# Thread: [Calc1] Four Calculus Questions.

1. ## [Calc1] Four Calculus Questions.

1. Find the values of a, b and c so that the curve y=ax^2+bx+c passes through the point (0,4) and such that the tangent to the curve at the point (1,2) is parallel to teh line y=x+1.

2. (Related Rates Question). Two sides of a triangle are 2m and 3m in length and teh angle between them is increasing at a rate of 4 rad/min. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/6.

3. (Hyperbolic Function Limit). Find limit as x-->0.
(sinhx)/(e^x)

4. Let f(x)=
-x+2 x≤1
x^2-2x+2 x>1

Is f differentiable at 1? Justify answer by using definition of differentiability at a point (Not supposed to do any derivates).

Don't just want the answers, actually want to learn how to do them.
+Rep .

2. 1.

y = ax^2 + bx + c
(dy/dx) = 2ax + b

Eqn of tangent: Y - y = (dy/dx)(X-x) --------- small y and x are points you sub in, big Y and X stay as X or Y. So you sub the point on tangent, (1,2) into this.
Y - 2 = (2ax + b)(1)
Y = 2ax + b + 2

Since this gradient is the same as the gradient of the line y = x+1 (which is 1),
2a = 1
a = 0.5

Sub a = 0.5 back into the eqn of tangent line
Y = x + b + 2 ----------- sub point on tangent into this eqn again (1,2)
2 = 1 + b + 2
b = -1

Sub a = 0.5 and b = -1 into the overall equation of Y=ax^2 + bx + c,
And also sub the point (0,4) that they say lies on this curve.
You get c = 4

[b]a = 0.5, b = -1, c = 4[/b
I really hope my answer is right.

---------- Post added at 01:29 AM ---------- Previous post was at 01:27 AM ----------

EDIT: Just wondering what sinhx means. What's sinh

3. sinh is the hyperbolic sine, and I'm surprised it's in your calc 1 course
from wikipedia:

sinh x = {e^x - e^{-x}} / {2} = {e^{2x} - 1} / {2e^x}

so sinh x / e^x = (e^2x - 1) / (2e^2x)
and lim as x-> 0 is equivalent to (e^2(0) - 2) / (2e^2(0))
anything to the zeroth power is 1, so this is (1 - 2) / (2), or -.5

Regarding four:
4. Let f(x)=
-x+2 for x≤1
x^2-2x+2 for x>1

The definition of differentiability at a point is that it's a smooth curve at that point
This wouldn't create a smooth curve, which you can see by drawing the two functions; you'll notice a sharp change from a downwards line to a upward curve
I don't know how to show that mathematically without taking derivatives, unfortunately

4. ## The Following User Says Thank You to Richy For This Useful Post:

Lmp (11-01-2012)

5. Show it graphically then.

Draw the graph. Straight downward sloping line till x = 1 then an increasing curve from the minimum point of a quadratic curve from x=1 onwards.
This shows it's not a smoove curve

I have not learnt sinh so can't help there.

The angle question is complex but angle of triagle = 0.5 ab sin c

6. Originally Posted by Shawn
1.

y = ax^2 + bx + c
(dy/dx) = 2ax + b

Eqn of tangent: Y - y = (dy/dx)(X-x) --------- small y and x are points you sub in, big Y and X stay as X or Y. So you sub the point on tangent, (1,2) into this.
Y - 2 = (2ax + b)(1)
Y = 2ax + b + 2

Since this gradient is the same as the gradient of the line y = x+1 (which is 1),
2a = 1
a = 0.5

Sub a = 0.5 back into the eqn of tangent line
Y = x + b + 2 ----------- sub point on tangent into this eqn again (1,2)
2 = 1 + b + 2
b = -1

Sub a = 0.5 and b = -1 into the overall equation of Y=ax^2 + bx + c,
And also sub the point (0,4) that they say lies on this curve.
You get c = 4

[b]a = 0.5, b = -1, c = 4[/b
I really hope my answer is right.

---------- Post added at 01:29 AM ---------- Previous post was at 01:27 AM ----------

EDIT: Just wondering what sinhx means. What's sinh
I understand C, which you can get with just plugging the point into the equation. Still don't get a&b.

"Y - 2 = (2ax + b)(1)"
Should the end of that be X-1?

---------- Post added at 11:48 AM ---------- Previous post was at 11:40 AM ----------

Originally Posted by Richy
sinh is the hyperbolic sine, and I'm surprised it's in your calc 1 course
from wikipedia:

sinh x = {e^x - e^{-x}} / {2} = {e^{2x} - 1} / {2e^x}

so sinh x / e^x = (e^2x - 1) / (2e^2x)
and lim as x-> 0 is equivalent to (e^2(0) - 2) / (2e^2(0))
anything to the zeroth power is 1, so this is (1 - 2) / (2), or -.5

Regarding four:
4. Let f(x)=
-x+2 for x≤1
x^2-2x+2 for x>1

The definition of differentiability at a point is that it's a smooth curve at that point
This wouldn't create a smooth curve, which you can see by drawing the two functions; you'll notice a sharp change from a downwards line to a upward curve
I don't know how to show that mathematically without taking derivatives, unfortunately
Quick question.
"(e^2(0) - 2)".
Where did that "-2" come from.

7. OMG x-1
I fucked up.

But yeah, it's something like that.

8. Quick question.
"(e^2(0) - 2)".
Where did that "-2" come from.
That was a typo, good catch! It should be -1, so (1 - 1) / 2 is 0

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