Exercises
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Exercises
Use implicit differentiation to find $dy/dx$ for the following functions.
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For each of the following equations, confirm that the given point is on the curve and find the equations of the lines that are tangent and normal (perpendicular) to the curve at that point.
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- In an earlier lecture, we used the formula for the derivative of an inverse function to derive formulas for the derivatives of inverse trig functions. We can also do this using implicit differentiation. Start with $y = \sin^{-1}x$, solve the function for $x$ then differentiate the result implicitly to confirm the formula for the derivative of the inverse sine function.
Confirm that the curves given by $x^2 = 3y^2$ and $x^2 + y^2 = 4$ meet orthogonally, i.e. their tangent lines are perpendicular at the points where the curves cross.
Explorations
Use the function $(x^2 + y^2)^2=a^2(x^2-y^2)$ with $a=1$ to answer the following questions.
- Use implicit differentiation to find y'.
- What are the equations of the vertical tangent lines?
- What are the equations of the horizontal tangent lines?
- Based on your results, try to sketch the graph of the curve.
Suppose you have two lines each of which is rotating around a fixed point. The curve sketched by their intersection is called the Trisectrix of Maclaurin. There's a good animation of this on the Wikipedia page but don't take a look until after you've tried to work out the following questions. The equation of the curve is $2x(x^2+y^2)=a(3x^2-y^2)$. Use this equation with $a=1$ to answer the following questions.
- Use implicit differentiation to find y'.
- At what points is the tangent to the graph horizontal?
- At what points is the tangent to the graph vertical?
Icons courtesy of icons8.com
Use implicit differentiation to find $dy/dx$ for the function $x^2-y^2=1$.
$$2x - 2yy' = 0$$ $$2yy' = 2x$$ $$y' = \frac{x}{y}$$
Use implicit differentiation to find $dy/dx$ for the function $x^2+2xy+y^2=0$.
$$2x + 2xy' + 2y + 2yy' = 0$$ $$2xy' + 2yy' = -2x - 2y$$ $$xy' + yy' = -x-y$$ $$y'(x + y) = -x-y$$ $$y'(x + y) = -(x + y)$$
You have to be careful at this step. You might be tempted to divide both sides of the equation by $x+y$ which would get you the right result but for the wrong reason. You can only do that division when $x + y$ doesn't equal 0 which it does for every pair of values that satisfy the original equation. (More on that in a minute.) The correct approach at this point would just be to observe that this equation is only true when $y' = -1$.
That may seem like an odd result but it makes sense if you work with the original equation a little bit:
$$x^2+2xy+y^2=0$$ $$(x+y)^2 = 0$$ $$x + y = 0$$ $$y = -x$$The second to the last line supports my previous claim that $x+y$ is always equal to 0 and the last line supports our result that the deriviative is, in fact, constant at -1.
Use implicit differentiation to find $dy/dx$ for the function $xy=\sec(xy)$.
$$xy' + y = (xy' + y)\sec(xy)\tan(xy)$$ $$xy' + y = xy'\sec(xy)\tan(xy) + y\sec(xy)\tan(xy)$$ $$xy' - xy'\sec(xy)\tan(xy) = y\sec(xy)\tan(xy) - y$$ $$y'(x - x\sec(xy)\tan(xy)) = y\sec(xy)\tan(xy) - y$$ $$y' = \frac{y\sec(xy)\tan(xy) - y}{x - x\sec(xy)\tan(xy)}$$
Use implicit differentiation to find $dy/dx$ for the function $\sin\left(\frac{x}{y}\right)=x$.
$$\cos\left(\frac{x}{y}\right)\frac{d}{dy}\left(\frac{x}{y}\right)=1$$ $$\cos\left(\frac{x}{y}\right) \cdot \frac{y - xy'}{y^2}=1$$ $$\cos\left(\frac{x}{y}\right) \cdot (y - xy')=y^2$$ $$y - xy'=y^2 \sec\left(\frac{x}{y}\right) $$ $$-xy'=y^2 \sec\left(\frac{x}{y}\right) - y $$ $$xy'=-y^2 \sec\left(\frac{x}{y}\right) + y $$ $$y'=\frac{-y^2 \sec\left(\frac{x}{y}\right) + y}{x} $$
For the equation $x^2-y^2=4$, confirm that the point $(2, 0)$ is on the curve and find the equations of the lines that are tangent and normal (perpendicular) to the curve at that point.
Since $2^2 - 0^2 = 4$, the point is clearly on the curve.
The slope of the tangent line to the curve is given by
$$2x - 2yy' = 0$$ $$2yy' = 2x$$ $$y' = \frac{x}{y}$$This means that the slope of the tangent line at $(2, 0)$ is $y' = \frac{2}{0}$ which is undefined so the tangent line at that point will be vertical. Given the coordinates of the point, the equation must be $y=0$. The normal line must therefore be horizontal so its equation will be $x = 2$.
For the equation $x^2+3x = e^y$, confirm that the point $(1, \ln 4)$ is on the curve and find the equations of the lines that are tangent and normal (perpendicular) to the curve at that point.
Since $1^2 + 3 \cdot 1 = 4$ and $e^{\ln 4} = 4$, the point is on the curve.
The slope of the tangent line to the curve is given by
$$2x + 3 = e^y y'$$ $$y' = \frac{2x+3}{e^y}$$This means that the slope of the tangent line at $(2, 0)$ is
$$y' = \frac{2\cdot 1 + 3}{e^{\ln 4}} = \frac{5}{4}$$This makes the equation of the tangent line $y - \ln 4 = \frac{5}{4}(x - 1)$ or $y = \frac{5}{4}x - \frac{5}{4} + \ln 4$.
