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Kernel: SageMath 10.0

Implicit Differentiation Assignment

Caution: Make sure you run the parts of each question in order!

Note

You may use −5<x<5 -5 < x < 5 and −5<y<5 -5 < y < 5 for each implicit_plot

Question 1

[2 points] Consider the curve defined by y4−4y2−x4+9x2=0 y^4-4y^2-x^4+9x^2=0

Part a

Calculate the derivative dydx\frac{dy}{dx}.

y=function ('y')(x)
derivative (y^4-4*y^2-x^2==0, x)
4*y(x)^3*diff(y(x), x) - 8*y(x)*diff(y(x), x) - 2*x == 0
show(_)

4 y(x)3∂∂xy(x)−8 y(x)∂∂xy(x)−2 x=0\displaystyle 4 \, y\left(x\right)^{3} \frac{\partial}{\partial x}y\left(x\right) - 8 \, y\left(x\right) \frac{\partial}{\partial x}y\left(x\right) - 2 \, x = 0

solve(derivative(y^4-4*y^2-x^4+9*x^2--0,x),derivative(y,x))
show(_)

4 y(x)3∂∂xy(x)−8 y(x)∂∂xy(x)−2 x=0\displaystyle 4 \, y\left(x\right)^{3} \frac{\partial}{\partial x}y\left(x\right) - 8 \, y\left(x\right) \frac{\partial}{\partial x}y\left(x\right) - 2 \, x = 0

Part b

Calculate the slope mm at the point (0.5888,1)(0.5888,1).

[Check: The slope should be approximately 2.445]

( (2*.5888^3) - (9*.5888))/ (2* (1^3-(2*1)))
2.44547161292800

Part c

Calculate the tangent line at the given point (x0,y0):TL(x)=y0+m⋅(x−x0)(x_0,y_0):\quad TL(x)=y_0+m\cdot(x-x_0).

var('x')
TL (x)=1+2.445*(x-.5888)
TL(x)
2.44500000000000*x - 0.439616000000000

Part d

Graph the original equation and the tangent line on the same window.

var ('x, y')
implicit_plot(y^4-4*y^2-x^4+9*x^2==0,(x,-5,5),
(y,-5,5), axes=True, frame=False)+plot(TL,xmin=-5,xmax=5,ymin=-5,color='red')+point([(1,1)],color='black',size=25)
Image in a Jupyter notebook

Question 2

[2 points] Consider the curve defined by x3+y3=9xy x^3+y^3=9xy

Part a

Calculate the derivative dydx\frac{dy}{dx}.

y=function ('y')(x)
solve (derivative(x^3+y^3==9*x*y,x), derivative(y,x))
[diff(y(x), x) == -(x^2 - 3*y(x))/(y(x)^2 - 3*x)]
show(_)

[∂∂xy(x)=−x2−3 y(x)y(x)2−3 x]\displaystyle \left[\frac{\partial}{\partial x}y\left(x\right) = -\frac{x^{2} - 3 \, y\left(x\right)}{y\left(x\right)^{2} - 3 \, x}\right]

Part b

Calculate the slope mm at the point (2,4)(2,4).

[Check: The slope should be 4/5]

-(2^2-(3*4))/(4^2-(3*2))
4/5

Part c

Calculate the tangent line at the given point (x0,y0):TL(x)=y0+m⋅(x−x0)(x_0,y_0): \quad TL(x)=y_0+m\cdot(x-x_0).

TL (x)=4+ (4/5)*(x-2)
TL (x)
4/5*x + 12/5

Part d

Graph the original equation and the tangent line on the same window.

var ('y')
implicit_plot (x^3+y^3==9*x*y, (x, -5,5),
(y, -5,5) ,axes=true, frame=false)+plot (TL (x) , xmin=-5, xmax=5, ymin=-5, ymax=5, color= 'red' ) +point ( ( ) , color= 'black' , size=25)
Image in a Jupyter notebook

Question 3

[2 points] Consider the curve defined by (x2+y2−1)3=x2y3\displaystyle (x^2+y^2-1)^3=x^2y^3

Part a

Calculate the derivative dydx\frac{dy}{dx}.

y=function ('y')(x)
solve (derivative((x^2+y^2-1)^3==x^2*y^3,x),derivative (y,x))
[diff(y(x), x) == -2/3*(3*x^5 + 3*x*y(x)^4 - x*y(x)^3 - 6*x^3 + 6*(x^3 - x)*y(x)^2 + 3*x)/(2*y(x)^5 - x^2*y(x)^2 + 4*(x^2 - 1)*y(x)^3 + 2*(x^4 - 2*x^2 + 1)*y(x))]
show(_)

