Long long question of differentiation.?
1.The standard equation for the position s of a body moving with a constant acceleration a along a coordination line is s= (a/2)t^2+v0t+s0 where v0 and s0 are the body's velocity and position at time t=0. Derive this equation by solving the initial value problem Differential equation:d^2s/dt^2=a Initial condition: ds/dt=v0 and s=s0 when t=0 2.For free fall near the surface of a planet where the acceleration due to gravity has a constant magnitude of g length-unit/sec^2, Equation (1) in question 1 takes the form s=-(1/2)gt^2+v0t+s0 where s is the body height above the surface. The equation has a minus sign because the acceleration act downward, in the direction of decreasing s. The velocity v0 is positive if the object is rising at time t=0 and negative if the object is falling. Instead of using using the result in question 1, you can derive equation (2) directly by solving an appropriate initial value problem? What initial value problem? Solve it to be sure you have the right one, exlaining the solution as u go along.
Question of differentiation?
Verify that,for all values of t,the point P ( at,at³) lies on the curve y= x^3 / a^2 x = at → t = x/a y = at³ = a(x/a)³ = ax³/a³ = x³/a² Find the equation of the tangent to the curve at this point. dy/dx = (dy/dt)/(dx/dt) = 3at²/a = 3t² So slope of tangent (m) at P ( at,at³) is 3t² Equation of tangent is given by y = mx + b = 3t² x + b Since P ( at,at³) lies on the tangent at³ = at*3t² + b So b = at³ - 3at³ = -2at³ ie the tangent is y = 3t² x - 2at³ If the tangent at P meets the x-axis and y axis at Q and R respectively,find area of triangle OQR At Q y = 0 So 3t² x -2at³ = 0 ie x = 2at³/(3t²) = 2at/3 ie Q ≡ (2at/3, 0) At R x = 0 So y = -2at³ ie R ≡ (0, -2at³) NOTE OQ _|_ OR ie ∆ OQR is right angled So Area = ½bh = ½|OQ|*|OR| Now length OQ = |2at/3| length OR = |-2at³| Thus Area ∆ OQR = ½|2at/3|*|-2at³| = ½|-4a²t^4|/3 = ⅔a²t^4 and the equation of locus midpoint of QR as P moves on original curve Let midpoint of QR be M M ≡ ½(2at/3, -2at³) ie M ≡ (at/3, -at³) So x = at/3 → t = 3x/a and y = -at³ = -a(3x/a)³ = -27x³/a² ie the locus of M is a²y = -27x³
Math Question About Differentiation?
First, you need to find f'(θ) which is easily accomplished. f'(θ) = -2sinθ + 2sin2θ by chain rule for the second term. Now, since you're bounded by 0 ≤ θ ≤ 2 pi, you're obviously searching for the increasing and decreasing intervals in it. But an interval changes from increasing to decreasing via a max or a min. So we must find where the function's critical points: f'(θ) = 0 And solve for theta this way. BUT, notice you have sin2θ in part of your derivative. So we use the trig identity sin2θ = 2sinθcosθ. -2sinθ + 2sin2θ = 0 -sinθ + sin2θ = 0 -sinθ + 2sinθcosθ = 0 -sinθ(1 - 2cosθ) = 0 Solve for thetas and you have your critical points. I think that's the formula your teacher was referring to above. To find the max's and min's of your critical points, use either the first derivative test or the second derivative test. For increasing and decreasing intervals, find when f'(θ) > 0 for increasing (positive slope) and f'(θ) < 0 for decreasing (negative slope). Lastly, the last one involves finding the second derivative. Setting the second derivative equal to 0 ( f''(θ) = 0 ) and solving for θ will give you your inflection points. Like the increasing/decreasing intervals, the function is concave up when f''(θ) > 0 and concave down when f''(θ) < 0. Hope that helps you! :)
How do I do these differentiation questions?
[math]y=\ln(\sqrt{1+x^2})[/math]This is the right time to use the chain rule.[math]f(g(x))'=f'(g(x))g'(x)[/math][math]f(u)=\ln(u)[/math][math]g(x)=\sqrt{1+x^2}[/math][math]f'(u)=\frac{1}{u}[/math][math]f'(g(x))=\frac{1}{\sqrt{1+x^2}}[/math]To find [math]g'(x)[/math], we use the chain rule again, here the functions being [math]a(b(x))=\sqrt{b(x)}, b(x)=1+x^2[/math]The differential of [math]a(u)[/math] is,[math]a'(u)=\frac{d}{dx}\sqrt{u}[/math]Use power rule, [math]nx^{n-1}[/math][math]a'(u)=\frac{1}{2\sqrt{u}}[/math][math]a'(b(x))=\frac{1}{2\sqrt{x^2+1}}[/math][math]b(x)=x^2+1[/math][math]b'(x)=2x[/math][math]a'(b(x))b'(x)=\frac{x}{\sqrt{x^2+1}}=g'(x)[/math]We multiply this with our initial result, and we get,[math]\frac{1}{\sqrt{x^2+1}} \times \frac{x}{\sqrt{x^2+1}}[/math][math]\frac{x}{x^2+1}[/math]Number 2:The graph has a slope/derivative of,[math]\frac{dy}{dx}(x-1+\frac{1}{x-1})[/math][math]1-0+\frac{dy}{dx}(\frac{1}{x-1})[/math]Use chain rule again,[math]x-1=u(x), \frac{1}{u}=v(u)[/math][math]v'(u)=-\frac{1}{u^2}[/math][math]u'(x)=1[/math][math]v'(x-1)=\frac{1}{(x-1)^2}[/math]Our final answer is when we multiply [math]u'(x)v'(x-1)[/math][math]\frac{1}{(x-1)^2}[/math]And, put it into our complete answer,[math]1-\frac{1}{x^2-2x+1}[/math]If we plugin [math]m[/math][math]=2,[/math][math]1-\frac{1}{x^2-2x+1}=2[/math][math]\frac{1}{x^2-2x+1}=-1[/math][math]x^2-2x+1=-1[/math]Of which only [math]\sqrt{-1}=+-i[/math] is a solution
How do you solve this question on differentiation?
