) dr

step 10

For part (c), we use linear approximation to estimate the maximum possible error in the volume calculation

step 11

The linear approximation formula is

\Delta V \approx \frac{\partial V}{\partial r} \Delta r + \frac{\partial V}{\partial s} \Delta s

, where

\Delta r

and

\Delta s

are the maximum possible errors in

r

and

s

step 12

We calculate the partial derivatives

\frac{\partial V}{\partial r}

and

\frac{\partial V}{\partial s}

and evaluate them at

r = 3

and

s = 2

step 13

The maximum possible error in

r

and

s

is 0.5, so

\Delta r = \Delta s = 0.5

step 14

We substitute

\Delta r

and

\Delta s

into the linear approximation formula to find the maximum possible error in the volume

[question 1] Answer

A

Key Concept

Volume of solids of revolution

Explanation

The volume of a solid formed by rotating a two-dimensional region around an axis can be calculated using the disk or shell method, depending on the shape of the region.

[question 2] Answer

B

Key Concept

Generalization of volume formulas

Explanation

The volume formulas can be generalized by replacing specific dimensions with variables to find the volume of similar solids with different dimensions.

[question 3] Answer

C

Key Concept

Linear approximation and error estimation

Explanation

Linear approximation can be used to estimate the maximum possible error in a measurement by considering the partial derivatives of the volume with respect to the measured dimensions and the maximum error in those dimensions.

AskSia

Plus