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Post a LessonAnswered on 06 Apr Learn Unit III: Calculus/Continuity and Differentiability
Sadika
To solve this problem, we will differentiate both sides of the given expression y=(tan−1x)2y=(tan−1x)2 with respect to xx.
Let's denote y=(tan−1x)2y=(tan−1x)2.
First, let's differentiate yy with respect to xx using the chain rule:
dydx=2tan−1x⋅11+x2dxdy=2tan−1x⋅1+x21
Now, let's express dydxdxdy in terms of xx and yy:
dydx=2y⋅11+x2dxdy=2y⋅1+x21
Now, let's differentiate dydxdxdy with respect to xx:
d2ydx2=ddx(2y⋅11+x2)dx2d2y=dxd(2y⋅1+x21)
Using the product rule:
d2ydx2=2⋅11+x2⋅dydx+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21⋅dxdy+2y⋅(−(1+x2)22x)
Substituting dydx=2y⋅11+x2dxdy=2y⋅1+x21:
d2ydx2=2⋅11+x2⋅2y⋅11+x2+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21⋅2y⋅1+x21+2y⋅(−(1+x2)22x)
d2ydx2=4y(1+x2)2−4x2y(1+x2)2dx2d2y=(1+x2)24y−(1+x2)24x2y
d2ydx2=4y−4x2y(1+x2)2dx2d2y=(1+x2)24y−4x2y
d2ydx2=4(y−x2y)(1+x2)2dx2d2y=(1+x2)24(y−x2y)
Now, let's substitute y=(tan−1x)2y=(tan−1x)2 back into the expression:
d2ydx2=4((tan−1x)2−x2(tan−1x)2)(1+x2)2dx2d2y=(1+x2)24((tan−1x)2−x2(tan−1x)2)
d2ydx2=4(tan−1x)2(1−x2)(1+x2)2dx2d2y=(1+x2)24(tan−1x)2(1−x2)
Now, let's manipulate the expression d2ydx2dx2d2y to match the left side of the given expression (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y′=2:
(x2+1)2y2+2x(x2+1)y′=(x2+1)2((tan−1x)2)2+2x(x2+1)dydx(x2+1)2y2+2x(x2+1)y′=(x2+1)2((tan−1x)2)2+2x(x2+1)dxdy
(x2+1)2y2+2x(x2+1)y′=(x2+1)2(tan−1x)4+2x(x2+1)⋅2y1+x2(x2+1)2y2+2x(x2+1)y′=(x2+1)2(tan−1x)4+2x(x2+1)⋅1+x22y
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2(1+x2)2+4x(x2+1)y1+x2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+1+x24x(x2+1)y
Now, let's simplify this expression:
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2+4x(x2+1)2(tan−1x)2(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)2(tan−1x)2
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2(1+x2+x)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2(1+x2+x)
(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan−1x)2(x2+1)(1+x2)2(x2+1)2y2+2x(x2+1)y′=(1+x2)24(x2+1)2(tan−1x)2(x2+1)
(x2+1)2y2+2x(x2+1)y′=4(x2+1)(tan−1x)2(x2+1)2y2+2x(x2+1)y′=4(x2+1)(tan−1x)2
(x2+1)2y2+2x(x2+1)y′=4(x2+1)y2(x2+1)2y2+2x(x2+1)y′=4(x2+1)y2
Thus, we have verified that (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y′=2.
Answered on 06 Apr Learn Unit III: Calculus/Continuity and Differentiability
Sadika
To verify Rolle's Theorem for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], we need to check three conditions:
Let's check each condition:
Continuity: The function y=x2+2y=x2+2 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the closed interval [−2,2][−2,2].
Differentiability: The function y=x2+2y=x2+2 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the open interval (−2,2)(−2,2).
Endpoints: y(−2)=(−2)2+2=4+2=6y(−2)=(−2)2+2=4+2=6 y(2)=(2)2+2=4+2=6y(2)=(2)2+2=4+2=6
So, y(−2)=y(2)y(−2)=y(2).
Since all three conditions of Rolle's Theorem are satisfied, there exists at least one cc in the open interval (−2,2)(−2,2) such that f′(c)=0f′(c)=0, where f(x)=x2+2f(x)=x2+2.