(MD)^2 = x^2 + 100 (MN)^2 = (15-x)^2 + y^2 (ND)^2 = (10-y)^2 + 225 since (ND)^2= (MN)^2 + (MD)^2 therefore (10-y)^2 + 225 = (15-x)^2 + y^2 + x^2 + 100 after simplification you will get x^2 - 15x+10y=0 therefore y=0.1* (15x-x^2) to get maximun value of y make dy/dx equals zero: dy/dx=15 - 2x=0 therefore x=7.5 & y=0.1 *(15*7.5 -(7.5)^2)= 5.625
Implicit Differentiation Question!?
>lamp located three units to the right of the y-axis Let lamp be at (3,a) >and a shadow created by an elliptical region x^2+4y^2 ≤ 5 Is the center of the shadow supposed to be translated from (0,0) → (3,a) ? >If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located? Well (-5,0) isn't on the edge of x^2+4y^2 ≤ 5 so I suppose we have to translate the shadow region. (0,0) → (3,a), thus (x,y) → (x+3,y+a) x^2+4y^2 ≤ 5 → (x+3)^2 + 4(y+a)^2 ≤ 5 If (-5,0) is on the edge of this region, it satisfies the equality: (x+3)^2 + 4(y+a)^2 = 5 (-5+3)^2 + 4(0+a)^2 = 5 (-2)^2 + 4a^2 = 5 4 + 4a^2 = 5 4a^2 = 1 a^2 = 1/4 a = +1/2 (not -1/2 because we were told lamp is ABOVE).
How can I differentiate this question: √ (5x³+4)?
Use the chain rule. This allows you to differentiate composite functions (functions nested within another function) with ease.Let [math]f(x) = \sqrt{5x^3+4}[/math][math]f(x) = (5x^3+4)^\frac{1}{2}[/math]Let [math]u = 5x^3+4[/math]The chain rule allows you to find the derivative of a composite function with respect to an independent variable [math]x[/math] by chaining together the derivative of the dependent variable [math]y[/math] with that of another intermediary variable [math]u[/math]. Confused? The mathematical expression for the above sentence is given below.[math]\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}[/math]Finding [math]\frac{dy}{du}[/math][math]y = u^\frac{1}{2}[/math][math]\frac{dy}{du} = \frac{1}{2}u^\frac{-1}{2}[/math]Finding [math]\frac{du}{dx} [/math][math]\frac{du}{dx} = 3(5x^2) = 15x^2[/math]Using the chain rule to bring them all together[math]\frac{dy}{dx} = \frac{1}{2}u^\frac{-1}{2}\times 15x^2[/math]Simplify and substitute back [math]u = 5x^3+4[/math][math]\frac{dy}{dx} = \frac{15}{2}x^2(5x^3+4)^\frac{-1}{2}[/math]
Question on differentiation please help!!!!?
a) y = (kx - x^2)^(1/2) y' = (1/2)(kx - x^2)^(-1/2)(k - 2x) when x = 2, y' = 2^(1/2) (1/2)(k*2 - 2^2)^(-1/2)(k - 2*2) = 2^(1/2) (1/2)(2k - 4)^(-1/2)(k - 4) = 2^(1/2) k - 4 = 2(2k - 4)^(1/2)(2^(1/2)) square both sides k^2 - 8k + 16 = 16k - 32 k^2 - 24k + 48 = 0 k = 12 + 4sqrt(6) = 21.8 b) the slope of the normal is -1/sqrt(2) when x = 2 y = (21.8*2 - 2^2)^(1/2) y = 6.3 the equation of the normal is (y - 6.3)/(x - 2) = -1/sqrt(2) -sqrt(2)y + 6.3sqrt(2) = x - 2 x + sqrt(2)y = -6.3sqrt(2) - 2 c = -6.3sqrt(2) - 2 = -10.9
How do I attempt this nice differentiation question?
the function is a^xso,a^(x+y)=a^x . a^ytry substituting logarithmic/exponential/trigonometric functions for a general approach to these type of differentials.proof:f(x+y)=f(x).f(y)put y=xf (2x )=f(x)^2now differentiate,2f'(2x)=2f(x).f'(x)put x=0 ,and f'(0)=2f(0)=1...........................(a)partially differentiate f(x+y)=f(x).f(y) with respect to yf'(x+y)=f(x).f'(y)put y=0f'(x)=f(x).2f'(x)/f(x) =2now integrate to get ln ( f(x) )=2x +cc=0 using result (a)so,f(x)= e^2x§