For the equation given below, evaluate dy/dx at the point (-2,1).?
We can expand through the LHS to yield: (2 + y)^2 + 4y = x + 33 ==> y^2 + 8y + 4 = x + 33. Taking derivatives: 2y(dy/dx) + 8(dy/dx) + 0 = 1 + 0 ==> 2y(dy/dx) + 8(dy/dx) = 1 ==> (dy/dx)(2y + 8) = 1 ==> dy/dx = 1/(2y + 8). Thus, the value of dy/dx at (-2, 1) is 1/[2(1) + 8] = 1/10. I hope this helps!
How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.
Implicit differentiation regarding an ellipse?
I am not sure this is 100% correct, but this is how I would answer it: First I would find the x-intercepts: (x)^2 -(x)(0)+(0)^2=16 x=-4 or x=4 Then I would differentiate both sides of the equation: (d/dx) (x^2-xy+y^2) = (16) (d/dx) 2x -(y + x(dy/dx)) + 2y(dy/dx)= 0 2x - y - x(dy/dx) + 2y(dy/dx)= 0 2x - y= x(dy/dx) -2y(dy/dx) 2x - y= (dy/dx) (x-2y) (dy/dx)= (2x-y)/ (x-2y) Now, plug in the two intercept points to the derivative (dy/dx)= (2x-y)/ (x-2y) (dy/dx)= (2(-4) -(0))/ ((-4) -2(0)) = -8/-4= 2 (dy/dx)= (2(4) -(0))/ ((4) -2(0)) = 8/4= 2 After that, plug the points and the slopes into the point slope formula y-0=2 (x -4) y=2x -8 (Answer) y-0=2 (x +4) y=2x +8 (Answer)
An equation of the tangent to a curve at the point (1,3) is y = x +2. If at any point (x,y) on the curve, y''(x) = 6x, how can I find an equation of the curve?
The first step is to take the integral with respect to x twice of the equation y''(x)=6x. You get a function with unknown constants, so you have to plug this function into the formula to give you the equation of the tangent at x=a which is y=f(a)+f'(a)(x-a), where we have the coordinate (1,3), so in this case we plug 1 in for a. You can simplify the expression that you get down into the form (f1(c,k))x+(f2(c,k)), then you know that f1(c,k)=1, because the coefficient on the x in the expression x+2 is 1, and you know that f2(c,k)=2 using similar logic, then you use systems of equations to solve for the c and k, then plug their values in the expression you got when you did the integrals, and then you're done.
How can one find the values of k for which the line 2x -k is tangent to the circle with the equation x^2 + y^2 = 5?
The simple way to do this is to clearly define what it means for tangent so that finding the k values is the easiest. Problem Specific AnswerWe have [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5[/math] and the line is tangent to the circle. What it means for a 2 things to be tangent is that one point and only one point satisfies both equations at the same time. That means that in the above two equations, only the point represented by [math](x,y)[/math] satisfies both equations. We get: [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5 [/math]Substituting, we get [math]x^2 + (2x-k)^2 = 5.[/math] This equation must only have one solution. [math]5x^2 -4kx +(k^2 - 5) = 0.[/math]This quadratic equation must have its determinant equal to 0 in order for the two to be tangent. [math]b^2 - 4ac = 0[/math][math]16k^2 - 20k^2+100 = 0[/math][math]4k^2 = 100[/math][math]\boxed{k = \pm 5}[/math]Generic AnswerWe can generalize this solution for any line and circle.Line: [math]y = mx+b[/math]Circle: [math]x^2 + y^2 = r^2[/math]Substituting we get: [math]x^2 + (mx+b)^2 = r^2[/math][math]\Rightarrow (m^2 + 1)x^2 + 2mbx + (b^2 - r^2) = 0[/math]Taking the discriminant and making it equal to 0, we get[math]4m^2b^2 - 4(m^2 + 1)(b^2 - r^2) = 0[/math][math]4m^2b^2 - 4m^2b^2 + 4m^2r^2 - 4b^2 + 4r^2 = 0[/math][math]m^2r^2 + r^2 = b^2[/math][math]m, r,[/math] and [math]b[/math] must satisfy these equations in order to be tangent. We can check out previous answer by plugging in here: [math]m = 2; r = \sqrt{5}; b = -k.[/math][math]20 + 5 = k^2.[/math]Therefore, [math]\boxed{k = \pm 5}[/math]
Consider the equation x^2-2xy+4y^2=64...the questions are below?
