What is the slope of the curve at t=3 assuming that the equations define x and y implicitly as differentiable functions $x=f\left(t\right)$, $y=g\left(t\right)$, and $x={t}^{5}+t$, $y+4{t}^{5}=4x+{t}^{4}$?

Hagman7v
2022-09-22
Answered

What is the slope of the curve at t=3 assuming that the equations define x and y implicitly as differentiable functions $x=f\left(t\right)$, $y=g\left(t\right)$, and $x={t}^{5}+t$, $y+4{t}^{5}=4x+{t}^{4}$?

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Adelaide Barr

Answered 2022-09-23
Author has **9** answers

If:

$x={t}^{5}+t$

$y+4{t}^{5}=4({t}^{5}+t)+{t}^{4}$

$y=-4{t}^{5}+4{t}^{5}+4t+{t}^{4}={t}^{4}+4t$

Now

$\frac{dx}{dt}=5{t}^{4}+1$

$\frac{dy}{dt}=4{t}^{3}+4$

But

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$

So

$\frac{dy}{dx}=\frac{4{t}^{3}+4}{5{t}^{4}+1}$

For t=3 the slope will be:

$\frac{dy}{dx}=\frac{4\cdot (27+4)}{5\cdot (81+1)}=\frac{124}{410}=\frac{62}{205}=3.31$

$x={t}^{5}+t$

$y+4{t}^{5}=4({t}^{5}+t)+{t}^{4}$

$y=-4{t}^{5}+4{t}^{5}+4t+{t}^{4}={t}^{4}+4t$

Now

$\frac{dx}{dt}=5{t}^{4}+1$

$\frac{dy}{dt}=4{t}^{3}+4$

But

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$

So

$\frac{dy}{dx}=\frac{4{t}^{3}+4}{5{t}^{4}+1}$

For t=3 the slope will be:

$\frac{dy}{dx}=\frac{4\cdot (27+4)}{5\cdot (81+1)}=\frac{124}{410}=\frac{62}{205}=3.31$

asked 2022-09-28

Given ${e}^{L}+KL=K{e}^{K}$, we are being asked to find $\frac{dL}{dK}$. I think O need to use implicit differentiation, but I am really not sure how to do it!

asked 2022-08-13

I'm trying to calculate this differential equation using implicit differentiation for $\frac{dy}{dx}$ in terms of x and y:

$2{x}^{3}y+5x+6y-sin(\pi y)=8$

What I have tried:

By differentiating with respect to x

$6{x}^{2}y+\frac{dy}{dx}2{x}^{3}+5+\frac{dy}{dx}6-\frac{dy}{dx}\pi \mathrm{cos}(\pi y)=0$

I'm not sure where to go from here.

$2{x}^{3}y+5x+6y-sin(\pi y)=8$

What I have tried:

By differentiating with respect to x

$6{x}^{2}y+\frac{dy}{dx}2{x}^{3}+5+\frac{dy}{dx}6-\frac{dy}{dx}\pi \mathrm{cos}(\pi y)=0$

I'm not sure where to go from here.

asked 2022-10-16

Find the equation of the tangent line to the curve (a lemniscate) $2({x}^{2}+{y}^{2}{)}^{2}=25({x}^{2}-{y}^{2})$ at the point (3,1). Write the equation of the tangent line in the form $y=mx+b.$

Every time that we do it we get a ridiculous number for the slope (150/362)

Every time that we do it we get a ridiculous number for the slope (150/362)

asked 2022-07-22

Find $\frac{dy}{dx}$ for ${x}^{2}=\frac{x-y}{x+y}$

I have solved this in two ways.

First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:

$\frac{1-3{x}^{2}-2xy}{{x}^{2}+1}$

So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:

$-\frac{x(y+x{)}^{2}-y}{x}$

Can anyone explain to me why I get different solutions ?

I have solved this in two ways.

First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:

$\frac{1-3{x}^{2}-2xy}{{x}^{2}+1}$

So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:

$-\frac{x(y+x{)}^{2}-y}{x}$

Can anyone explain to me why I get different solutions ?

asked 2022-09-15

Find partial derivative of $z$ with respect to the partial of $y$ using the result of the chain rule.

$\mathrm{ln}({x}^{2}+{y}^{2})+x\mathrm{ln}(z)-\mathrm{cos}(xyz)=3.$

I would use regular implicit differentiation for this problem but what does "using result of the chain rule" mean?

$\mathrm{ln}({x}^{2}+{y}^{2})+x\mathrm{ln}(z)-\mathrm{cos}(xyz)=3.$

I would use regular implicit differentiation for this problem but what does "using result of the chain rule" mean?

asked 2022-10-08

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

asked 2022-08-25

Say you are given the equation

$p(V-b)=RT{e}^{-\frac{a}{RVT}}$

Where p, b, R and a are all constants.

And you are asked to find an expression for $\frac{dV}{dT}$, would the left side be a chain rule with a product rule with an implicit differential?So far I believe that the right hand side differentiates to:

$\frac{a}{TV}{e}^{\frac{-a}{RVT}}+\frac{dV}{dT}\ast \frac{a}{{V}^{2}}{e}^{\frac{-1}{RVT}}$

Is that fully correct?

$p(V-b)=RT{e}^{-\frac{a}{RVT}}$

Where p, b, R and a are all constants.

And you are asked to find an expression for $\frac{dV}{dT}$, would the left side be a chain rule with a product rule with an implicit differential?So far I believe that the right hand side differentiates to:

$\frac{a}{TV}{e}^{\frac{-a}{RVT}}+\frac{dV}{dT}\ast \frac{a}{{V}^{2}}{e}^{\frac{-1}{RVT}}$

Is that fully correct?