The slope of the normal or perpendicular line would be $-\frac{4}{5}$ and, since it passes through the same point, its equation would be $y=-\frac{5}{4}x + \frac{5}{4} + \ln 4$.
In an earlier lecture, we used the formula for the derivative of an inverse function to derive formulas for the derivatives of inverse trig functions. We can also do this using implicit differentiation. Start with $y = \sin^{-1}x$, solve the function for $x$ then differentiate the result implicitly to confirm the formula for the derivative of the inverse sine function.
$$y = \sin^{-1}x$$ $$\sin y = x$$ $$y' \cos /y = 1$$ $$y' = \frac{1}{\cos y}$$
Looking back at the original equation, if the sine of $y$ equals $x$ then $y$ is an angle and, if we think of it as being an angle in a right triangle, then the ratio of the side opposite it and the triangle's hypotenuse is $x/1$. This makes the adjacent leg's length $\sqrt{1-x^2}$ according to the Pythagorean Theorem so the cosine of the angle must be
$$\cos y = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$$Substituting that into the derivative result gives us
$$y' = \frac{1}{\sqrt{1 - x^2}}$$Which is the same result we saw earlier.
Given $(x^2 + y^2)^2=a^2(x^2-y^2)$ with $a=1$, find y'
$$(x^2 + y^2)^2=(1)^2(x^2-y^2)$$ $$(x^2 + y^2)^2=x^2-y^2$$ $$2(x^2 + y^2) \cdot D_x(x^2 + y^2)= 2x - 2yy'$$ $$2(x^2 + y^2)(2x + 2yy')= 2x - 2yy'$$ $$(x^2 + y^2)(2x + 2yy')= x - yy'$$ $$2x^3 + 2x^2yy' + 2xy^2 + 2y^3y' = x-yy'$$ $$2x^2yy' + 2y^3y' + yy' = x - 2xy^2 - 2x^3$$ $$(2x^2y + 2y^3 + y)y' = x - 2xy^2 - 2x^3$$ $$y' = \frac{x - 2xy^2 - 2x^3}{2x^2y + 2y^3 + y}$$
What are the equations of the vertical tangent lines?
There are several technical considerations to keep in mind here.
- You can find the potential locations of the vertical tangent lines by setting the denominator equal to 0.
- When you evaluate the derivative at a point, if it evaluates to $\frac{a}{0}$ where $a e 0$ then you've got a horizontal tangent at that point. If it evaluates to $\frac{0}{0}$ then the derivative doesn't exist at that point and the graph doesn't have a single tangent line there.
- An expression that looks like $(\text{some expression}^2 + 1$$ is always strictly greater than 1, i.e. it can never equal 0.
What are the equations of the vertical tangent lines?
First, recall from the previous question that $$y' = \frac{x - 2xy^2 - 2x^3}{2x^2y + 2y^3 + y}$$.
These occur wherever the denominator of the derivative is equal to 0.
$$2x^2y + 2y^3 + y = 0$$ $$y(2x^2 + 2y^2 + 1) = 0$$Clearly this occurs when $y = 0$. We don't get any solutions from the $2x^2 + 2y^2 + 1$ factor since that's never less than 1.
To find the corresponding x-values, I'll substitute $y=0$ back into the original equation.
$$(x^2 - 0^2)^2 = x^2 - 0^2$$ $$x^4 = x^2$$ $$x^4 - x^2 = 0$$ $$x^2(x^2 - 1) = 0$$ $$x^2(x-1)(x+1) = 0$$So we end up with three points $(-1, 0)$, $(0, 0)$ and $(1, 0)$. At this point, we need to go back and double check what's going on with the derivative. If we substitute those values back into the formula for $y'$, we get
| $$(-1, 0)$$ $$y' = \frac{1}{0}$$ | $$(0, 0)$$ $$y' = \frac{0}{0}|$$ | $$(1, 0)$$ $$y' = \frac{-1}{0}$$ |
This tells us that the tangents are vertical at $x=\pm1$ and that, at $x=0$ the slope of the tangent line is undefined or indeterminate. This makes the vertical tangents, $$x=1$$ and $$x = -1$$.
Use implicit differentiation to find y' given $2x(x^2+y^2)=a(3x^2-y^2)$.
$$2x(x^2+y^2)=3x^2-y^2$$ $$2[(x^2 + y^2)\frac{d}{dx}x + x\frac{d}{dx}(x^2 + y^2)] = 6x - 2yy'$$ $$2[(x^2 + y^2) + x(2x + 2yy')] = 6x - 2yy'$$ $$2x^2 + 2y^2 + 4x^2 + 4xyy' = 6x - 2yy'$$ $$4xyy' + 2yy' = 6x - 2x^2 - 2y^2 - 4x^2$$ $$y'(4xy + 2y) = 6x - 6x^2 - 2y^2$$ $$y' = \frac{6x - 6x^2 - 2y^2}{4xy + 2y}$$ $$y' = \frac{3x - 3x^2 - y^2}{2xy + y}$$
At what points is the tangent to the graph vertical?
$(3/2, 0)$ is the only point where this happens. The derivative is indeteriminate (0/0) at the point $(0, 0)$ and there are no $y$ values that correspond to $x=-1/2$.
