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Student123

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Find the tangent slope at point (3,4) of y=sqrt(25-x2).

I rationalized the numerator, but I'm bad at algebra so afterwards I got lost and I got to:
lim h->0 (-h2-6h-38)/(h(sqrt(25-(3+h2))+4). What do I do?
 
Student123 said:
Find the tangent slope at point (3,4) of y=sqrt(25-x2).

I rationalized the numerator, but I'm bad at algebra so afterwards I got lost and I got to:
lim h->0 (-h2-6h-38)/(h(sqrt(25-(3+h2))+4). What do I do?
Click to expand...
"I am bad at algebra" - can't be and shouldn't be an excuse for Calculus students!!

Have you been taught to "differentiate" yet?
 
[MATH]f(x + h) - f(x) = \sqrt{25 - (x + h)^2} - \sqrt{25 -x^2} =[/MATH]
[MATH]\dfrac{\sqrt{25 - (x + h)^2} - \sqrt{25 - x^2}}{1} * \dfrac{\sqrt{25 - (x + h)^2} + \sqrt{25 -x^2}}{\sqrt{25 - (x + h)^2} + \sqrt{25 -x^2}}=[/MATH]
[MATH]\dfrac{25 - x^2 - 2hx - h^2 - (25 - x^2)}{\sqrt{25 - (x + h)^2} + \sqrt{25 -x^2}} =[/MATH]
[MATH]- \dfrac{2hx + h^2}{\sqrt{25 - (x + h)^2} + \sqrt{25 -x^2}} \implies[/MATH]
[MATH]\dfrac{f(x + h) - f(x)}{h} = - \dfrac{2x + h}{\sqrt{25 - (x + h)^2} + \sqrt{25 -x^2}}.[/MATH]
[MATH]\lim_{h \rightarrow 0} \sqrt{25 - (x + h)^2} + \sqrt{25 - x^2} = \text {WHAT?}[/MATH]
Where is that limit = 0?

Are those values relevant to this problem?

[MATH]\lim_{h \rightarrow 0} (2x + h) = \text {WHAT?}[/MATH]
[MATH]\therefore \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h} = \dfrac{\text {WHAT}}{\text {WHAT}}?[/MATH]
What is that equal if x = 3.
 
Student123 said:
Find the tangent slope at point (3,4) of y=sqrt(25-x2).

I rationalized the numerator, but I'm bad at algebra so afterwards I got lost and I got to:
lim h->0 (-h2-6h-38)/(h(sqrt(25-(3+h2))+4). What do I do?
Click to expand...
The -38 shouldn't be there; (it should be 0).
Also, your bottom line should read [MATH]h(\sqrt{25-(3+h)^2}+4)[/MATH]After these two corrections, you should be able to work on, simplify and find the limit.
 
Last edited:
Beer soaked opinion follows.
Student123 said:
Find the tangent slope at point (3,4) of y=sqrt(25-x2).

I rationalized the numerator, but I'm bad at algebra so afterwards I got lost and I got to:
lim h->0 (-h2-6h-38)/(h(sqrt(25-(3+h2))+4). What do I do?
Click to expand...
You could avoid dealing with limits and differentiation (if you're allowed) altogether considering that the problem falls under the slopes of perpendicular lines section of any Precalculus text.
20210429_151218.jpg
 
On the top line, for the constant, you did 25 - 3 = (22) instead of [MATH]25-3^2[/MATH]= (16) and somehow with the 22 and the final -16, ended up with -38 instead of the correct [MATH]16 - 16 = 0[/MATH]
[MATH] \hspace3ex \tfrac{\sqrt{25 - (3+h)^2} \; - \; \sqrt{25-3^2}}{h}\\ =\frac{\sqrt{25 - (3^2+6h+h^2)} \; - \; \sqrt{25-9}}{h}\\ =\frac{\sqrt{25 - 9-6h-h^2} \; - \; 4}{h}\\ =\frac{\sqrt{16-6h-h^2} \; - \; 4}{h}\\ =\frac{(\sqrt{16-6h-h^2} \; - \; 4)(\sqrt{16-6h-h^2} \; + \; 4)}{h(\sqrt{16-6h-h^2} \; + \; 4)}\\ =\frac{(16-6h-h^2) \; - \; 4^2}{h(\sqrt{16-6h-h^2} \; + \; 4)}\\ ...\\[/MATH]
 
jonah2.0 said:
You could avoid dealing with limits and differentiation (if you're allowed) altogether considering that the problem falls under the slopes of perpendicular lines section of any Precalculus text.
Click to expand...
Very nice. So much simpler!
 
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