1) You'd need to differentiate it explicitly: d/dx (x² - 2xy + 4y²) = d/dx(64) 2x - 2(xy' + y) + 8yy' = 0 2x - 2xy' - 2y + 8yy' = 0 x - xy' - y + 4yy' = 0 y'(4y - x) = y - x y' = (y - x)/(4y - x) 2) At x=2: (2)² - 2(2)y + 4y² = 64 4 - 4y + 4y² = 64 y² - y - 15 = 0 y = (1 ± √61)/2 Messy...then you'd have to plug each of those y values into the derivative function above to get the slopes, then use the point-slope equation to get lines... Really messy... unless I miessed up somewhere... 3) Differentiate the first derivative: d²y/dx² = d²/dx² (y - x)/(4y - x) = [(4y - x)(y' - 1) - (y - x)(4y' - 1)]/(4y - x)² And y'(0) = (4 - 0)/(16 - 0) = 1/4 Plugging in (0,4): =[(16 - 0)(1/4 - 1) - (4 - 0)(1 - 1)]/(16 - 0)² = -12/256 = -3/64 Since a_math_guy and I now agree, I think this is right. Really messy though....
How can I find the equation of the tangent to [math] x ^ 4 + 16 * y ^4 = 32 [/math] at (2, 1)?
We’ll do it using middle school math.Let [math]f(x,y)=x^4+16y^4 -32[/math] so our curve is [math]f(x,y)=0.[/math]We’re after the tangent at [math](r,s)[/math] so we want to rewrite [math]f[/math] as a polynomial in [math]x-r[/math] and [math]y-s.[/math][math]f(x,y)=f(r+(x-r), s+(y-s)) = (r+(x-r))^4 + 16(s+(y-s))^4 -32[/math][math]f(x,y) = r^4 + 4r^3(x-r) + 6r^2(x-r)^2 + 4r(x-r)^3 + (x-r)^4 [/math][math]\qquad +\ 16(s^4 + 4s^3(y-s) + 6s^2(y-s)^2 + 4s(y-s)^3 + (y-s)^4 ) -32[/math]Let’s sort the terms by increasing degree[math]f(x,y) = (r^4 + 16s^4 - 32) [/math][math] \qquad+\ 4r^3(x-r) + 64s^3(y-s) [/math][math] \qquad +\ 6r^2(x-r)^2 + 96s^2(y-s)^2 [/math][math]\qquad +\ 4r(x-r)^3 + 64s(y-s)^3[/math][math]\qquad +\ (x-r)^4 + 16(y-s)^4[/math]The constant term [math]r^4 + 16s^4 - 32[/math] is [math]f(r,s)[/math] which is [math]0[/math] because we’re assuming [math](r,s)[/math] is on our curve.In the neighborhood of [math](r,s)[/math] we know [math]x-r[/math] and [math]y-s[/math] are small and their powers are smaller still. Our tangent line is the best linear approximation to [math]f[/math] gotten by dropping the higher order terms and setting the whole thing to zero:[math]0 = 4r^3(x-r) + 64s^3(y-s) [/math]When [math](r,s)=(2,1)[/math] we get our tangent line[math]0 = 32(x-2)+64(y-1)[/math][math]x+ 2y = 4[/math]Tangents of algebraic curves using middle school math. Who needs calculus?We can do higher order tangents just as easily. It’s a pain with calculus. Let’s generate three tangent conics at [math](-2,1)[/math] and three tangent cubics at [math](-2,-1)[/math]The tangent conics are gotten by including one or both of the squared terms:[math]0= 4r^3(x-r) + 64s^3(y-s) + 6r^2(x-r)^2 + 96s^2(y-s)^2[/math]At [math](r,s)=(-2,1)[/math] that’s[math]0= -32(x+2) + 64(y-1) + 24(x+2)^2 + 96(y-1)^2[/math]The other two are [math]0= -32(x+2) + 64(y-1) + 24(x+2)^2 [/math] and [math]0= -32(x+2) + 64(y-1) + 96(y-1)^2[/math]Tangent cubic at (-2,-1)[math]0= -32(x+2) - 64(y+1) + 24(x+2)^2 + 96(y+1)^2 + -8(x+2)^3 - 64(y+1)^3[/math]Let’s plot:One tangent line, three tangent conics and three tangent cubics, all with middle school math.