Now, let's find the derivative of y=x2+2y=x2+2:
y′=ddx(x2+2)=2xy′=dxd(x2+2)=2x
To find the critical point(s), we set y′=0y′=0: 2x=02x=0 x=0x=0
So, f′(0)=0f′(0)=0.
Thus, Rolle's Theorem is verified for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], and there exists at least one point cc in the open interval (−2,2)(−2,2) where the derivative f′(c)f′(c) is equal to zero.
Answered on 06 Apr Learn Unit III: Calculus/Applications of Derivatives
Sadika
Let's denote:
According to the Pythagorean theorem, for any position of the ladder, we have x2+y2=52x2+y2=52, since the ladder is always 5 meters long.
Now, we differentiate both sides of this equation with respect to time tt, considering that xx is changing with time: ddt(x2+y2)=ddt(52)dtd(x2+y2)=dtd(52) 2xdxdt+2ydydt=02xdtdx+2ydtdy=0
Given that the ladder is being pulled away from the wall at a rate of dxdt=2dtdx=2 cm/s, and we want to find dydtdtdy when x=4x=4 m, we can plug in these values into the equation:
2(4)(2)+2ydydt=02(4)(2)+2ydtdy=0 16+4ydydt=016+4ydtdy=0 4ydydt=−164ydtdy=−16 dydt=−164ydtdy=−4y16
Now, we need to find yy when x=4x=4. Using the Pythagorean theorem: 42+y2=5242+y2=52 16+y2=2516+y2=25 y2=9y2=9 y=3y=3
Now, we can find dydtdtdy when y=3y=3: dydt=−164⋅3dtdy=−4⋅316 dydt=−43dtdy=−34
So, the height of the ladder on the wall is decreasing at a rate of 4334 meters per second when the foot of the ladder is 4 meters away from the wall.
Answered on 06 Apr Learn Unit III: Calculus/Integrals
Sadika
To integrate ∫sin3(x)cos2(x) dx∫sin3(x)cos2(x)dx, we can use the substitution method.
Let's denote u=sin(x)u=sin(x), then du=cos(x) dxdu=cos(x)dx. However, the substitution u=sin(x)u=sin(x) is not directly useful since we don't have a cos2(x)cos2(x) term.
We can use a trigonometric identity to rewrite cos2(x)cos2(x) in terms of sin(x)sin(x):
cos2(x)=1−sin2(x)cos2(x)=1−sin2(x)
Now, let's rewrite the integral in terms of uu:
∫u3(1−u2) du∫u3(1−u2)du
Now, we can expand this expression:
∫(u3−u5) du∫(u3−u5)du
Now, we can integrate term by term:
∫u3 du−∫u5 du∫u3du−∫u5du
=u44−u66+C=4u4−6u6+C
Finally, we need to replace uu back in terms of xx, which gives us:
sin4(x)4−sin6(x)6+C4sin4(x)−6sin6(x)+C
So, the antiderivative of sin3(x)cos2(x)sin3(x)cos2(x) with respect to xx is:
sin4(x)4−sin6(x)6+C4sin4(x)−6sin6(x)+C
where CC is the constant of integration.
Answered on 06 Apr Learn Unit III: Calculus/Integrals
Sadika
To integrate ∫11+tan(x) dx∫1+tan(x)1dx using the substitution method, we'll make the substitution:
u=1+tan(x)u=1+tan(x)
Then, we find dudu:
dudx=sec2(x)dxdu=sec2(x)
Now, we need to express dxdx in terms of dudu. Since dudx=sec2(x)dxdu=sec2(x), we can rewrite it as:
dx=dusec2(x)dx=sec2(x)du
dx=duu2dx=u2du
Substituting uu and dxdx into the integral, we get:
∫1u du∫u1du
Now, we can integrate 1uu1 with respect to uu:
∫1u du=ln∣u∣+C∫u1du=ln∣u∣+C
Finally, we need to substitute back u=1+tan(x)u=1+tan(x):
ln∣1+tan(x)∣+Cln∣1+tan(x)∣+C
So, the antiderivative of 11+tan(x)1+tan(x)1 with respect to xx is:
ln∣1+tan(x)∣+Cln∣1+tan(x)∣+C
where CC is the constant of integration.
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