[∂∂xy(x)=−2 (3 x5+3 xy(x)4−xy(x)3−6 x3+6 (x3−x)y(x)2+3 x)3 (2 y(x)5−x2y(x)2+4 (x2−1)y(x)3+2 (x4−2 x2+1)y(x))]\displaystyle \left[\frac{\partial}{\partial x}y\left(x\right) = -\frac{2 \, {\left(3 \, x^{5} + 3 \, x y\left(x\right)^{4} - x y\left(x\right)^{3} - 6 \, x^{3} + 6 \, {\left(x^{3} - x\right)} y\left(x\right)^{2} + 3 \, x\right)}}{3 \, {\left(2 \, y\left(x\right)^{5} - x^{2} y\left(x\right)^{2} + 4 \, {\left(x^{2} - 1\right)} y\left(x\right)^{3} + 2 \, {\left(x^{4} - 2 \, x^{2} + 1\right)} y\left(x\right)\right)}}\right]

Part b

Calculate the slope mm at the point (1,1)(1,1).

[Check: The slope should be -4/3]

-2* (3*1^5+ (3*1*1^4) -(1*1^3)-6*1^3+6*(1^3-1) *1^2+3*1) / (3* (2*1^5-(1^2*1^2)+4*(1^2-1)*1^3+2*(1^4-2*1^2+1) *1))
-4/3
y=function ('y')(x)
solve(derivative ((x^3-3*y^2) ==4*x+2*y), derivative (y, x))
[diff(y(x), x) == 1/2*(3*x^2 - 4)/(3*y(x) + 1)]
1/2* (3*4^2-4)/((3*3.68053* (4)+1))
0.487088178015674

Part c

Calculate the tangent line at the given point (x0,y0):TL(x)=y0+m⋅(x−x0)(x_0,y_0): \quad TL(x)=y_0+m\cdot(x-x_0).

TL(x)=1+-4/3*(x-1)
TL(x)
-4/3*x + 7/3

Part d

Graph the original equation and the tangent line on the same window.

var ('y')
implicit_plot((x^2+y^2-1)^3 == x^2*y^3, (x, -5, 5), (y, -5, 5), axes=True, frame=False) + plot(TL, xmin=-5, xmax=5, ymin=-5, ymax=5, color='red') + point((), color='black', size=25)
Image in a Jupyter notebook

Question 4

[3 points] Consider the curves defined by y2=x3y^2=x^3 and 2x2+3y2=52x^2+3y^2=5.

Part a

Find dydx\frac{dy}{dx} for the first curve.

y=function ('y')(x)
solve(derivative(y^2==x^3,x),derivative(y,x))
[diff(y(x), x) == 3/2*x^2/y(x)]
show(_)

[∂∂xy(x)=3 x22 y(x)]\displaystyle \left[\frac{\partial}{\partial x}y\left(x\right) = \frac{3 \, x^{2}}{2 \, y\left(x\right)}\right]

Part b

Find the tangent line to the first curve at (1,1)(1,1).

3*1^2/(2*1)
3/2

Part c

Find dydx\frac{dy}{dx} for the second curve.

y=function ('y')(x)
solve(derivative(2*x^2+3*y^2==5,x),derivative(y,x))
[diff(y(x), x) == -2/3*x/y(x)]
show(_)

[∂∂xy(x)=−2 x3 y(x)]\displaystyle \left[\frac{\partial}{\partial x}y\left(x\right) = -\frac{2 \, x}{3 \, y\left(x\right)}\right]

Part d

Find the tangent line to the second curve at (1,1)(1,1).

[Caution: Make sure you give this tangent line a different name than the tangent line in Part b, like TL2(x).]

-2*1/(3*1)
-2/3
TL2(x)=1+(-2/3)*(x-1)
TL2(x)
-2/3*x + 5/3

Part e

Graph the two curves and the two tangent lines on the same axes (use red for the tangent lines).

[Notice that the two tangent lines are perpendicular; their slopes should be 3/2 and -2/3.]

var('y')
implicit_plot(y^2==x^3,(x, -5,5),(y,-5,5),axes=True, frame=False)+implicit_plot(2*x^2+3*y^2==5,(x, -5,5),(y, -5,5), axes=True, frame=False) + plot(TL, xmin=-5, xmax=5, ymin=-5, ymax=5, color='red')+ point((1, 1), color='black',size=25) + plot(TL2, xmin=-5, xmax=5, ymin=-5, ymax=5, color='red')
Image in a Jupyter notebook

Question 5

[1 point] Graph the curve given by 1x2−1y2+1z2=0\displaystyle\frac{1}{x^2} - \frac{1}{y^2} + \frac{1}{z^2}=0 with −5<x<5-5 < x < 5, −5<y<5-5 < y < 5, and −5<z<5-5 < z < 5.

var('y,z')
implicit_plot3d((1/x^2)-(1/y^2)+(1/z^2)==0, (x, -5,5), (y,-5,5), (z,-5,5))