If the line y=x is tangential to the curve y=b^x, with b>1, what's the value of the constant b and the point of contact of the tangent?
dy/dx=(b^x)*ln(b). dy/dx=1 at the point of tangency since dy/dx=1 for y=x. Thus (b^x)*ln (b)=1. Thus for any b, the only value of x where a tangent is possible satisfies ln((b^x)*ln(b))=ln (1)=0. ln(b)*x+ln(ln(b))=0. x=-ln(ln(b))/ln(b). Also at the tangent point the value of the function equals x, so b^x=x. Substituting the value of x at which the slope equals one, we have: b^(-ln(ln(b))/ln(b))=-ln(ln(b))/ln(b). Taking the natural logarithm of both sides, we get: ln(b^(-ln(ln(b))/ln(b)))=ln(-ln(ln(b))/ln(b)), or -ln(ln (b))*ln(b)/ln(b)=ln(-ln(ln(b)))-ln(ln(b)). -ln(ln(b))=ln(-ln(ln(b)))-ln(ln(b)). 0=ln(-ln(ln(b))). e^0=e^ln(-ln(ln(b))). 1=-ln(ln(b)). e^ -1=e^ln(ln(b)). e^-1=ln(b). e^(e^-1)=e^ln(b). b=e^(1/e). Thus y=(e^(1/e))^x is tangent to y=x. We know that for the point of tangency, x=-ln(ln(e^(1/e)))/ln(e^(1/e)). x=-ln(1/e)/(1/e). x=e. Then y=(e^(1/e))^e=e^(e/e)=e. So the point of tangency is (e, e). This point lies on the line y=x as desired. Now check the slope of our function at this point: dy/dx=((e^(1/e))^x)*ln(e^(1/e))=((e^(1/e))^x)/e. At x=e, we have dy/dx=((e^(1/e))^e)/e=(e^(e/e))/e=e/e=1 as desired.
Differentation word problem?
a) R(t) = [ (3t^2 + 16) * 2 - 2t * 6t ] / [ (3t^2 + 16)^2 ] R(t) = [ (6t^2 + 32) - 12t^2 ] / [ (3t^2 + 16)^2 ] R(t) = [ 6t^2 + 32 - 12t^2 ] / [ (3t^2 + 16)^2 ] R(t) = [ -6t^2 + 32 ] / [ (3t^2 + 16)^2 ] b) after one hour: R(1) = [ -6 * 1^2 + 32 ] / [ (3 * 1^2 + 16)^2 ] R(1) = (26) / ( 19^2 ) R(1) ≈ 0.07 <==== increasing c) let's find the interval by finding the first derivative and equal to zero as the following: 0 = [ -6t^2 + 32 ] / [ (3t^2 + 16)^2 ] 0 = -6t^2 + 32 6t^2 = 32 t^2 = 32/6 t^2 = 16/3 t = √(16/3) ≈ 2.31 ( critical point ) <===== this is the time when it started to decline. d) to find the interval, we are going to check before and after the critical point into the first derived equation: R(t) = [ -6t^2 + 32 ] / [ (3t^2 + 16)^2 ] +++++++0------------ ________________ . . . . . .2.31.. . .. decline as ( √(16/3) , ∞ )
Quick Calculus Question, please!?
you have the right idea here. use equation x^2+y^2=z^2 what we know: x=24, dx/dt=0, y=32 meters, dy/dt=90, z=40, dz/dt is what you are trying to find. x^2+y^2=40^2 x^2+y^2=z^2 24^2=z^2-y^2 576=2zz'-2yy' find the derivative which I already did. plug in what you know. 576=2(40)*dz/dt-2(32)(90) 576+5760=80*dz/dt 6336=80*dz/dt dz/dt=79.2. So the distance between the observer and the balloon is increasing at a rate of 79.2 meters per second. This makes sense because the distance between the observer and balloon is getting bigger. Here, as you can see the dx/dt isn't important here because dx/dt=0. Think of this as a triangle.